Add Margining Capability to a DC/DC Converter

Introduction

A similar version of this article appeared in the September 18, 2008 issue of EDN magazine.

You can easily add margining capability to a DC/DC converter (digital adjustment of the output voltage) by making a single connection to the existing circuit as shown by the dotted line in Figure 1. The extra IC is a 2- or 4-channel, I²C-adjustable current DAC (DS4402 or DS4404). Because each DAC output is 0mA at power up, the extra circuitry is essentially transparent to the system until a command is written via the I²C bus.

As an example, assume V_IN = 3V to 5.5V, V_OUT = 1.8V (the desired nominal output voltage), and V_FB = 0.6V (not to be confused with V_REF of the DS4404). You can obtain the V_FB value from the DC/DC converter datasheet, being sure to verify that it is within the OUTx voltage range specified in the current DAC datasheet (specified as V_OUT:SINK and V_OUT:SOURCE depending on whether you are sinking or sourcing current). It's also important to verify the input impedance of the DC/DC converter's FB pin (the circuit shown assumes a high impedance).

Assume we want to add a ±20% margining capability to the DC/DC converter output (V_OUT):

V_OUTMAX = 2.16V
V_OUTNOM = 1.8V
V_OUTMIN = 1.44V

First, determine the necessary relationship between R_TOP and R_BOTTOM that yields the nominal output (V_OUTNOM) when I_DS4404 = 0A:

Solving for R_TOP,

For our example,

The current (I_DS4404) required to make V_OUT increase to V_OUTMAX is derived by summing currents at the FB node:

This equation can be simplified by solving Equation 1 for R_BOTTOM and substituting, which yields:

In terms of margin percentage:

where margin = 0.2, to implement ±20% margining in this case. Before you can use this relationship to calculate R_TOP and R_BOTTOM, you must select the full-scale current I_FS.

According to the DS4404 datasheet, the full-scale current (specified as I_OUT:SINK or I_OUT:SOURCE) must be between 0.5mA and 2.0mA, to guarantee the specifications for accuracy and linearity. Unfortunately, no formula is available for calculating the ideal full-scale current. That value is influenced by the desired number of steps, the step size, and the values for R_TOP and R_BOTTOM. Another factor affecting the full-scale current value would be the requirement that a particular register setting correspond to a particular margin percentage.

In any case, your selection of a full-scale current will likely require several iterations, in which you select an arbitrary value (within the range), and then calculate R_TOP, R_BOTTOM, R_FS, and step size. When you've determined an acceptable full-scale current value, you may want to further adjust it (or some of the resistor values) to ensure that the resistor values finally specified are commonly available.

To calculate R_TOP for the original example, we choose IFS = IDS4404, which gives us 31 equal increments (steps) from V_OUTNOM to V_OUTMAX, as well as 31 steps from V_OUTNOM to V_OUTMIN. This resolution is more than adequate for our example. We could, for instance, begin by arbitrarily choosing I_FS in the center (1.25mA) of the specified range, and then performing all the calculations. Instead, for illustrative purposes we perform calculations for the endpoints of the range (0.5mA, 2.0mA).

So, for I_FS = I_DS4404 = 0.5mA: Using Equation 3 and solving for R_TOP,

To calculate R_FS, use the formula in the DS4404 datasheet plus the V_REF value also found in that datasheet:

Finally, for completeness we determine the DS4404 output current as a function of register setting: I_OUT(register setting) = step size × register setting.

Note that the register setting above does not include the sign bit, which is used to select sink or source. The DS4404 sinks current when the sign bit = 0, making V_OUT increase to V_OUTMAX. It sources current when the sign bit = 1, making V_OUT decrease towards V_OUTMIN.

For I_FS = I_DS4404 = 2.0mA:

Comparing R_TOP and R_BOTTOM for the two cases (I_FS = 0.5mA vs. 2.0mA), you can see that I_FS = 0.5mA is more attractive because the resistances are higher.