# Why is a transistor amplifier gain less than expected?

I'm trying to build a voltage amplifier using a transistor but am getting a lower gain than I expect.

The circuit I have is here:

simulate this circuit – Schematic created using CircuitLab

Resistor values are:

• R1 = 330
• R2 = 570
• Re = 1000
• Rc = 100

Electret datasheet is here: https://resources.kitronik.co.uk/pdf/3310-pcb-mount-microphone-insert-datsheet.pdf

My power supply, Vcc, is 4.46V. All other values are measured at:

• Vb = 2.75
• Ve = 2.11
• Vc = 4.26

I expect the amplification to have an upper limit of Vcc = 4.46V, and a lower limit of the transistor diode drop, (Vb-Ve) = 0.56V.

I built the transistor to have a voltage gain of -10. I thought the gain equation was G=R2/R1, but didn't get the results I expected (a unity gain), interchanged the resistors and got something more reasonable. The resistor values above are for my final configuration.

My question is, why has the collector voltage not been pushed up to Vcc ?

I was expecting a gain of 10 would attempt to give a collector voltage of 10*Vb, so 27.5V, and failing this would hit the upper rail of Vcc, so 4.46V.

An explanation of where my misunderstanding is, along with an explanation as to why, would be appreciated.

Tips on whether I had the correct gain equation would also be helpful.

• Which transistor are you using, is it really a PNP? Sep 2, 2023 at 19:15
• Thanks for noting - I have updated the post. Sep 2, 2023 at 19:19
• Is your signal source really an electret microphone capsule? If so, the circuit for making it work is completely wrong so the amplifier is irrelevant. Sep 2, 2023 at 19:21
• @user10709800 Please use the built-in schematic editor to draw a correct schematic rather than just pointing out in the question that it’s wrong. You can access this from the icon at the top of the editor. Sep 2, 2023 at 19:36
• The emitter voltage tracks the input voltage. The resulting current can be seen by dividing through the emitter resistor value. The output voltage is then the voltage drop caused by this current across the collector resistor. You got Re and Rc mixed up if you want voltage amplification. Sep 2, 2023 at 19:47

Because your microphone has its own internal JFET pre-amplifier, you should connected the positive terminal to a constant voltage supply (perhaps the same as used in your BJT amplifier, but not necessarily), through a resistor. You should also connect the positive terminal of your microphone to the BJT amplifier through a capacitor. The following schematic illustrates these connections and components.

simulate this circuit – Schematic created using CircuitLab

The purpose of these components, and this connection method is to allow the microphone to have a different DC bias than the BJT. Because capacitors "block" DC, the DC bias on each side of the capacitor can be different.

As GodJihyo points out in their answer, your values for Rc and Re are reversed. In a common emitter amplifier with a non-trivial emitter resistor, the voltage gain is approximately Rc/Re. So, you want Rc to be 10 times bigger than Re to get a gain near 10.

As periblebsis points out, the pre-amp in the microphone capsule cannot output a great deal of current. The values of R1 and R2 in your question are very low, and will make the microphone seem "weak". That is, conversion of physical sound waves to electrical signals will not be very "efficient". Try increasing their values so that R2 is at least 15 k$$\\Omega\$$. (R1 will have to be increased accordingly).

• Yeah, but the loading of R1 and R2 on the jfet of the electret requires some care. I would much prefer to at least see a bootstrapped design, at minimum. Better yet would be a multi-BJT (use low-noise BJTs) class-A amplifier with global NFB where the base current that the electret is exposed to is in the region of a few tens of nanoamps. This is readily achieved. (See figure 4 here.) Sep 2, 2023 at 20:48
• @periblepsis Yes, the BJT amps input impedance affects the JFET pre-amp. But bootstrapping, which involves positive feedback requires care as well. The easiest, though not the least noisy, is to use large values for the base biasing network. It is unlikely the OP is designing for HiFi. The OP needs to learn to walk before running. Sep 2, 2023 at 21:15
• Albert Einstein's Principle of Simplicity says, "Everything should be made as simple as possible, but not simpler." Too much simplification doesn't characterize, but instead becomes a caricature. Sep 2, 2023 at 21:18
• @periblepsis You are skipping a few points in the learning curve... The above (apart from missing the values) is what is needed to make the original design sound. Your proposal is something else entirely. Quite nice, but different operating parameters and circuit. Sep 2, 2023 at 21:21
• @periblepsis given that there is no stated spec for noise, the simplest way to increase input impedance of OP's BJT amplifier is to use larger R1 and R2. Sep 2, 2023 at 21:21

You have the gain equation mixed up. It is not R2 / R1, it is RC / RE. With that, your resistors are swapped, you should have RC = 1k and RE = 100 $$\\Omega\$$ for a gain of -10. That and what others pointed out about the electret mic power should get you closer to a working circuit.

• The circuit might work, but a gain of 10x for electrect might not be sufficient. Maybe a gain of 100x could be more closer to what is required. Sep 2, 2023 at 19:48
• @Justme I think you are underselling the output level of a properly biased electret microphone capsule. Sep 3, 2023 at 0:55

Electret mic preamps today use an audio opamp for as much signal gain that you want with very low distortion. To get a signal gain of 100 or more from a single transistor then the distortion will be severe.

Here is a transistor with a signal gain of about 122 times and with a signal gain of 6.4 times: