# How to calculate the gain of a BJT Differential Amplifier (Long Tailed Pair)

I'm currently reading Designing Audio Power Amplifiers by Bob Cordell. After seeing the "Basic 50W Amplifier" that he shows in the beginning of chapter 2, I now am trying to use the same building blocks to design an amp to specifications I've chosen to help me understand things.

I've chosen an output power of 5W through an 8ohm load (Which should at least be audible I hope), this means I am going to need a peak output voltage of about 9V. I'm assuming my input signal is coming from something like a phone, and I'm making an educated guess that its peak voltage is 1v. This means I'm going to need a total gain of about 9.

The building blocks of his circuit are as follows:

Stage 1 of 3 (Input Amplification Stage): Constant Current Driven Differential Long Tailed Pair

• Rejects noise well, and negative feedback is easy to achieve from the final output of the amplifier with a simple voltage divider

Stage 2 of 3 (Voltage Amplification Stage): Constant Current Driven Common Emitter Amplifier

• Easy to get a really high gain from what I understand

Stage 3 of 3 (Output Amplification Stage): Triple Emitter Follower

• High input impedance, with a decent amount of output capability.

This is basically an op-amp I think, but with higher power capabilities. What I'm really struggling with is the input stage.

What I've been taught in my Electronics 1 class so far about BJTs is that for these amplifier configurations, the gain usually is a ratio of the resistors used (and maybe includes the base-emitter resistance). I have no idea where the gain is coming from in this schematic. I basically just fiddled with the resistor values until I got an output that looked alright-ish.

Is a source out there that goes through the design process of this type of configuration, or can anyone can help me out? Thanks!

• But in the Bob's book, you can find all the equations you need to use. And in this case, the gain is $A_V = \frac{R_C}{2(re + RE)}$ – G36 Mar 10 at 6:58
• R2 is way too big (the LED current is too small). Also,electronics.stackexchange.com/questions/347501/… try read this electronics.stackexchange.com/questions/343833/… – G36 Mar 10 at 7:02
• So your circuit the gain cannot be larger than: $A_V = \frac{5k\Omega}{2 * 300\Omega} = 8.3 V/V$ – G36 Mar 10 at 7:06
• Thank you for such a fast reply! I guess I should read more closely. I just went back to the diff-amp part of chapter 1 and that equation is definitely there, I need more sleep! I do have one more question though, how do you choose how much current to source/sink through the LTP? Regarding the constant current source I have up top, are you sure that resistance is too high? I'm getting right around my target current value in the simulation. The schematic doesn't say it but its a green LED with a 2.1V drop, so 1.4V across R1 gives about 3.5mA. – Michael Beckwith Mar 10 at 7:21
• Nevermind you are right about the resistor, I'm only getting 2.9mA out, which is actually kind of far off from 3.5. Thank you for that advice. – Michael Beckwith Mar 10 at 7:33