# NPN Transistor Parameter Determination Gain Question

In the interest of making an oscillator, I made the above circuit after reading about gain, current flow, active region, and how an NPN transistor works.

I then put the circuit in LTSpice and note the input and output voltages.

The input is in blue, the output is in green.

I see that there is gain. For the oscillator, where all I think I need is some amout of positive gain, this should probably work. I then started thinking about microphoens and speakers and how to dial in an exact amount of gain given a mic.

So, my question is this, lets say I have a signal that oscillates += .005 volts and I want it amplified to swing between +9 volts and 0 volts, what equations would get me the appropriate resistor values. I would set the middle of the active region to be Vcc/2 as a Vce reference. This would be 4.5 volts. How do I best determine the resistors to get to this level?

Lets say, I then change the input swing between += .2, how do I reset the resistors to get to the appropriate swings in the output region?

When I say swing, I mean above and below a DC bias voltage on the base.

I feel I should be able to adjust this to exactly to whats needed by choosing the appropriate values of resistors. Thanks, Jeff

• en.wikipedia.org/wiki/Bipolar_transistor_biasing Aug 8, 2017 at 18:04
• Microphones often have special requirements. So you will want a pre-amp stage. You need to specify the microphone type, at least. Also, a speaker similarly has it's own special requirements. So you will want a power output stage for the speaker. In between these, you could consider your simple common emitter voltage stage, I suppose. But you aren't going to get all this work done with a single BJT.
– jonk
Aug 8, 2017 at 18:33
• What do you mean by pre-amp? Something that takes a .05 volt swing and turns it into a .8 volt swing. This being followed by another stage that takes a .8 volt swing and turns it into a 2.5 volt swing. I was wondering if it can be done with one bjt with the appropriate resistors. I was playing with values and suspect that to do this the resistances become very large and the collector current because extremely small Aug 8, 2017 at 19:52
• I think they wrote a song about blue on black. Very hard to see. Oct 28, 2017 at 19:35
• @JeffreyEdwardMessikian Didn't see your comment until today. You didn't use my name in the comment. Sorry about that. The pre-amp is designed for the transducer itself. Transducers have a wide variation of physical behaviors -- their purpose is to convert some physical phenomena into electrical signals (or the reverse, as the case may be.) Special considerations are often required when designing a circuit, which must take into account the physical details of the transducer. The pre-amplifier takes note of these details, appropriately. It's not just a matter of gain.
– jonk
Feb 17, 2020 at 7:36

The maximum voltage gain from a single bipolar is VDD / 0.026.

Thus your 9v example has max gain of 9/0.026 or 9 * 39 ~~ 360.

Yet to convert 0.005 volt into 9vpp you need gain of 9/0.005 = 9*200 = 1,800.

Summary: a single bipolar is not the answer.

After some digging I found that the following link helped me out the most.

http://www.electronics-tutorials.ws/amplifier/emitter-resistance.html

In particular, the article describes the gain of the above circuit to be Rc/R4. The article related collector current to base current to emitter resistor size to gain.

Saying that the gain is Rc/R4 in my above circuit is all I think I needed.

A single transistor with a voltage gain more than about 100 will have horribly high distortion. 50 years ago two transistors made a reasonable mic preamp. Today an opamp is used.