Worst-case power consumed by a circuit with MOSFETs

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zenterix
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Homework Statement
Consider the circuit depicted below. Use the Switch-Resistor model for the MOSFETs.

Assume ##R_6=R_7=10\mathrm{k\Omega}##, the threshold voltage for the MOSFETs is ##\mathrm{2V}##, and the resistance of the MOSFET in its ON state is ##\mathrm{1k\Omega}##.

Compute the worst-case power consumed by this circuit.
Relevant Equations
##P=iV##
Here is the circuit.
Circuit schematic of two MOSFET branches from Vs: left branch has R6, transistor A, a parallel pair of transistors B and an always-on transistor, with OUT node; right branch has R7, node V2, and transistor C to ground. Currents i and i2 are marked in red.


Note that no current flows between the left and right sides of the circuit: their only relationship happens through the MOSFET that is parallel to B.

There are eight cases to consider: all the combinations of ON/OFF for the three MOSFETs.

Here is a summary of the eight cases according to my calculations

Handwritten table with columns A, B, C, Vout, i, i2 and P and eight numbered rows listing ON (1) and OFF (0) states of the three MOSFETs with corresponding voltage, current and power expressions, such as 25/6 V, 5/(12·10^3) A and 5·5/(11.5·10^3) W.


It seems that the highest power consumption occurs when all MOSFETs are on.

Here is more in depth explanation of how the table above was created.

It seems that whether MOSFET C is on or off, the voltage at the node right above it (beneath resistor R7) is higher than ##V_T=\mathrm{2V}##.

Thus, the MOSFET in parallel with B is always on.

Let's consider just the left side of the circuit for now.

If ##A## is off, then no current flows so ##i=0##.

Suppose A is on.

Then we have two cases:
B on, B off.

If B is on, then the left side of the circuit becomes

Hand-drawn circuit schematic of a vertical series chain: a 5V source Vs at top with an arrow for current i pointing down, then resistors labeled 10 (top), 1 (middle) and 1/2 (bottom), with a node Vout tapped between the top two resistors and ground at the bottom.


##i=\frac{5}{11.5\cdot 10^3}\text{A}## and ##v_{OUT}=\frac{10}{11.5}\cdot 5\text{V}##.

Power consumed by this subcircuit is ##i\cdot V_S=5\cdot\frac{5}{11.5\cdot 10^3}\mathrm{\frac{J}{S}}##.

If, on the other hand, B is off then the only difference is that the resistance that is ##1/2\mathrm{k\Omega}## above becomes ##1\mathrm{k\Omega}##.

Thus, ##i=\frac{5}{12\cdot 10^3}\mathrm{A}## and ##v_{OUT}=\frac{10}{12}\cdot 5=\frac{25}{6}\text{V}##.

Now consider the right side of the circuit.

Suppose ##C## is on. Then the current on the right side (call it ##i_2##) will be ##i_2=\frac{5\text{V}}{11\cdot 10^3}\text{A}##.

If ##C## is off, then no current flows on the right side.

Finally, to obtain power consumption I simply computed ##P=i\cdot V_S+i_2\cdot V_S## for all eight cases.

Is this the correct calculation?