### Abstract

To minimize the power dissipation in a single-supply op amp, driving a ground-referenced load, connect a pullup resistor, with value equal to the load resistor between the output and the positive supply voltage. This addition enables the op amp to operate at higher ambient temperatures and drive lower-resistance loads, limited only by its maximum ratings for output voltage and current (rather than the package power dissipation).

To maximize signal swing, the output of a single-supply op amp is usually biased at half the supply voltage (Figure 1a). For ground-referenced loads, however, this configuration causes maximum power dissipation in the IC.

The solution is simple and effective: connect a pullup resistor, with value equal to the load resistor, between the output and the positive supply voltage (Figure 1b). This addition enables the op amp to operate at higher ambient temperatures and drive lower-resistance loads, limited only by its maximum ratings for output voltage and current (rather than the package power dissipation).

For example, consider the MAX4220 quad op amp, in which each output drives a 30Ω resistor to ground. For V_{CC} = 5V, the device exceeds its package power rating. Connecting 30Ω pullups at each output, however, minimizes the IC's power dissipation because each op-amp's output current is zero. Power is now dissipated in the pull-up resistors and not in the op amps.

Calculating power dissipation for the op amp in Figure 1a is straightforward:

P_{DC} = (V_{CC} - V_{OUT}) V_{OUT}/R.

Solving the differential equation dP_{DC}/dV_{OUT} = 0 for V_{OUT} shows that the op amp's maximum power dissipation (V_{CC}²/4R) occurs when V_{OUT} = V_{CC}/2.

The corresponding power calculation for the circuit with load and pullup resistors (Figure 1b) is simpler if you convert the load circuit to its Thevenin equivalent (Figure 1c):

P_{DC} = (V_{CC} - V_{OUT})(V_{OUT} - ½V_{CC})/½R (for V_{OUT} ≥ ½V_{CC}),

and P_{DC} = V_{OUT}(½V_{CC} - V_{OUT})/½R (for V_{OUT} ≤ ½V_{CC}).

Again, solving dP_{DC}/dV_{OUT} = 0 for these two equations shows that the maximum power dissipation (V_{CC}²/8R) occurs for V_{OUT} = ¾V_{CC} and for V_{OUT} = ¼V_{CC}. Note that this maximum power level is only half of that with no pullup resistor. The amplifier with no pullup resistor delivers maximum output current at the V_{CC}/2 quiescent point, but with a pullup resistor (Figure 2) the op amp delivers no output current at all!

Similar power advantages accrue for AC applications. Consider a sinusoidal signal superimposed on a DC level of V_{CC}/2: V_{OUT} = ½V_{CC} + V_{p}sinωt, where V_{p} is the peak value of the sinusoid. The resulting waveforms are shown in Figure 3. The simplest method for calculating power dissipation in the op amp is to solve a power-balance equation, in which supply power equals the sum of power dissipated in the load and in the op amp. Thus, the op amp dissipation equals supply power minus load power.

For the case of Figure 1a, supply power equals the average supply current (V_{CC}/2R) times V_{CC}, i.e., V_{CC}²/2R. Power in the load is the sum of the DC and AC components: (1/R)(½V_{CC})² + (1/R)(V_{p}/2^{½})². Thus, supply power minus load power for the Figure 1a circuit is P_{AC} = V_{CC}²/4R - V_{p}²/2R, as shown in Figure 4.

For the Figure 1b circuit, supply power equals the average supply current 2V_{p}/πR times V_{CC} as shown in Figure 3, i.e., 2V_{p}V_{CC}/πR. Power in the load is 2(V_{p}/2^{½})²/R, and the supply power minus load power is P_{AC} = 2V_{CC}V_{p}/πR - V_{p}²/R as shown in Figure 4. Solving the equation dP_{AC}/dV_{p} = 0 for V_{p} shows that the Figure 1b op amp's maximum power dissipation (V_{CC}/π)²/R) occurs when V_{p} = V_{CC}/π.

Thus, you can minimize the power dissipation in a single-supply op amp, driving a ground-referenced load, by adding a single external resistor.

A similar version of this article appeared in the July 23, 2001 issue of *Electronic Design *magazine.