# Calculating the efficiency of a voltage regulator made with an op-amp

I have the following circuit:

which can be re-drawn as:

This is supposed to be a voltage regulator. Note: the diode is ideal.

I'm asked to find resistor values for $$\R_0\$$, $$\R_1\$$ and $$\R_2\$$ to obtain $$\5\ \mathrm{V} \pm 0.1\ \mathrm{V}\$$ when the input voltage varies between $$\10\ \mathrm{V}\$$ and $$\20\ \mathrm{V}\$$, and the circuit efficiency associated to these values. Both questions come together, one depends on the other, so I can't split this question in 2.

The zener diode has a $$\v_z=2.5\ \mathrm{V}\$$ and $$\R_z=1\ \Omega\$$. This means that the zener can be replaced by a $$\1\ \Omega\$$ resistor in series with a $$\5\ \mathrm{V}\$$ voltage supply.

I understand the op-amp is in a non-inverting configuration, so $$\v_{outopamp} = v^+(1 + \frac{R_1}{R_2})\$$.

Since $$\v_z = 2.5\ \mathrm{V}\$$ and $$\R_z = 1\ \Omega\$$, I calculate $$\v^+\$$ as:

$$v^+ = v_{IN} - v_{R_0} = v_{IN} - R_0I_0 = v_{IN} - R_0\frac{v_{IN} - v_z}{R_0 + R_z}$$

With the above formula and the non-inverting op-amp one, I obtain: $$v_{OUT} = (v_{IN} - R_0\frac{v_{IN} - v_z}{R_0 + R_z})(1 + \frac{R_1}{R_2})$$

With all the above, I choose $$\R_0 = R_1 = R_2 = 100\ \mathrm{k\Omega}\$$. With those values, I substitute the values above in $$\v^+\$$ and $$\v_{OUT}\$$ and I obtain the right output voltages for $$\10\ \mathrm{V}\$$ and $$\20\ \mathrm{V}\$$.

Are the calculations above correct? I would like to confirm this.

If so, I need now to calculate the efficiency of this circuit with $$\K = 0.95\$$ and a load resistor $$\R_L\$$ connected at $$\v_{OUT}\$$:

Note: I must ignore the power consumed by the op-amp itself, but not the power consumed as a result of the current going out of its output terminal.

What I do is:

Assuming $$\I\$$ is the current flowing through $$\R_0\$$, find $$\I\$$ and $$\i\$$:

$$I = \frac{v_{IN} - v_Z}{R_0 + R_z}$$

$$i = \frac{v_{OUT} - v^+}{R_1} + \frac{v_{OUT}}{R_L}$$

Then, I calculate the power consumed by the input stage, by the op-amp (power rails are not visible, but they exist) and by the output stage:

$$P_{input} = v_{IN}(Ki + I)$$

$$P_{opamp} = v_{OUT}(1-K)i$$

$$P_{output} = \frac{(v_{OUT} - v^+)^2}{R_1} + \frac{(v^+)^2}{R_2}$$

$$P_{load} = \frac{v_{OUT}^2}{R_L}$$

Then, the ratio is the power consumed by the output divided by the total power consumed, so:

$$Ratio = \frac{P_{load}}{P_{input} + P_{opamp} + P_{load} + P_{output}}$$

And still get the wrong result. Could I get some help please?

• Don't forget the power consumed by the zener diode circuitry.
– qrk
Dec 7, 2023 at 20:49

EDIT, I made a mistake in the answer I gave. I fixed it now!

Well, let's analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_\text{z}+\text{I}_0&=\text{K}\cdot\text{I}_1\\ \\ \text{I}_2+\text{K}\cdot\text{I}_1&=\text{I}_1 \end{alignat*} \end{cases}\tag1

When we use and apply Ohm's law, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_\text{z}&=\frac{\displaystyle\text{V}_\text{z}-\text{V}_+}{\displaystyle\text{R}_\text{z}}\\ \\ \text{I}_\text{z}&=\frac{\displaystyle\text{V}_+-\text{V}_\text{i}}{\displaystyle\text{R}_0}\\ \\ \text{I}_1&=\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\left(\text{V}_1-\text{V}_2\right)}{\displaystyle\eta\text{k}\text{T}}\right)-1\right)\\ \\ \text{I}_1&=\frac{\displaystyle\text{V}_2-\text{V}_-}{\displaystyle\text{R}_1}\\ \\ \text{I}_1&=\frac{\displaystyle\text{V}_--0}{\displaystyle\text{R}_2} \end{alignat*} \end{cases}\tag2

Where the Shockley diode equation, gives the relation between the voltage across and the current through a diode with $$\\displaystyle\text{I}_1\$$ is the diode current, $$\\displaystyle\text{I}_\text{S}\$$ is the reverse bias saturation current, $$\\displaystyle\text{V}_1-\text{V}_2\$$ is the voltage across the diode, $$\\displaystyle\text{q}\$$ is the electron charge, $$\\displaystyle\text{k}\$$ is the Boltzmann constant, $$\\text{T}\$$ is the temperature and $$\\displaystyle\eta\$$ is the ideality factor.

When working with an ideal opamp we know that:

$$\text{V}_+=\text{V}_-:=\text{V}\tag3$$

So, we can rewrite $$\(2)\$$:

\begin{cases} \begin{alignat*}{1} \text{I}_\text{z}&=\frac{\displaystyle\text{V}_\text{z}-\text{V}}{\displaystyle\text{R}_\text{z}}\\ \\ \text{I}_\text{z}&=\frac{\displaystyle\text{V}-\text{V}_\text{i}}{\displaystyle\text{R}_0}\\ \\ \text{I}_1&=\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\left(\text{V}_1-\text{V}_2\right)}{\displaystyle\eta\text{k}\text{T}}\right)-1\right)\\ \\ \text{I}_1&=\frac{\displaystyle\text{V}_2-\text{V}}{\displaystyle\text{R}_1}\\ \\ \text{I}_1&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_2} \end{alignat*} \end{cases}\tag4

Now, the effiency of your circuit is given by:

$$\eta_\text{circuit}:=\frac{\displaystyle\text{P}_\text{out}}{\displaystyle\text{P}_\text{in}}=\frac{\displaystyle\text{V}_2\cdot\text{I}_1}{\displaystyle\text{V}_\text{i}\cdot\text{I}_0}\tag5$$

So, we need to solve the equations for $$\\displaystyle\text{I}_0\$$, $$\\displaystyle\text{V}_2\$$ and $$\\displaystyle\text{I}_1\$$.

I used Mathematica to do so, and found:

$$\text{I}_0=\frac{\displaystyle\text{V}_\text{z}\left(\text{R}_2-\text{K}\text{R}_0\right)-\text{V}_\text{i}\left(\text{R}_2+\text{K}\text{R}_\text{z}\right)}{\displaystyle\text{R}_2\left(\text{R}_0+\text{R}_\text{z}\right)}\tag6$$ $$\text{V}_2=\frac{\displaystyle\left(\text{R}_1+\text{R}_2\right)\left(\text{V}_\text{i}\text{R}_\text{z}+\text{V}_\text{z}\text{R}_0\right)}{\displaystyle\text{R}_2\left(\text{R}_0+\text{R}_\text{z}\right)}\tag7$$ $$\text{I}_1=\frac{\displaystyle\text{V}_\text{i}\text{R}_\text{z}+\text{V}_\text{z}\text{R}_0}{\displaystyle\text{R}_2\left(\text{R}_0+\text{R}_\text{z}\right)}\tag8$$

So, we end up with:

$$\eta_\text{circuit}=\frac{\displaystyle\text{V}_\text{i}\text{R}_\text{z}+\text{V}_\text{z}\text{R}_0}{\displaystyle\text{R}_2\left(\text{R}_0+\text{R}_\text{z}\right)}\cdot\frac{\displaystyle\left(\text{R}_1+\text{R}_2\right)\left(\text{V}_\text{i}\text{R}_\text{z}+\text{V}_\text{z}\text{R}_0\right)}{\displaystyle\text{V}_\text{i}\left(\text{V}_\text{z}\left(\text{R}_2-\text{K}\text{R}_0\right)-\text{V}_\text{i}\left(\text{R}_2+\text{K}\text{R}_\text{z}\right)\right)}\tag9$$

So, we find when using your values:

$$\eta_\text{circuit}=\frac{\displaystyle2\left(250000+\text{V}_\text{i}\right)^2}{\displaystyle100001\text{V}_\text{i}\left(250000 \left(\text{K}-1\right)+\text{V}_\text{i}\left(100000+\text{K}\right)\right)}\tag{10}$$

So, for example when $$\\displaystyle\text{V}_\text{i}=15\space\text{V}\$$ and $$\\displaystyle\text{K}=10\$$ we get:

$$\eta_\text{circuit}=\frac{2500300009}{112505625045}\approx2.22238\space\text{%}\tag{11}$$

I used the following Mathematica code:

Clear["Global*"];
Vp = V;
Vm = V;
FullSimplify[
Solve[{Iz + I0 == k1*I1, I2 + k1*I1 == I1, Iz == (Vz - Vp)/Rz,
Iz == (Vp - Vi)/R0,
I1 == (Is*(Exp[(q*(V1 - V2))/(\[Eta]*k*T)] - 1)),
I1 == (V2 - Vm)/R1, I1 == (Vm - 0)/R2}, {Iz, I0, I1, I2, V, V1,
V2}]]
`

Where $$\\displaystyle k1=\text{K}\$$, $$\\displaystyle Vm=\text{V}_-\$$ and $$\\displaystyle Vp=\text{V}_+\$$.

• As you say, the power efficiency of the circuit is Pout / Pin, not Pout / (Pin + Pout). My confusion came from the fact that my problem's description specified that the efficiency was the second option; it was an errata. Thanks for your answer. Dec 9, 2023 at 11:19
• @Martel You're more than welcome! Dec 9, 2023 at 11:28
• @Martel I also added the mathematica code I used. Dec 9, 2023 at 11:33

It looks like in your model, K is the current gain of the transistor. However, for that to be Ki, i would have to be the base current, not the emitter current...unless K=Hfe/(Hfe+1). Then you have it labeled correctly.

However, I think your main issue is that you neglected the power dissipated by the pass tranistor. The power this device has to dissipate is roughly (Vin-Vout)*i or, if you're going to be especially picky, (Vin-Vout)*Ki. IRL, to be unnecessarily precise, you'd have to include a term for the miniscule base current times the BE voltage, but your ideal diode means that term will drop out anyway.

I don't think you should model the zener as a 2.5V source with 1 ohm in series. Rather a source with 1Ω in series that adds up to 2.5V at the recommended current.

Otherwise your feedback divider will be wrong. If you've been told otherwise then the feedback divider will have to be adjusted but the calcuations will not change much.

Ignoring the op-amp quiescent supply current (if you wanted to include it, input power would have another term Iop*Vin, but you're specifically instructed to ignore it).

Input power is then just $$\P_{IN} = V_{IN} (\frac {(V_{IN}-V_Z)}{R_0} + i)\$$

where $$\i = \frac{5V}{(R_L||(R_1+R_2)}\$$

The input current at voltage Vi is composed of the current that flows to the zener diode and $$\R_0\$$ and the current that flows entirely to the 5V net (through the op-amp and through the BJT).

And output power is $$\P_{OUT} = \frac{(5V)^2}{R_L}\$$

Efficiency is $$\\frac{P_{OUT}}{P_{IN}}\$$

The BJT gain doesn't matter for the purposes of efficiency calculation because the current all comes from Vin and it all goes to the 5V net. The gain just determines how much the op-amp vs. the transistor must handle.