# Basic transistor Guitar Amp

#### calx

May 15, 2012
10
Cheers, that looks achievable, and I think I just about understand how it works

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Bob, with all due respect, you may want to recalculate.

#### BobK

Jan 5, 2010
7,682
Why don't you show me where I am wrong?

Bob

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Hello Bob.
My calculations indicate that when RL is dissipating 1W then RS is dissipating 2.18W.

Chris.

#### BobK

Jan 5, 2010
7,682
Except that 9V is the wrong voltage to use. The junction of the two emitters in my circuit sits at 4.5V and swings up and down from there, so the Vce is never as high as the calucution you are using. And on top of that, each transsitor is hard off for half the cycle and it is the average power that matters when calculating dissapation. And, for a sine wave (which is what my calculation is based on), the power is not constant over time.

The true answer depends on the waveform. For instance, what is the power dissapation in the transistors for a push-pull circuit with 9V supply and an 4V p-p square wave?

Edited to add: Which is 2 Watts into an 8 Ohm speaker.

Bob

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#### calx

May 15, 2012
10
Right I have been trying some stuff out just on the basic circuit whist I am waiting for some more bits to arrive. I have a question now to do with the biasing potential divider circuit. I was trying to bias it with roughly 3V (that's about enough right?) and I used a 22k and a 47k to start with (will alter them for impedence later) wich gave me just less than 3V when I calculated it from a 9V supply, and worked out about right in practice. However, When i added the transistor in, the drop across the lower went to the 0.7Veb and the upper resistor went to the remaining 8V ish. Now I know the transistor set the 0.7V itself, but why does this then alter the divider voltages? I know this is a simple question, but I can't work this one out -_-
Any suggestions would be great,
Cheers, Callum

#### BobK

Jan 5, 2010
7,682
The base is always about 0.7V above the emitter. If the emitter is tied to ground, this is difficult to control. Using an emitter resistor allows you to calculate the bias point much more easily, provides negative feedback to keep it in range, and prevents bad things from happening if the transistor heats up.

For maximum output, you want to bias it so the collector is at 1/2 the supply voltage, 4.5V in the case of a 9V battery.

Here is how you do it.

1. Choose a collector resistor.
2. The emitter resistor will be 1/G times the collector resitor, where G is the desired voltage gain.
3. Determine what current will drop 1/2 the supply voltage across this resistor.
4. Determine what voltage the emitter will be at, based on the approximation that the emitter current is the same as the collector current.
5. Add 0.7 V to the result of step 3.
6. This gives you the voltage you need at the middle of the biasing divider.
7. Choose resistors to give a little more than that voltage, and also so that their parallel combination of gives you a current which is a little bigger than the collector current divided by the beta (current gain) of the transistor.

Then, when it doesn't come out exactly right, adjust the resistors until it does

Bob

Edited to add step to compute the emitter resistor.

#### calx

May 15, 2012
10
thanks Bob. I was looking at this from the wrong angle - trying to sort out divider network before the rest which, as I now see, wasn't really going to work
Callum

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
The power dissapation in the output transistors is not the same as the output power. <snip>
Bob

Bob, my original comment was in response to this statement and as for the Vcc not being 9V.. what's this?

https://www.electronicspoint.com/at...3150-basic-transistor-guitar-amp-audioamp.jpg

My concept of what constitutes a guitar amplifier is obviously far different from yours. I don't see 1W as being in that category. I would fully expect output stages using 2N3055's or similar and a Vcc => 24V. We're not going to have a meeting of the minds, so I'll just leave it at that.

Chris

#### BobK

Jan 5, 2010
7,682
Here is what I said way back. We are not in disagreement.

The output stage will depend on how much power you want to produce. If you are looking at <1W something like 2n2222 / 2n2907 pair would work fine. For more power you will be looking for power darlington compilmetary pairs.

And the circuit I posted is just one I use to drive a little speaker on my bench. It was not proposed as a serious guitar amp, I said it was "a guitar amp only your neighbors could love". I posted it as an exmaple of a push-pull output stage. It maxes out at about 1/4 W.

Bob

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