# Power Dissipation of Op-Amp

If we look at the output stage of the MP111 op-amp it's just 2 MOSFETs in series. I'm curious why the power dissipation of the device isn't simply $$\R_{\text{ds(on)}} \times I_{\text{out}}\$$, but rather $$\I_{\text{out}} \times (\text{supply voltage} - \text{output voltage})\$$

If we had a MOSFET in a box, the power dissipated by that 'box' would still just be $$\R_{\text{ds(on)}} \times I_{\text{out}}\$$ wouldn't it?

• Rds(on) is specified when the MOSFET is fully on and has << Vgs(th) from drain to source which is not the case here. What is the equivalent resistance of the drain to source if the voltage across it is actually Vsupply - Vout, and the current is Iout? Given Ohm's law.. – Spehro Pefhany Nov 5 '18 at 2:57

$$\R_{\text{ds(on)}}\$$ is relevant when the MOSFET is fully on, in which case the channel resistance can be considered to be constant, but in that case the power dissipation would actually be $$\R_{\text{ds(on)}} \times {I_{\text{ds}}}^2\$$, which you can derive from Ohm's law and $$\P = IV\$$.
However, this is not relevant when the MOSFET is operated in the linear region, especially when there is feedback that regulates the voltage across the MOSFET, as in an op-amp output stage. Here channel resistance is a limiting factor in the maximum output voltage range with respect to the power supply as a function of output current, but it is not a useful parameter in determining power dissipation in the part. This is because the equivalent channel resistance changes as the surrounding circuity manipulates the gate voltage to hold the circuit's output voltage (which is directly related to the MOSFETs' $$\V_{\text{ds}}\$$) constant. This is not really an accurate way to understand MOSFET operation, but it's good enough to understand why $$\R_{\text{ds}}\$$ is not useful here.
Since $$\V_{\text{ds}}\$$ is controlled, we can simply multiply that by $$\I_{\text{ds}}\$$ and get the power dissipation in each MOSFET in the output stage. Note that this isn't quite the same thing as saying that the power dissipation is $$\I_{\text{out}} \times (\text{supply voltage} - \text{output voltage})\$$, because we have to consider the direction of current flow. If the circuit is sourcing current into a load, than the current is flowing from the positive rail through the high side transistor, and so $$\P = (V_{S+}-V_{\text{out}}) \times I_{\text{out}}\$$. Conversely, if the circuit is sinking current, then current is flowing out of the load, through the low side transistor, and into the negative rail, so $$\P = (V_{S-}-V_{\text{out}}) * I_{\text{out}}\$$. Thus if the output voltage is close to the positive supply then the device will dissipate more power when sinking current than when sourcing current.