MAXREFDES1012

Miniature, 5.3V/2A, Synchronous, No-Opto Flyback DC-DC Converter with 91.4% Efficiency Using MAX17690 and MAX17606

SAMPLE PRODUCTS

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Design & Integration File

Figure 2. Isolated flyback converter topology with typical waveforms.

Designed, Built, Tested

Board pictured here has been fully assembled and tested.

Overview

Design Resources

Design & Integration File

Schematic
Bill of Materials
PCB Layout

Download Design Files 2.72 M

Description

Due to its simplicity and low cost, the flyback converter is the preferred choice for low-to-medium isolated DC-DC power-conversion applications. However, the use of an optocoupler or an auxiliary winding on the flyback transformer for voltage feedback across the isolation barrier increases the number of components and design complexity. The MAX17690 eliminates the need for an optocoupler or auxiliary transformer winding and achieves ±5% output voltage regulation over line, load, and temperature variations.

The MAX17690 implements an innovative algorithm to accurately determine the output voltage by sensing the reflected voltage across the primary winding during the flyback time interval. By sampling and regulating this reflected voltage when the secondary current is close to zero, the effects of secondary-side DC losses in the transformer winding, the PCB tracks, and the rectifying diode on output voltage regulation can be minimized.

The MAX17690 also compensates for the negative temperature coefficient of the rectifying diode.

Features & Benefits

4.5V to 60V Input Voltage Range
Programmable Switching Frequency from 50kHz to 250kHz
Programmable Input Enable/UVLO Feature
Adjustable Soft-Start
2A/4A Peak Source/Sink Gate Drive Capability
Hiccup Mode Short-Circuit Protection
Fast Cycle-by-Cycle Peak Current Limit
Thermal Shutdown Protection
Space-Saving, 16-Pin, 3mm x 3mm TQFN Package
-40°C to +125°C Operating Temperature Range

Important Notice and Disclaimer

Please read the Reference Design Disclaimer for Analog Devices design information and resources.

Parts Used

MAX17690

Details Section

Introduction

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The MAX17690 also compensates for the negative temperature coefficient of the rectifying diode.

Other features include the following:

4.5V to 60V Input Voltage Range
Programmable Switching Frequency from 50kHz to 250kHz
Programmable Input Enable/UVLO Feature
Programmable Input Overvoltage Protection
Adjustable Soft-Start
2A/4A Peak Source/Sink Gate Drive Capability
Hiccup Mode Short-Circuit Protection
Fast Cycle-by-Cycle Peak Current Limit
Thermal Shutdown Protection
Space-Saving, 16-Pin, 3mm x 3mm TQFN Package
-40°C to +125°C Operating Temperature Range

Hardware Specification

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An isolated no-opto flyback DC-DC converter using the MAX17690 and MAX17606 is demonstrated for a 5.3V DC output application. The power supply delivers up to 2A at 5.3V. Table 1 shows an overview of the design specification.

Table 1. Design Specification
PARAMETER	SYMBOL	MIN	MAX
Input Voltage	V_IN	8V	20V
Frequency	f_SW	143.5kHz
Peak Efficiency at Full Load	η_MAX	0.914
Efficiency at Minimum Load	η_MIN	0.6
Output Voltage	V_OUT	5.3V
Output Voltage Ripple	∆V_O	50mV
Maximum Output Current	I_OUT	2A
Maximum Output Power	P_OUT	10.6W

Designed–Built–Tested

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This document describes the hardware shown in Figure 1. It provides a detailed systematic technical guide to designing an isolated no-opto flyback DC-DC converter using Maxim's MAX17690 controller. The power supply has been built and tested.

The Isolated No-Opto Flyback Converter

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One of the drawbacks encountered in most isolated DC-DC converter topologies is that information relating to the output voltage on the isolated secondary side of the transformer must be communicated back to the primary side to maintain output voltage regulation. In a regular isolated flyback converter, this is normally achieved using an optocoupler feedback circuit or an additional auxiliary winding on the flyback transformer. Optocoupler feedback circuits reduce overall power-supply efficiency, and the extra components increase the cost and physical size of the power supply. In addition, optocoupler feedback circuits are difficult to design reliably due to their limited bandwidth, nonlinearity, high CTR variation, and aging effects. Feedback circuits employing auxiliary transformer windings also exhibit deficiencies. Using an extra winding adds to the flyback transformer's complexity, physical size, and cost, while load regulation and dynamic response are often poor.

The MAX17690 is a peak current-mode controller designed specifically to eliminate the need for optocoupler or auxiliary transformer winding feedback in the traditional isolated flyback topology, therefore reducing size, cost, and design complexity. It derives information about the isolated output voltage by examining the voltage on the primary-side winding of the flyback transformer.

Other than this uniquely innovative method for regulating the output voltage, the no-opto isolated flyback converter using the MAX17690 follows the same general design process as a traditional flyback converter. To understand the operation and benefits of the no-opto flyback converter, it is useful to review the schematic and typical waveforms of the traditional flyback converter (using the MAX17595), shown in Figure 2.

The simplified schematic in Figure 2 illustrates how information about the output voltage is obtained across the isolation barrier in traditional isolated flyback converters. The optocoupler feedback mechanism requires at least 10 components including an optocoupler and a shunt regulator, in addition to a primary-side bias voltage, V_BIAS, to drive the photo-transistor. The error voltage FB2 connects to the FB pin of the flyback controller.

The transformer feedback method requires an additional winding on the primary side of the flyback transformer, a diode, a capacitor, and two resistors to generate a voltage proportional to the output voltage. This voltage is compared to an internal reference in a traditional flyback controller to generate the error voltage.

By including additional innovative features internally in the MAX17690 no-opto flyback controller, Maxim has enabled power-supply designers to eliminate the additional components, board area, complexity, and cost associated with both the optocoupler and transformer feedback methods. Figure 3 illustrates a simplified schematic and typical waveforms for an isolated no-opto flyback DC-DC converter using the MAX17690.

By comparing Figure 3 with Figure 2, it is evident that there is no difference in the voltage and current waveforms in the traditional and no-opto flyback topologies. The difference is in the control method used to maintain V_OUT at its target value over the required load, line, and temperature range. The MAX17690 achieves this with minimum components by forcing the voltage V_FLYBACK during the conduction period of DFR to be precisely the voltage required to maintain a constant V_OUT. When Q_P turns off, DFR conducts and the drain voltage of Q_P, rises to a voltage V_FLYBACK above V_IN. After initial ringing due to transformer leakage inductance and the junction capacitance of DFR and output capacitance of Q_P, the voltage V_FLYBACK is given by:

$V_{F L Y B A C K} = V_{I N} + \frac{{(V}_{O U T} + V_{D F R} (T) + I_{L S} (t) \times R_{S} (T))}{n_{S P}}$

where:

V_FLYBACK is the Q_P drain voltage relative to primary ground

V_DFR(T) is the forward voltage drop of DFR, which has a negative temperature coefficient

I_LS(t) is the instantaneous secondary transformer current

R_S(T) is the total DC resistance of the secondary circuit, which has a positive temperature coefficient

n_SP is the secondary to primary turns ratio of the flyback transformer

The voltage of interest is (V_FLYBACK - V_IN) since this is a measure of V_OUT. An internal voltage to current amplifier generates a current proportional to (V_FLYBACK - V_IN). This current then flows through R_SET to generate a ground referenced voltage, V_SET, proportional to (V_FLYBACK - V_IN). This requires that:

$\frac{V_{F L Y B A C K} - V_{I N}}{R_{F B}} = \frac{V_{S E T}}{R_{S E T}}$

Combining this equation with the previous equation for V_FLYBACK, we have:

$V_{O U T} = V_{S E T} \times (\frac{R_{F B}}{R_{S E T}}) \times n_{S P} - V_{D F R} (T) - I_{L S} (t) \times R_{S} (T)$

Figure 3. Isolated no-opto flyback converter topology with typical waveforms.

Temperature Compensation

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We need to consider the effect of the temperature dependence of V_DFR and the time dependence of I_LS on the control system. If V_FLYBACK is sampled at a time when I_LS is very close to zero, then the term I_LS(t) × R_S(T) is negligible and can be assumed to be zero in the previous expression. This is the case when the flyback converter is operating in, or close to, discontinuous conduction mode. It is very important to sample the V_FLYBACK voltage before the secondary current reaches zero since there is a very large oscillation on V_FLYBACK due to the resonance between the primary magnetizing inductance of the flyback transformer and the output capacitance of Q_P as soon as the current reaches zero in the secondary. The time at which V_FLYBACK is sampled is set by resistor R_VCM.

The V_DFR term has a significant negative temperature coefficient that must be compensated to ensure acceptable output voltage regulation over the required temperature range. This is achieved by internally connecting a positive temperature coefficient current source to the V_SET pin. The current is set by resistor R_TC connected to ground. The simplest way to understand the temperature compensation mechanism is to think about what needs to happen in the control system when temperature increases. In an uncompensated system, as the temperature increases, V_DFR decreases due to its negative temperature coefficient. Since V_DFR decreases, V_OUT increases by the same amount, therefore V_FLYBACK remains unchanged. Since V_SET is proportional to V_FLYBACK, V_SET also remains unchanged. Since there is no change in V_SET there is no change in duty cycle demand to bring V_OUT back down to its target value. What needs to happen in the temperature compensated case is, when V_OUT increases due to the negative temperature coefficient of V_DFR, V_SET needs to increase by an amount just sufficient to bring V_OUT back to its target value. This is achieved by designing V_SET with a positive temperature coefficient. Expressed mathematically as:

$\frac{δ V_{D F R}}{δ T} + \frac{1}{n_{S P}} \times \frac{δ V_{T C}}{δ T} \times \frac{R_{F B}}{R_{T C}} = 0$

where:

δV_DFR/δT is the diodes forward temperature coefficient

δV_TC/δT = 1.85mV/°C

V_TC = 0.55V is the voltage at the TC pin at +25°C

Rearranging the above expression gives:

$R_{T C} = - R_{F B} \times n_{S P} \times \frac{δ T}{δ V_{D F R}} \times \frac{δ V_{T C}}{δ T}$

The effect of adding the positive temperature coefficient current, I_TC, to the current in R_FB is equivalent to adding a positive temperature coefficient voltage in series with V_DFR on the secondary side of value:

$\frac{V_{T C}}{R_{T C}} \times R_{F B} \times n_{S P}$

Substituting from the previous expression, this becomes:

$- V_{T C} \times \frac{δ V_{D F R}}{δ T} \times \frac{δ T}{δ V_{T C}}$

We can now substitute this expression into the expression for V_OUT as follows:

$V_{O U T} = V_{S E T} \times (\frac{R_{F B}}{R_{S E T}}) \times n_{S P} - V_{D F R} - V_{T C} \times \frac{δ V_{D F R}}{δ T} \times \frac{δ T}{δ V_{T C}}$

and finally solve for R_FB:

$R_{F B} = \frac{R_{S E T}}{n_{S P} \times V_{S E T}} \times (V_{O U T} + V_{D F R} + V_{T C} \times \frac{δ V_{D F R}}{δ T} \times \frac{δ T}{δ V_{T C}})$

Values for R_SET, V_SET, and δV_TC/δT can be obtained from the MAX17690 data sheet as follows:

R_SET = 10kΩ

V_SET = 1V

δV_TC/δT = 1.85mV/°C

Values for V_DFR and δV_DFR/δT can be obtained from the output diode data sheet, and n_SP is calculated when the flyback transformer is designed.

The value of R_TC can then be calculated using the expression from earlier, restated below:

$R_{T C} = - R_{F B} \times n_{S P} \times \frac{δ T}{δ V_{D F R}} \times \frac{δ V_{T C}}{δ T}$

The calculated resistor values for R_FB and R_TC should always be verified experimentally and adjusted, if necessary, to achieve optimum performance over the required temperature range. Note that the reference design described in this document has only been verified at room temperature.

Finally, the internal temperature compensation circuitry requires a current proportional to V_IN. R_RIN should be chosen as approximately:

$R_{R I N} = 0.6 \times R_{F B}$

Setting the VFLYBACK Sampling Instant

The MAX17690 generates an internal voltage proportional to the on-time volt-second product. This enables the device to determine the correct sampling instant for V_FLYBACK during the Q_P off-time. The R_VCM resistor is used to scale this internal voltage to an acceptable internal voltage limit in the device.

Design Procedure

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Design Procedure for No-Opto Flyback Converter Using MAX17690

Now that the principle difference between a traditional isolated flyback converter using optocoupler or auxiliary transformer winding feedback and the isolated no-opto flyback converter using the MAX17690 is understood, a practical design example can be illustrated. The converter design process can be divided into three parts: the power stage design, the setup of the MAX17690 no-opto flyback controller, and closing the control loop. This document is intended to complement the information contained in the MAX17690 data sheet.

The following design parameters are used throughout this document:


SYMBOL	FUNCTION
V_IN	Input voltage
V_UVLO	Undervoltage turn-on threshold
V_OVI	Overvoltage turn-off threshold
t_SS	Soft-start time
V_OUT	Output voltage
ΔV_O	Steady-state output ripple voltage
I_OUT	Output current
P_OUT	Output power
η_(MAX)	Target efficiency at maximum load
η_(MIN)	Target efficiency at minimum load
P_IN	Input power
f_SW	Switching frequency
D	Duty cycle
n_SP	Secondary-primary turns ratio

Throughout the design procedure reference is made to the schematic. See the Design Resources section.

Part I: Designing the Power Components

Step 1: Calculate the Minimum Turns Ratio for the Flyback Transformer

The secondary-primary turns ratio, n_SP, and the duty cycle, D, for the flyback converter are related by the flyback DC gain function as follows:

$n_{S P} = \frac{V_{O U T}}{V_{I N}} \times (\frac{1 - D}{D})$

The converter's absolute minimum input voltage is the undervoltage lockout threshold (V_IN falling), which is programmed with a resistor-divider for the MAX17690. At this voltage and at maximum output power, D should be less than or equal to 66% (the maximum duty cycle at which the MAX17690 can operate) to ensure reliable converter operation. For the current design, the undervoltage lockout threshold (V_IN falling) occurs at 6.4V, so with D set at 66%, the absolute minimum turns ratio, n_SP(MIN), for the flyback transformer is calculated as follows

$n_{S P (M I N)} = 0.43$

This transformer turns ratio assumes that there are no DC voltage drops in the primary and/or secondary circuits. In practice, a larger transformer turns ratio must be chosen to account for these DC voltage drops. For the current design, a transformer turns ratio n_SP = 0.5 was chosen.

Step 2: Estimate the Maximum and Minimum Duty Cycle Under Normal Operating Conditions

Normal input voltage operating conditions are defined as V_IN(MIN) and V_IN(MAX)on page 1. By using the flyback DC gain function again, the duty cycle is estimated as:

$D = \frac{1}{1 + n_{S P} \times (\frac{V_{I N}}{V_{O U T}})}$

n_SP and V_OUT are fixed, D_MAX occurs when V_IN is a minimum, i.e., at V_IN(MIN). For the current design, V_IN(MIN) = 8V, so:

$D_{M A X} = 0.57$

The MAX17690 derives the current, ΔI_LP, in the primary magnetizing inductance by measuring the voltage, ΔV_RCS, across the current-sense resistor (R_CS) during the on-time of the primary-side MOSFET. So:

$Δ I_{L P} = \frac{Δ V_{R C S}}{R_{C S}}$

ΔI_LP is a maximum at D_MAX and V_IN(MIN) and a minimum at D_MIN and V_IN(MAX), so:

$\frac{V_{I N (M I N)}}{L_{P}} = \frac{Δ V_{R C S (M I N)} \times f_{S W}}{R_{C S} \times D_{M A X}}$

and

$\frac{V_{I N (M A X)}}{L_{P}} = (\frac{η_{M A X}}{η_{M I N}} \times \frac{Δ V_{R C S (M I N)}}{R_{C S}}) \times \frac{f_{S W}}{D_{M I N}}$

Solving these two equations:

$D_{M I N} = D_{M A X} \times \frac{η_{M A X}}{η_{M I N}} \times \frac{V_{I N (M I N)}}{V_{I N (M A X)}} \times \frac{Δ V_{R C S (M I N)}}{Δ V_{R C S (M A X)}}$

where ΔV_RCS(MIN) and ΔV_RCS(MAX) correspond to the minimum current limit threshold (20mV) and the maximum current limit threshold (100mV) of the MAX17690, respectively. So, for V_IN(MIN) = 8V, V_IN(MAX) = 20V, and D_MAX = 0.57, we have:

$D_{M I N} = 0.057$

Step 3: Calculate the Maximum Allowable Switching Frequency

The isolated no-opto flyback topology requires the primary side MOSFET to constantly maintain switching, otherwise there is no way to sense the reflected secondary-side voltage at the drain of the primary-side MOSFET. The MAX17690 achieves this by having a critical minimum on-time, t_ON(CRIT), for which it drives the MOSFET. At a given switching frequency, t_ON(MIN) corresponds to D_MIN. From the MAX17690 data sheet, the critical minimum on-time t_ON(CRIT) for the NDRV pin is 235ns. We can therefore calculate the maximum allowable switching frequency to ensure that t_ON(MIN) > t_ON(CRIT) as follows:

$f_{S W (M A X)} = \frac{D_{M I N}}{t_{O N (C R I T)}} \approx 242.5 k H z$

Since D_MIN is fixed by ΔV_RCS(MIN), ΔV_RCS(MAX), D_MAX, V_IN(MIN), and V_IN(MAX), then t_ON(MIN) can be chosen arbitrarily larger than t_ON(CRIT) so that f_SW is less than f_SW(MAX). With t_ON(MIN) = 400ns, the switching frequency is as follows:

$f_{S W} = \frac{D_{M I N}}{t_{O N (M I N)}} \approx 143 k H z$

Note that the MAX17690 should always be operated in the 50kHz to 250kHz switching frequency range and t_ON(MIN) must be chosen accordingly to ensure that this constraint is met.

Step 4: Estimate the Primary Magnetizing Inductance

Maximum input power is given by:

$P_{I N (M A X)} = \frac{P_{O U T (M A X)}}{η_{M A X}} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X}}$

For the discontinuous flyback converter all the energy stored in the primary magnetizing inductance, L_P, during the MOSFET on-time is transferred to the output during the MOSFET off-time, i.e., the full power transfer occurs during one switching cycle. Therefore, because E = P x t, we have:

$E_{I N (M A X)} = P_{I N (M A X)} \times τ_{S W} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times f_{S W}}$

The maximum input energy must be stored in L_P during the on-time of the MOSFET, so:

$E_{I N (M A X)} = \frac{1}{2} \times L_{P} \times Δ I_{L P (M A X)}^{2}$

We also know that the peak current in L_P, ΔI_LP(MAX) occurs at input voltage V_IN(MIN) and MOSFET on-time t_ON(MAX), so:

$Δ I_{L P (M A X)}^{2} = \frac{V_{I N (M I N)}^{2} \times t_{O N (M A X)}^{2}}{L_{P}^{2}}$

and substituting:

$E_{I N (M A X)} = \frac{V_{I N (M I N)}^{2} \times t_{O N (M A X)}^{2}}{2 L_{P}}$

Combining with the original P x t equation gives:

$\frac{V_{I N (M I N)}^{2} \times t_{O N (M A X)}^{2}}{2 L_{P}} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times f_{S W}}$

Finally, by rearranging we have an expression for the primary magnetizing inductance L_P:

$L_{P} = \frac{η_{M A X} \times V_{I N (M I N)}^{2} \times D_{M A X}^{2}}{2 \times V_{O U T} \times I_{O U T} \times f_{S W}}$

Estimating the power converter efficiency at 90% and with V_IN(MIN) = 8V, D_MAX = 0.57, V_OUT = 5.3V, and f_SW = 143kHz, then:

$L_{P (M A X)} \approx 4.4 μ H$

This inductance represents the maximum primary magnetizing inductance since it sets the current-limit threshold. Choosing a larger inductance sets the current-limit threshold at a lower value and could cause the converter to go into current limit at a value lower than I_OUT, which would be undesirable. Assuming a ±10% tolerance for the primary magnetizing inductance gives:

$L_{P} \approx 4 μ H \pm 10 %$

Step 5: Recalculate D_MAX, D_MIN, and t_ON(MIN) Based on the Selected Value for L_P

Rearranging the L_P equation in Step 4 gives an expression for D_MAX as follows:

$D_{M A X} = \sqrt{\frac{2 \times L_{P} \times V_{O U T} \times I_{O U T} \times f_{S W}}{η_{M A X} \times V_{I N (M I N)}^{2}}} = 0.460$

Referring to Step 2:

$D_{M I N} = D_{M A X} \times \frac{η_{M A X}}{η_{M I N}} \times \frac{V_{I N (M I N)}}{V_{I N (M A X)}} \times \frac{Δ V_{R C S (M I N)}}{Δ V_{R C S (M A X)}} = 0.055$

and:

$t_{O N (M I N)} = \frac{D_{M I N}}{f_{S W}} = 384 n s$

Step 6: Calculate the Peak and RMS Currents in the Primary Winding of the Flyback Transformer

The peak primary winding current occurs at V_IN(MIN) and D_MAX according to the following equation:

$Δ I_{L P (M A X)} = \frac{V_{I N (M I N)} \times D_{M A X}}{L_{P} \times f_{S W}} \approx 6.40 A$

The RMS primary winding current can be calculated from ΔI_LP(MAX) and D_MAX as follows:

$I_{L P (R M S)} = Δ I_{L P (M A X)} \times \sqrt{\frac{D_{M A X}}{3}} \approx 2.51 A$

Step 7: Calculate the Peak and RMS Currents in the Secondary Winding of the Flyback Transformer

The peak current in the flyback transformer’s secondary-side winding can be established by considering that the entire energy transferred from the primary-side winding to the secondary-side winding is delivered to the load during one switching period. Again, since E = P x t:

$E_{O U T} = \frac{1}{2} \times L_{S} \times Δ I_{L S (M A X)}^{2} = P_{O U T} \times τ_{S W}$

Substituting:

$P_{O U T} \times τ_{S W} = \frac{V_{O U T} \times I_{O U T}}{f_{S W}}$

And rearranging:

$Δ I_{L S (M A X)} = \sqrt{\frac{2 \times V_{O U T} \times I_{O U T}}{f_{S W} \times L_{P} \times n_{S P}^{2}}} = 12.15 A$

Current flows in the secondary-side winding of the flyback transformer during the time the secondary-side rectifying device is conducting. This conduction time, t_ON(SEC), is calculated using the inductor volt-second equation:

$V = L \times \frac{d I}{d t}$

where V = V_OUT, L = L_S, dI = ΔI_LS(MAX), and dt = t_ON(SEC), so:

$t_{O N (S E C)} = L_{S} \times \frac{Δ I_{L S (M A X)}}{V_{O U T}} = L_{P} \times n_{S P}^{2} \times \frac{Δ I_{L S (M A X)}}{V_{O U T}}$

The maximum duty cycle of the secondary-side rectifying device, D_S(MAX), can now be calculated:

$D_{S M A X} = \frac{t_{O N (S E C)}}{τ_{S W}} = t_{O N (S E C)} \times f_{S W} = 0.32$

Finally, the RMS secondary winding current can be calculated from ΔI_LS(MAX) and D_S(MAX) as follows:

$I_{L S (R M S)} = Δ I_{L S (M A X)} \times \sqrt{\frac{1 - D_{S (M A X)}}{3}} = 5.77 A$

Step 8: Summarize the Flyback Transformer Specification

All the critical parameters for the flyback transformer have been calculated and are summarized below. Using these parameters, a suitable transformer can be designed.


PARAMETER	SYMBOL	VALUE
Primary Magnetizing Inductance	L_P	4µH ±10%
Primary Peak Current	∆I_LP(MAX)	6.4A
Primary RMS Current	I_LP(RMS)	2.51A
Turns Ratio (N_S/N_P)	n_SP	0.5
Secondary Peak Current	∆I_LS(MAX)	12.15A
Secondary RMS Current	I_LS(RMS)	5.77A

Step 9: Calculate Design Parameters for Secondary-Side Rectifying Device

Depending on the output voltage and current, a choice can be made for the secondary-side rectifying device. Generally, for output voltages above 12V at low currents (less than 1A), Schottky diodes are used, and for voltages less than 12V synchronous rectification (MOSFET) is used. The current design is a 5.3V/2A output converter, so we outline a procedure for selecting a suitable MOSFET for the synchronous switch.

Figure 4 shows a simplified schematic with the synchronous MOSFET, Q_S. The MAX17606 is a secondary-side synchronous driver and controller specifically designed for the isolated flyback topology operating in Discontinuous Conduction Mode (DCM) or Border Conduction Mode (BCM).

Figure 4. Simplified no-opto flyback schematic with synchronous rectification.

The important parameters to consider for the synchronous rectifying device are the same as those of a regular rectifying diode: peak instantaneous current, RMS current, voltage stress, and power losses. Since Q_S and L_S are in series, they experience the same peak and RMS currents, so:

$I_{Q S (R M S)} = I_{L S (R M S)} = 5.77 A$

and

$I_{Q S (M A X)} = Δ I_{L S (M A X)} = 12.15 A$

When Q_S is off, V_IN reflected to the secondary side of the flyback transformer plus (V_OUT + V_QS(SAT)) is applied across the drain-source of Q_S, so:

$V_{Q S (M A X)} = n_{S P} \times V_{I N (M A X)} + V_{O U T} + V_{Q S (S A T)}$

$= 0.5 \times 20 V + 5.3 V + (5.77 A \times 6.1 m Ω)$

$\approx 15.4 V$

Q_S has both conduction losses due to its R_DS(ON) and switching losses. Allowing for reasonable design margin, we chose the Infineon BSZ040N04LSG for this design with the following specifications:


PARAMETER	VALUE
Maximum Drain-Source Voltage	40V
Continuous Drain Current	40A
Drain-Source Resistance	6.1mΩ
Minimum V_GS Threshold V_GSTH	1.2V
Typical V_GS Plateau V_GSPL	3.0V
Maximum Q_G(T)	64nC
Typical Q_GD	4.9nC
Total Output Capacitance C_OSS	1100pF

The power losses in the Q_S can be approximated as follows:

$P_{T O T} = P_{C O N} + P_{C D S} + P_{S W} \approx 239 m W$

where:

P_CON is the loss due to I_QS(RMS) flowing through the drain-source on resistance of Q_S:

$P_{C O N} = I_{Q S (R M S)}^{2} \times R_{D S (O N)} \approx 203 m W$

P_CDS is the loss due to the energy in the drain-source output capacitance being dissipated in Q_S at turn-on:

$P_{C D S} = \frac{1}{2} \times f_{S W} \times C_{O S S} \times V_{Q S (M A X)}^{2} \approx 19 m W$

and P_SW is the turn-on voltage-current transition loss that occurs as the drain-source voltage decreases and the drain current increases during the turn-on transition:

$P_{S W} = \frac{1}{2} \times f_{S W} \times I_{Q S (t - O N)} \times [\frac{V_{G S (P L)} - V_{G S (T H)}}{V_{G S (P L)}} \times (\frac{Q_{G (T)} + Q_{G D}}{I_{D R V}})] \approx 18 m W$

where I_DRV is the maximum drive current capability of the MAX17606’s GATE output and I_QS(t-ON) is the instantaneous current in Q_S at turn-on. I_QS(t-ON) is equal to I_QS(MAX).

Step 10: Calculate Design Parameters for Primary-Side MOSFET

The important parameters to consider for the primary-side MOSFET (Q_P) are peak instantaneous current, RMS current, voltage stress, and power losses. Because Q_P and L_P are in series they experience the same peak and RMS currents, so from Step 6:

$I_{Q P (M A X)} = Δ I_{L P (M A X)} \approx 6.40 A$

and

$I_{Q P (R M S)} = I_{L P (R M S)} \approx 2.51 A$

When Q_P turns off, V_OUT reflected to the primary side of the flyback transformer plus V_IN(MAX) is applied across the drain source of Q_P. In addition, until Q_S starts to conduct, there is no path for the leakage inductance energy to flow through. This causes the drain-source voltage of Q_P to rise even further. The factor of (1.5) in the equation below represents this additional voltage rise; however, this factor can be higher or lower depending on the transformer and PCB leakage inductances:

$V_{Q P (M A X)} \approx 1.5 \times (\frac{V_{O U T} + V_{Q S (S A T)}}{n_{S P}}) + V_{I N (M A X)} \approx 37 V$

Allowing for reasonable design margin, we chose the Vishay SQJA96EP for this design with the following specifications:


PARAMETER	VALUE
Maximum Drain-Source Voltage	80V
Continuous Drain Current	30A
Drain-Source Resistance	35.3mΩ
Minimum V_GS Threshold V_GSTH	2.5V
Typical V_GS Plateau V_GSPL	4V
Maximum Q_G(T)	25nC
Typical Q_GD	3nC
Total Output Capacitance C_OSS	625pF

The power losses in the Q_P can be approximated as follows:

$P_{T O T} = P_{C O N} + P_{C D S} + P_{S W} \approx 284 m W$

where:

P_CON is the loss due to I_QP(RMS) flowing through the drain-source on resistance of Q_P:

$P_{C O N} = I_{Q P (R M S)}^{2} \times R_{D S (O N)} \approx 222 m W$

P_CDS is the loss due to the energy in the drain-source output capacitance being dissipated in Q_P at turn-on:

$P_{C D S} = \frac{1}{2} \times f_{S W} \times C_{O S S} \times V_{Q P (M A X)}^{2} \approx 62 m W$

And P_SW is the turn-on voltage-current transition loss that occurs as the drain-source voltage decreases and the drain current increases during the turn-on transition:

$P_{S W} = \frac{1}{2} \times f_{S W} \times I_{Q P (t - O N)} \times [\frac{V_{G S (P L)} - V_{G S (T H)}}{V_{G S (P L)}} \times (\frac{Q_{G (T)} + Q_{G D}}{I_{D R V}})] \approx 0 m W$

where I_DRV is the maximum drive current capability of the MAX17690’s NDRV output and I_QP(t-ON) is the instantaneous current in Q_P at turn-on. Because the flyback converter is operating in DCM, I_QP(t-ON) is zero and so is P_SW.

Step 11: Select the RCD Snubber Components

Referring to Figure 5, when Q_P turns off, I_LP charges the output capacitance, C_OSS, of Q_P. When the voltage across C_OSS exceeds the input voltage plus the reflected secondary to primary voltage, the secondary-side diode (or synchronous switch) turns on. Since the diode (or synchronous switch) is now on, the energy stored in the primary magnetizing inductance is transferred to the secondary; however, the energy stored in the leakage inductance will continue to charge C_OSS since there is nowhere else for it to go. Since the voltage across C_OSS is the same as the voltage across Q_P, if the energy stored in the leakage inductance charges C_OSS to a voltage level greater than the maximum allowable drain-source voltage of Q_P, the MOSFET fails.

One way to avoid this situation arising is to add a suitable RCD snubber across the transformer’s primary winding. In Figure 5, the snubber is labeled R_SN, C_SN, and D_SN. In this situation, when Q_P turns off, the voltage at Node A i:

$V_{N O D E A} = V_{C S N} + V_{I N}$

When the secondary-side diode (or synchronous switch) turns on, the voltage at Node B is:

$V_{N O D E B} = V_{I N} + \frac{V_{O U T} + V_{D F R}}{n_{S P}}$

So, the voltage across the leakage inductance is:

$V_{L L K} = V_{C S N} + V_{I N} - (V_{I N} + \frac{V_{O U T} + V_{D F R}}{n_{S P}})$

$= V_{C S N} - (\frac{V_{O U T} + V_{D F R}}{n_{S P}}) = L_{L K} \times \frac{Δ I_{S N}}{Δ t_{S N}}$

So:

$Δ t_{S N} = \frac{L_{L K} \times Δ I_{S N}}{V_{C S N} - (\frac{V_{O U T} + V_{D F R}}{n_{S P}})}$

The average power dissipated in the snubber network is:

$P_{S N} = V_{C S N} \times \frac{Δ I_{S N} \times Δ t_{S N}}{2 \times τ_{S W}}$

Substituting Δt_SN into this expression we have:

$P_{S N} = \frac{1}{2} \times L_{L K} \times Δ I_{S N}^{2} \times \frac{V_{C S N}}{V_{C S N} - (\frac{V_{O U T} + V_{D F R}}{n_{S P}})} \times f_{S W}$

The leakage inductance energy is dissipated in R_SN, so from:

$P_{S N} = \frac{V_{C S N}^{2}}{R_{S N}}$

We can calculate the required R_SN as follows:

$R_{S N} = \frac{V_{C S N}^{2}}{\frac{1}{2} \times L_{L K} \times Δ I_{S N}^{2} \times \frac{V_{C S N}}{V_{C S N} - (\frac{V_{O U T} + V_{D F R}}{n_{S P}})} \times f_{S W}}$

Over one switching cycle we must have:

$I_{S N} = \frac{V_{C S N}}{R_{S N}} = C_{S N} \times \frac{Δ V_{C S N}}{τ_{S W}}$

So, we can calculate the required C_SN as follows:

$C_{S N} = \frac{V_{C S N}}{Δ V_{C S N} \times R_{S N} \times f_{S W}}$

Generally, ΔV_CSN should be kept to approximately 10% to 30% of V_CSN. Figure 6 illustrates V_CSN, ΔI_SN, and Δt_SN. The voltage across the snubber capacitor, V_CSN, should be selected so that:

$V_{C S N} < V_{D S M A X (Q P)} - V_{I N (M A X)}$

Choosing too large a value for V_CSN causes the voltage on the Q_P drain to get too close to its maximum allowable drain-source voltage, while choosing too small a value results in higher power losses in the snubber resistor. A reasonable value should result in a maximum drain voltage on Q_P that is approximately 75% of its maximum allowable value. The worst-case condition for the snubber circuit occurs at maximum output power when:

$Δ I_{S N} = Δ I_{L P (M A X)}$

Assuming the leakage inductance is 1.5% of the primary inductance, choosing V_CSN = 39V and ΔV_CSN = 7V, we get the following approximate values:

$P_{S N} = 272 m W$

$R_{S N} = 6.2 k Ω$

$C_{S N} = 6.9 n F$

Finally, we consider the snubber diode, D_SN. This diode should have at least the same voltage rating as the MOSFET, Q_P. Although the average forward current is very low, it must have a peak repetitive current rating greater than ΔI_LP(MAX).

Step 12: Calculate the Required Current-Sense Resistor

From Step 4 we have the maximum input power given by:

$P_{I N (M A X)} = \frac{P_{O U T (M A X)}}{η_{M A X}} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X}}$

For the DCM flyback converter all the energy stored in the primary magnetizing inductance, L_P, during the MOSFET on-time is transferred to the output during the MOSFET off-time, i.e., the full power transfer occurs during one switching cycle. Therefore, since E = P x t, we have:

$E_{I N (M A X)} = P_{I N (M A X)} \times τ_{S W} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times f_{S W}}$

The maximum input energy must be stored in L_P during the primary-side MOSFET on-time, so:

$E_{I N (M A X)} = \frac{1}{2} \times L_{P} \times Δ I_{L P (M A X)}^{2}$

Substituting the equations above:

$\frac{1}{2} \times L_{P} \times Δ I_{L P (M A X)}^{2} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times f_{S W}}$

and

$Δ I_{L P} = \sqrt{\frac{2 \times V_{O U T} \times I_{O U T}}{η_{M A X} \times L_{P} \times f_{S W}}}$

From Step 2 we have:

$R_{C S} = Δ V_{R C S} \times \sqrt{\frac{η_{M A X} \times L_{P} \times f_{S W}}{2 \times V_{O U T} \times I_{O U T}}} = 16 m Ω$

A standard 16mΩ resistor was chosen for R_CS.

Figure 6. RCD snubber circuit waveforms.

Step 13: Calculate and Select the Input Capacitors

Figure 7 shows a simplified schematic of the primary side of the flyback converter and the associated current waveforms. In steady-state operation, the converter draws a pulsed high-frequency current from the input capacitor, C_IN. This current leads to a high-frequency ripple voltage across the capacitor according to the following expression:

$I_{C I N} = C_{I N} \times \frac{Δ V_{C I N}}{Δ t}$

It is the ripple voltage arising from the amp-second product through the input capacitor.

During the Q_P on-time interval from t₀ to t₁, the capacitor is supplying current to the primary inductance L_P of the flyback transformer and its voltage is decreasing. During the Q_P off-time time interval from t₁ to t₂, no current is flowing in L_P, and current is being supplied to the capacitor from the input voltage source. According to the charge balance law, the decrease in capacitor voltage during time t₀ to t₁ must equal the increase in capacitor voltage during time t₁ to t₂. So:

$I_{C I N [t_{1} - t_{2}]} = C_{I N} \times \frac{Δ V_{C I N}}{(t_{2} - t_{1})} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times V_{I N (M I N)}}$

And finally, since:

$\frac{1}{(t_{2} - t_{1})} = \frac{f_{S W}}{(1 - D_{M A X})}$

we have:

$C_{I N} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times V_{I N (M I N)}} \times \frac{1}{Δ V_{C I N}} \times \frac{(1 - D_{M A X})}{f_{S W}}$

For maximum high-frequency ripple voltage requirement ΔV_CIN, we can now calculate the required minimum C_IN. There is high-frequency AC current flowing in C_IN, as shown in the center waveform of Figure 7. The selected capacitor must be specified to tolerate the maximum RMS current, I_CIN(RMS). From the simplified schematic:

$I_{L P} = I_{I N} + I_{C I N}$

Therefore:

$I_{C I N (R M S)} = \sqrt{I_{L P (R M S)}^{2} - I_{I N (R M S)}^{2}}$

where:

$I_{I N (R M S)} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times V_{I N (M I N)}}$

and from Step 6:

$I_{L P (R M S)} = Δ I_{L P (M A X)} \times \sqrt{\frac{D_{M A X}}{3}}$

So:

$I_{C I N (R M S)} = \sqrt{\frac{D_{M A X}}{3} \times Δ I_{L P (M A X)}^{2} - \frac{V_{O U T}^{2} \times I_{O U T}^{2}}{η_{M A X}^{2} \times V_{I N (M I N)}^{2}}} \approx 2.03 A_{R M S}$

An additional high-frequency ripple voltage is present due to this RMS current flowing through the ESR of the capacitor. Ceramic capacitors are generally used for limiting high-frequency ripple due to their high AC current capability and low ESR.

Figure 7. Primary-side circuit and currents.

In addition to using a ceramic capacitor for high-frequency input ripple-voltage control as described above, an electrolytic capacitor is sometimes inserted at the input of a flyback converter to limit the input voltage deviation when there is a rapid output load change. A 100% load change gives rise to an input current transient of:

$Δ I_{I N} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times V_{I N (M I N)}}$

During this transient, there is a voltage drop across any series stray inductance, L_IN(STRAY), that exists between the input voltage source and the input capacitor of the power supply. So from:

$\frac{1}{2} \times C_{I N} \times Δ V_{C I N}^{2} = \frac{1}{2} \times L_{I N (S T R A Y)} \times Δ I_{I N}^{2}$

we have:

$C_{I N} = L_{I N (S T R A Y)} \times \frac{Δ I_{I N (M A X)}^{2}}{Δ V_{C I N}^{2}}$

We now have two values for C_IN. One for input high-frequency ripple-voltage control:

$C_{I N (C E R)} = \frac{V_{O U T} \times I_{O U T}}{η_{M A X} \times V_{I N (M I N)}} \times \frac{1}{Δ V_{C I N}} \times \frac{(1 - D_{M A X})}{f_{S W}}$

and a second for transient input voltage control:

$C_{I N (E L E)} = L_{I N (S T R A Y)} \times \frac{Δ I_{I N (M A X)}^{2}}{Δ V_{C I N}^{2}}$

If C_IN(ELE) > C_IN(CER), both ceramic and electrolytic capacitors must be used at the input of the power supply and ΔV_CIN should be limited to approximately 75mV to keep the AC current in the ESR of the electrolytic capacitor within acceptable limits. Otherwise, C_IN(ELE) is not required. In this case, the value of C_IN(CER) can be significantly reduced since there is no longer any requirement to limit ΔV_CIN to less than 75mV. Based on the current design specification with L_IN(STRAY) approximated at 50nH:

C_{I N (C E R)} \approx 73.9 μ F

and

C_{I N (E L E)} \approx 22.5 μ F

Since C_IN(ELE) < C_IN(CER), an electrolytic capacitor is not required. We can now recalculate C_IN(CER) based on a ΔV_CIN = 280mV:

C_{I N (C E R)} \approx 19.8 μ F

Allowing for a ±10% capacitor tolerance and a further reduction of capacitance of 48% due to the DC bias effect (operating a 50V ceramic capacitor at 20V), the final nominal value of input capacitance required is:

$C_{I N (C E R)} = \frac{19.8 μ F}{90 % \times 52 %} \approx 42.3 μ F$

This is achieved by using four 10µF ceramic capacitor (Murata GRM32ER71H106KA12). The AC current in the capacitor is:

I_{C I N (R M S)} \approx 0.5 A_{R M S}

which is well within specification for the selected capacitor.

Step 14: Calculate and Select the Output Capacitor

High-frequency ripple voltage requirements are also used to determine the value of the output capacitor in a flyback converter. Figure 8 shows a simplified schematic of the secondary side of the flyback converter and the associated current waveforms.

In steady-state operation, the load draws a DC current from the secondary side of the flyback converter. By examining the secondary current waveforms, we see that C_O is supplying the full output current I_OUT to the load during the time interval from t₂ to t₃. During this time interval, the voltage across C_O decreases. At time t₃, Q_P has just turned off and the secondary rectifying diode DFR (or the secondary synchronous MOSFET Q_S) starts to conduct supplying current to the load and to C_O. The charging and discharging of C_O leads to a high-frequency ripple voltage at the output according to the following expression:

$I_{C O} = C_{O} \times \frac{Δ V_{C O}}{Δ t}$

Again, as with the input capacitor, this is the ripple voltage arising from the amp-second product through the output capacitor.

By the capacitor charge balance law, the decrease in capacitor voltage during time t₂ to t₃ must equal the increase in capacitor voltage during time t₁ to t₂. When the capacitor is discharging, we have:

$I_{C O [t_{2} - t_{3}]} = C_{O} \times \frac{Δ V_{C O}}{(t_{3} - t_{2})} = I_{O U T}$

Finally, since:

$\frac{1}{(t_{3} - t_{2})} = \frac{f_{S W}}{(1 - D_{S (M A X)})}$

We have:

$C_{O} = I_{O U T} \times \frac{1}{Δ V_{C O}} \times \frac{(1 - D_{S (M A X)})}{f_{S W}}$

For maximum high-frequency ripple voltage requirement ΔV_CO, we can now calculate the required minimum C_O.

C_O ≈ 169.7µF.

As with the input capacitor, an additional high-frequency ripple voltage occurs at the output due to the output capacitor’s ESR and can be minimized by choosing a capacitor with low ESR.

Also, as with the input capacitor, there is high-frequency AC current flowing in C_O as shown in the center waveform of Figure 8.. The selected capacitor must be specified to tolerate this maximum RMS current, I_CO(RMS). From the simplified schematic:

$I_{L S} = I_{O U T} + I_{C O}$

Therefore:

$I_{C O (R M S)} = \sqrt{I_{L S (R M S)}^{2} - I_{O U T (R M S)}^{2}}$

where:

$I_{O U T (R M S)} = I_{O U T}$

and from Step 7:

$I_{L S (R M S)} = Δ I_{L S (M A X)} \times \sqrt{\frac{1 - D_{S (M A X)}}{3}}$

so:

$I_{C O (R M S)} = \sqrt{\frac{1 - D_{S (M A X)}}{3} \times Δ I_{L S (M A X)}^{2} - I_{O U T}^{2}} \approx 5.41 A$

If we allow for a ±20% capacitor tolerance and a further reduction of capacitance of 57% due to the DC bias effect (operating a 6.3V ceramic capacitor at 5V), our final nominal value is:

$C_{O} = \frac{169.7 μ F}{80 % \times 43 %} \approx 493 μ F$

We can achieve this by placing five 100µF ceramic capacitors (Murata GRM32ER60J107ME20) in parallel. The minimum output capacitance using the above combination is 172µF. The AC current in each capacitor is:

$\frac{I_{C O (R M S)}}{5} \approx 1.08 A$

which is well within specification for the selected capacitor.

Figure 8. Secondary-side circuit and currents.

Step 15: Summarize the Power Component Design

A first pass at calculating the power components in the no-opto flyback converter using MAX17690 has been completed. Referring to the schematic, a summary of the power components is listed below:


POWER COMPONENT	QTY	DESCRIPTION
Flyback Transformer	1	PRI. INDUCTANCE = 4µH SEC-PRI TURNS RATIO = 0.5 PEAK. PRI CURRENT = 6.4A PRI. RMS CURRENT = 2.51A PEAK SEC. CURRENT = 12.15A SEC. RMS CURRENT = 5.77A SWITCHING FREQ. = 143kHz
Input Capacitor	4	CAPACITOR; SMT (1210); CERAMIC CHIP 10µF; 50V; 10%; X7R Murata GRM32ER71H106KA12
Output Capacitors	5	CAPACITOR; SMT (1210); CERAMIC CHIP 100µF; 6.3V; 20%; X7R Murata GRM32ER60J107ME20
Primary MOSFET	1	MOSFET; NCH; I-(30A); V-(80V) Vishay SQJA96EP
Synchronous MOSFET	1	MOSFET; NCH; I-(40A); V-(40V) Infineon BSZ040N04LSG

Part II: Setting Up the MAX17690 No-Opto Flyback Controller

Step 16: Setting Up the Switching Frequency

The MAX17690 can operate at switching frequencies between 50kHz and 250kHz (subject to the considerations in Step 3). A lower switching frequency optimizes the design for efficiency, whereas increasing the switching frequency allows for smaller inductive and capacitive components sizes and costs. A switching frequency of 143kHz was chosen in Step 3. R9 sets the switching frequency according to the following expression:

$R 9 = \frac{5 \times 10^{6}}{f_{S W}} \approx 34.8 k Ω$

where R9 is in kΩ and f_SW is in Hz.

Step 17: Setting Up the Soft-Start Time

The capacitor C6 connected between the SS pin and SGND programs the soft-start time. A precision internal 5µA current source charges the soft-start capacitor C6. During the soft-start time, the voltage at the SS pin is used as a reference for the internal error amplifier during startup. The soft-start feature reduces inrush current during startup. Since the reference voltage for the internal error amplifier is ramping up linearly, so too is the output voltage during soft-start. The soft-start capacitor is chosen based on the required soft-start time (10ms) as follows:

$C 6 = 5 \times t_{S S} \approx 50 n F$

where C6 is in nF and t_SS is in ms. We chose a standard 47nF capacitor.

Step 18: Setting Up the UVLO and OVI Resistors

A resistor-divider network of R1, R3, and R2 from V_IN to SGND sets the input undervoltage lockout threshold and the output overvoltage inhibit threshold. The MAX17690 does not commence its startup operation until the voltage on the EN/UVLO pin (R3/R2 node) exceeds 1.215V (typ). When the voltage on the OVI pin (R1/R3 node) exceeds 1.215V (typ), the MAX17690 stops switching, thus inhibiting the output. Both pins have hysteresis built in to avoid unstable turn-on/turn-off at the UVLO/EN and OVI thresholds. After the device is enabled, if the voltage on the UVLO/EN pin drops below 1.1V (typ), the controller turns off; after the device is OVI inhibited, it turns back on when the voltage at the OVI pin drops below 1.1V (typ). Whenever the controller turns on, it goes through the soft-start sequence. For the current design R1 = 10kΩ, R2 = 140kΩ, and R3 = 20kΩ give rise to an UVLO/EN threshold of 6.9V and an OVI threshold of 20.7V.

Step 19: Placing Decoupling Capacitors on V_IN and INTVCC

The MAX17690 no-opto flyback controller compares the voltage V_FLYBACK to V_IN. This voltage difference is converted to a proportional current that flows in R5. The voltage across R5 is sampled and compared to an internal reference by the error amplifier. The output of the error amplifier is used to regulate the output voltage. The V_IN pin should be directly connected to the input voltage supply. For robust and accurate operation, a ceramic capacitor (C2 = 1µF) should be placed between V_IN and SGND as close as possible to the IC.

V_IN powers the MAX17690’s internal low dropout regulator. The LDO’s regulated output is connected to the INTV_CC pin. A ceramic capacitor (C3 = 2.2µF min) should be connected between the INTV_CC and PGND pins for the stable operation over the full temperature range. Place this capacitor as close as possible to the IC.

Step 20: Setting Up the Feedback Components

R_SET (R5), R_FB (R4, R18), R_RIN (R8), R_VCM (R6), and R_TC (R7) are critically important to achieving optimum output voltage regulation across all specified line, load and temperature ranges.

R_SET resistor (R5): This resistor value is optimized based on the IC’s internal voltage to current amplifier and should not be changed.

R 5 = R_{S E T} = 10 k Ω

R_FB resistor (R4, R18): The feedback resistor is calculated according to the following equation:

$R_{F B} = \frac{R_{S E T}}{n_{S P} \times V_{S E T}} \times (V_{O U T} + V_{D F R} + V_{T C} \times \frac{δ V_{D F R}}{δ T} \times \frac{δ T}{δ V_{T C}})$

Since we are using a synchronous MOSFET for rectification on the secondary side, we can assume that δV_DFR / δT = 0, so:

$R_{F B} = \frac{R_{S E T}}{n_{S P} \times V_{S E T}} \times (V_{O U T} + V_{D F R}) \approx 107 k Ω$

From the MAX17690 data sheet, V_SET = 1V. The two resistors R4 = 90.9kΩ and R18 = 18kΩ form R_FB. Using one high value resistor and one low value resistor in series allows slight adjustment to the series resistance combination so that the output voltage can be fine-tuned to its required value, if necessary.

R_RIN resistor (R8): The internal temperature compensation circuitry requires a current proportional to V_IN. R_RIN establishes this current and is calculated according to the following equation:

R_{R I N} \approx 0.6 \times R_{F B}

R_VCM resistor (R6): The MAX17690 generates an internal voltage proportional to the on-time volt-second product. This enables the device to determine the correct sampling instant for V_FLYBACK during the Q_P off-time. Resistor R6 is used to scale this internal voltage to an acceptable internal voltage limit in the device. To calculate the resistor, we must first calculate a scaling constant as follows:

$K_{C} = \frac{(1 - D_{M A X}) \times 10^{8}}{3 \times f_{S W}} = 125$

After K_C is calculated, the R6 value can be selected from the table below by choosing the resistance value that corresponds to the next largest K_C.


K_C	R6
640	0Ω
320	75kΩ
160	121kΩ
80	220Ω
40	Open

In the present case, R6 = 121kΩ.

R_TC resistor (R7): The value of R_TC can then be calculated using the previous expression, restated below:

$R_{T C} = - R_{F B} \times n_{S P} \times \frac{δ T}{δ V_{D F R}} \times \frac{δ V_{T C}}{δ T}$

Since we are using a synchronous MOSFET for secondary-side rectification, we can assume that the temperature coefficient δV_DFR/δT is zero so R_TC should be infinite (i.e., open circuit).

This completes the setup of the MAX17690 no-opto flyback controller.

Part III: Closing the Control Loop

Step 21: Determine the Required Bandwidth

The bandwidth of the control loop determines how quickly the converter can respond to changes at its input and output. If we have a step change in output current, the voltage across the output capacitor decreases as shown in Figure 9.

The control loop detects this reduction in output voltage and increases the duty cycle of Q_P to supply more current to the output capacitor. The amount of time required by the control loop to increase the duty cycle from its minimum value to its maximum value is the response time, τ_RES, of the control loop. For the MAX17690 we have:

$τ_{R E S} \approx {\frac{1}{3 \times f_{C}} + \frac{1}{f_{S W}}}$

where f_C is the bandwidth of the power converter. If we apply a switching load step of amplitude ΔI_STEP, at a frequency of (1/τ_RES) and a 50% duty cycle then, to limit the output voltage deviation to ±ΔV_OUT(STEP) we must have a minimum output capacitance of:

$C_{O (M I N)} = \frac{Δ I_{O U T (S T E P)} \times (\frac{τ_{R E S}}{2})}{Δ V_{O U T (S T E P)}}$

Combining the two previous equations, we have:

$f_{C} = \frac{f_{S W} \times Δ I_{O U T (S T E P)}}{3 \times f_{S W} \times C_{O (M I N)} \times Δ V_{O U T (S T E P)} - Δ I_{O U T (S T E P)}}$

It is normal to specify ΔV_OUT(STEP) for a load step from 50% to 100% of the maximum output current. We have already calculated C_O(MIN) = 172µF in Step 14, f_SW = 143kHz, so based on a 3% maximum ΔV_OUT(STEP):

f_{C} \approx 7 k H z

Step 22: Calculate the Loop Compensation

The MAX17690 uses peak current-mode control and an internal transconductance error amplifier to compensate the control loop. The control loop is modeled, as shown in figure 10, by a power modulator transfer function G_MOD(S), an output-voltage feedback transfer function G_FB(S), and an error amplifier transfer function G_EA(S).

Figure 10. Simplified model of control loop.

The power modulator has a pole located at f_P(MOD) determined by the impedance of the output capacitor C_O and the load impedance R_L. It also has a zero at f_Z(MOD) determined by the impedance of C_O and the ESR of C_O. The DC gain of the power modulator is determined by the peak primary current ΔI_LP and the current-sense resistor R_CS. So:

$G_{M O D (D C)} = \frac{1}{Δ I_{L P} \times R_{C S}}$

$f_{P (M O D)} = \frac{1}{2 π \times C_{O} \times R_{L}} = \frac{I_{O U T}}{2 π \times C_{O} \times V_{O U T}}$

and:

$f_{Z (M O D)} = \frac{1}{2 π \times C_{O} \times E S R_{C O}}$

The output voltage feedback transfer function G_FB(S) is independent of frequency and has a DC gain determined by V_IN, V_FLYBACK, and V_SET as follows:

$G_{F B (D C)} = \frac{V_{S E T}}{V_{F L Y B A C K} - V_{I N}} = \frac{V_{S E T} \times n_{S P}}{V_{O U T} + V_{D F R}}$

The MAX17690’s transconductance error amplifier should be set up in a configuration to compensate for the pole at f_P(MOD) and the zero at f_Z(MOD) of the modulator. This can be achieved by a Type II transconductance error amplifier compensation shown in Figure 11.

This compensation scheme type has a low frequency pole at f_P-LF(EA) due to the very large output resistance R_O (30MΩ to 50MΩ) of the operational transconductance amplifier (OTA). It has a zero at f_Z(EA) determined by C_Z and R_Z of the compensation network, and it has an additional pole at f_P(EA) determined by C_P and R_Z of the compensation network. So:

$f_{P - L F (E A)} = \frac{1}{2 π \times C_{Z} \times (R_{O} + R_{Z})}$

$f_{Z (E A)} = \frac{1}{2 π \times C_{Z} \times R_{Z}}$

Figure 11. Type II compensation for OTA.

and:

$f_{P (E A)} = \frac{1}{2 π \times C_{P} \times R_{Z}}$

To achieve stable operation, we must ensure that:

$f_{P (M O D)} ≪ f_{C} < \frac{f_{S W}}{20}$

Set the closed-loop gain at f_C equal to 1:

$G_{M O D (f_{C})} \times G_{F B (f_{C})} \times G_{E A (f_{C})} = 1$

Place the zero in the error amplifier network at the same frequency as the pole in the power modulator transfer function:

$f_{Z (E A)} = f_{P (M O D)}$

$\frac{1}{2 π \times C_{Z} \times R_{Z}} = \frac{I_{O U T}}{2 π \times C_{O} \times V_{O U T}}$

The frequency f_Z(MOD) at which the zero occurs in the power modulator transfer function depends on the ESR of C_O. If ceramic capacitors are used for C_O, f_Z(MOD) will be generally much higher than f_C. However, if the ESR of C_O is large, f_Z(MOD) could be lower than f_C. This is a very important point since both the gain of the power modulator at f_C, and the gain of the error amplifier at f_C depend on whether f_Z(MOD) is greater than or less than f_C. This is illustrated in Figure 12.

By examining the gain plots in Figure 12, we see that for f_Z(MOD) > f_C:

$G_{M O D (f_{C})} = G_{M O D (D C)} \times (\frac{f_{P (M O D)}}{f_{C}})$

$G_{E A (f_{C})} = g_{m (E A)} \times R_{Z}$

and for f_Z(MOD) < f_C:

$G_{M O D (f_{C})} = G_{M O D (D C)} \times (\frac{f_{P (M O D)}}{f_{Z (M O D)}})$

$G_{E A (f_{C})} = g_{m (E A)} \times (\frac{f_{Z (M O D)}}{f_{C}}) \times R_{Z}$

For the current design, we have:

$f_{P (M O D)} = \frac{I_{O U T}}{2 π \times V_{O U T} \times C_{O}} \approx 349 H z$

$f_{Z (M O D)} = \frac{1}{2 π \times E S R_{C O} \times C_{O}} \approx 4.6 M H z$

Since f_Z(MOD) > f_C:

$G_{M O D (f_{C})} = G_{M O D (D C)} \times (\frac{f_{P (M O D)}}{f_{C}}) = \frac{1}{Δ I_{L P} \times R_{C S}} \times (\frac{f_{P (M O D)}}{f_{C}})$

$G_{E A (f_{C})} = g_{m (E A)} \times R_{Z}$

and since G_FB is independent of frequency, we have:

$G_{F B (f_{C})} = G_{F B (D C)} = \frac{V_{S E T} \times n_{S P}}{V_{O U T} + V_{D F R}}$

We can now set the closed-loop gain equal to 1 as follows:

$G_{M O D (f_{C})} \times G_{F B (D C)} \times G_{E A (f_{C})} = 1$

$\frac{1}{Δ I_{L P} \times R_{C S}} \times (\frac{f_{P (M O D)}}{f_{C}}) \times \frac{V_{S E T} \times n_{S P}}{V_{O U T} + V_{D F R}} \times g_{m (E A)} \times R_{Z} = 1$

Rearranging we can calculate:

$R_{Z} = \frac{1}{g_{m (E A)}} \times \frac{V_{O U T} + V_{D F R}}{V_{S E T} \times n_{S P}} \times (\frac{f_{C}}{f_{P (M O D)}}) \times R_{C S} \times Δ I_{L P}$

Substituting ΔI_LP from Step 12:

$R_{Z} = \frac{1}{g_{m (E A)}} \times \frac{V_{O U T} + V_{D F R}}{V_{S E T} \times n_{S P}} \times (\frac{f_{C}}{f_{P (M O D)}}) \times R_{C S} \times \sqrt{\frac{2 \times V_{O U T} \times I_{O U T}}{η_{M A X} \times L_{P} \times f_{S W}}} = 13.8 k Ω$

Finally, we can calculate the remaining components, C_Z and C_P, in the error amplifier compensation network as follows:

$C_{Z} = \frac{1}{2 π \times f_{P (M O D)} \times R_{Z}} = 36 n F$

and:

$C_{P} = \frac{1}{2 π \times f_{Z (M O D)} \times R_{Z}} = 3 p F$

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Design & Integration File

MAXREFDES1012 Design Files
7/9/2019

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Circuit Note

MAXREFDES1012: Miniature, 5.3V/2A, Synchronous, No-Opto Flyback DC-DC Converter with 91.4% Efficiency Using MAX17690 and MAX17606
7/9/2019

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Overview

Design Resources

Design & Integration File

Description

Features & Benefits

Parts Used

Details Section

Introduction

Hardware Specification

Designed–Built–Tested

The Isolated No-Opto Flyback Converter

Temperature Compensation

Setting the VFLYBACK Sampling Instant

Design Procedure

Design Procedure for No-Opto Flyback Converter Using MAX17690

Part I: Designing the Power Components

Step 1: Calculate the Minimum Turns Ratio for the Flyback Transformer

Step 2: Estimate the Maximum and Minimum Duty Cycle Under Normal Operating Conditions

Step 3: Calculate the Maximum Allowable Switching Frequency

Step 4: Estimate the Primary Magnetizing Inductance

Step 5: Recalculate DMAX, DMIN, and tON(MIN) Based on the Selected Value for LP

Step 6: Calculate the Peak and RMS Currents in the Primary Winding of the Flyback Transformer

Step 7: Calculate the Peak and RMS Currents in the Secondary Winding of the Flyback Transformer

Step 8: Summarize the Flyback Transformer Specification

Step 9: Calculate Design Parameters for Secondary-Side Rectifying Device

Step 10: Calculate Design Parameters for Primary-Side MOSFET

Step 11: Select the RCD Snubber Components

Step 12: Calculate the Required Current-Sense Resistor

Step 13: Calculate and Select the Input Capacitors

Step 14: Calculate and Select the Output Capacitor

Step 15: Summarize the Power Component Design

Part II: Setting Up the MAX17690 No-Opto Flyback Controller

Step 16: Setting Up the Switching Frequency

Step 17: Setting Up the Soft-Start Time

Step 18: Setting Up the UVLO and OVI Resistors

Step 19: Placing Decoupling Capacitors on VIN and INTVCC

Step 20: Setting Up the Feedback Components

Part III: Closing the Control Loop

Step 21: Determine the Required Bandwidth

Step 22: Calculate the Loop Compensation

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Step 5: Recalculate D_MAX, D_MIN, and t_ON(MIN) Based on the Selected Value for L_P

Step 19: Placing Decoupling Capacitors on V_IN and INTVCC