# Request guidance on calculating R values in this amp

Discussion in 'The Projects Forum' started by logans-electronics, Dec 5, 2009.

1. ### logans-electronics Thread Starter Member

Sep 1, 2009
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Hello,

I would like some guidance or advice on R1, R2, and R4 on how the are or should be calculated and the purpose of them.

I have experimented in multisim with different values and this is the best I could get out of it.

Thanks to forum member SGT WOOKIE for the starting point and template!

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2. ### logans-electronics Thread Starter Member

Sep 1, 2009
36
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Just wondering if anyone had a chance to review and offer some advice on how to calculate the resistor vales and the purpose of them...

3. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Embedding your schematics in a Word document makes it more difficult for people to read them; and many don't like to download such files because they can contain viruses.

I've extracted your circuit and attached it in .png format.

I've also attached a version that was posted previously.

I don't have the transistors you used available in any simulators that I have installed.

You have installed R5 backwards. The input signal can be shorted to ground if R5's resistance is turned way down. The signal input should be connected to the end of the pot furthest from ground, and the amplifier input taken from the wiper of the pot.

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4. ### Audioguru New Member

Dec 20, 2007
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The values of R1, R2 and R4 are calculated from the max amount of base current you need for the transistors considering their minimum hFE.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Well, maybe this helps
Vcc=Vr+Ve+Vbe+VL
Vcc - supply voltage
Vr - voltage drop on R1+R2
VL - maximum voltage on the load (speaker)
Ve- voltage on emitter (typical 0.5Vcc)
then
VL=√(2*Pout*RL)
IL=VL/RL

The maximum value of the load current is equal:
IL_max = ( Vcc - Ve - Vbe ) / ( (R1+R2)/(β+1) + RL )
then
$R1+R2=(\frac{Vcc-Ve-Vbe}{IL}-RL)*(\beta_{min}+1)$

For example
Vcc=28V and output power Pout=7W; β_min=50; Vbe=0.7V

VL=√(2*7W*6Ω)=9.2V

IL=9.2V/6Ω=1.54A

R1+R2=( ( 28V - 14V - 0.7V / 1.54A ) - 6Ω) * 51= 130Ω

In the calculation I ignore the bootstrap circuit.

Last edited: Dec 7, 2009
6. ### logans-electronics Thread Starter Member

Sep 1, 2009
36
0
This is making sense now....

Did you mean that the bootstrapping is part of R4?.....I came up with this high value experimenting in Multisim and received the best results.

7. ### logans-electronics Thread Starter Member

Sep 1, 2009
36
0
R1+R2=( ( 28V - 14V - 0.7V / 1.54A ) - 6Ω) * 51= 130Ω

Is the 51 a constant or a solved value from the circuit somewhere?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Bootstrap circuit is C3 and R2, R1

51 is not a constant its a BJT minimum current gain β=Hfe=Ic/Ib