# How to calculate the gain of a non-inverting opamp with a transistor?

I have a question about this circuit. How do i rearrange the formula to calculate the voltage gain of this opamp with a transistor: I know the formula for the gain of a non-inverting opamp like in the next picture : and it is, Av = 1 + (R1/R2) (the gain).

But in the first circuit the resistor connected to ground is not there anymore, it is switched to the other side of R2.

So is the formula for gain now, Av = 1+(R2/0), so that the gain is 1 ?

and does the opamp in the first example now act as a voltage follower, so that the voltage is the same on Vout of the opamp ? So that the base of the transistor gets a current of : I = (5V/R1).

Like this circuit : • Consider the transistor as an additional stage in the opamp. Multiply the opamp's open loop gain by the open loop gain of an emitter follower, then proceed as normal. Dec 19, 2021 at 15:56
• In the schematic from the book, it is WRONGLY written backwards as " Av = 1 + (R1/R2) (the gain)." R1 and R2 are swapped. Dec 19, 2021 at 17:56
• No (signal) current flows into the '-' input, so no current flows through R2, so there is no voltage drop across R2. Now it should be clearer that VA = VUb. The purpose of the transistor is to supply more current to the load than the opamp is capable of. Dec 20, 2021 at 0:10

But in the first circuit the resistor connected to ground is not there anymore, it is switched to the other side of R2.

It isn't switched anywhere; it becomes $$\\infty\text{ }\Omega\$$

So is the formula for gain now, Av = 1+(R2/0), so that the gain is 1 ?

It is a gain of 1 but you made the resistor zero in your formula. It should be made $$\\infty\text{ }\Omega\$$ not zero.

and does the opamp in the first example now act as a voltage follower, so that the voltage is the same on Vout of the opamp ?

Not quite; once R2 has extended to the emitter of the BJT, the emitter becomes the "new" output and, this means that the voltage across RL equals Ub: - And, because the op-amp inputs are very high impedance, you can also short out R2 (doesn't affect the gain formula).

In other words, you have made a unity gain buffer amplifier that can supply a lot more load current into RL because of the transistor. Any BJT imperfections (such as volt drop from base the emitter) are "inside" (or within) the feedback loop and the op-amp tries its very best to make the voltage across RL equal to Ub.

• Thanks for your anwser, but i still have a question about something. What happens with the voltage drop at the Vout of the opamp ? Does the voltage drop to 0 V at the base of the transistor ? So that you get I = (5-0,6)/R1 ?At the collector of the transistor i have 12 V so then the transistor is in the active region, so there is a current flowing from Ic to Ie, but i dont really understand what happens with the voltages at this point, and i am still a little confused about why the voltage across RL is the same as Ub. Dec 19, 2021 at 22:10
• @Daan_1 I think you are telling me that Ub is 5 volts, yes? On that basis, the voltage across RL will also be 5 volts due to the negative feedback action of the op-amp forcing the voltage across RL to equal Ub. So, if that 5 volt appears on the emitter relative to 0 volts then, the base voltage will be 5 volts +0.6 volts (just an approximation for the base-emitter volt drop because it acts like a diode). This means the op-amp output will be 5.6 volts. But, if the base-emitter region volt drop were (say) 0.8 volts (not unheard of) then the op-amp output will naturally be 5.8 volts. Dec 20, 2021 at 0:03

Instead of trying to jump to some random formula for gain, think about what you know. What is the voltage at the inverting input of the op amp? What is the current through R2? What is the voltage at node A? Keep going in this manner until you get the result you need.