Audio amplifier design, basic questions.


Joined Nov 9, 2007
The TIP31/32 have metal backplates that are electrically connected to the collector.

The heatsink would normally be earthed to prevent hum and other pickup, and for safety.

So they need to be mounted in an isolated manner from the heatsink.

They are designed for a single screw fixing and this needs an isolating bush/washer to pass through when the assembly is screwed to the heatsink.
Also the metal pad on the transistor needs an isolating shim/washer to prevent contact between the transistor and the heatsink.

See what you can dig up, on the net then come back and ask questions. I think there should be a section in the Ebook about this.

Thread Starter


Joined Dec 18, 2014
ok, thank you. will have a look as soon as i have found some suitable matched pair transistors for all the others that farnell have in stock.
proving surprisingly difficult.

Thread Starter


Joined Dec 18, 2014

Thread Starter


Joined Dec 18, 2014
sorry guys, don't think i explained clearly.
was going to use tip 31/32 A as the output pair as they are in stock.
these are what i have used for calculations so far.
i was looking for transistors suitable for the other transistors in my current design, its pretty much the one in post 88. ie- q1, q2, q3, q4, q6 and the rubber zener.

can a t092 be used for the rubber zener if it is to be glued to one of the outputs? i'm assuming its casing would be relatively heatproofed

Thread Starter


Joined Dec 18, 2014
i have realised the complete error i made when discussing r7 earlier....
"calcs so far.

10 watts into 4 Ohms. 10W x 4 Ohms = 40 = V^2 sqrt 40 = 6.324Vrms 6.324 x sqrt 2 = 8.94Vpk 8.94/4ohms = 2.235A Ipk

Vcc required = 8.94 x 2 = 17.88V so lets say 20 giving 1.06V overhead for output stage.

aiming for minimal current waste at output 2.235/100 (paired output) = 0.02235A/22.35mA

20V x 0.02235 = 0.447W power consumption by output.

using 0.5 watt output resistors rms @ each output = 2.235 /sqrt 2 =1.58A
P = I^2 x R so p/I^2 = R = 0.5/1.58^2 = 0.20 Ohms

I checked farnell and i can get 0.2Ohm 0.5W power resistors, hurrah!

I settled upon a TIP 31A for the output transistor

using this i think i can get away with 80 mA Ic

so 80mA x 0.20 Ohms = 0.016V@r9

The Vbe required is about 0.62V giving me 0.016V + 2 x 0.62 = 1.256V @ rubber zener."

this is for q6 and q7? with q3 supplying the voltage for q5 and q6? so i need to select a smaller transistor for q6 that will run of Vbe 0.62V, then calculate r7 to give 0.62V at the base of q7 using a Ve of my original 8.94V-0.016V =8.904V? or is this just miles off?"

R7 is to source current for the collector at q6, since the base of q6 can only sink it. is this the mythical pull up resistor?

still don't think i've landed on how its ti calculate R7, keep getting values that seem way too low...



Joined Nov 30, 2010
R7 and R8 merely keep Q7 and Q5 turned off while not in use (half the time).

Q4 and Q6 will get more leaky from collector to emitter when hot, usually in the neighborhood of less than 1 ma, so 870 ohms seems good, considering I didn't look anything up or do any math.

R = .55Vbe /0.001A
550 ohms.
Yep. Seems good to me.
Just allow an ma or 2 in case Q4 and Q6 get good and hot. That's all there is to it.
I mean, it's not like you're wasting a significant part of 2.7 amps.

Thread Starter


Joined Dec 18, 2014
is the value for R5 considered in parallel with my input impedance for q6?
it has to hold most of my voltage (-volltage for output pair?) at 0.023A but its value seems low if just this is considered.


Joined Feb 17, 2009
I was very busy at work, so I do not have time much time. But ok. Now let me try to show you some example.
The circuit diagram look like
Now let as assume Pout = 10V for RL = 4Ω.
This means that UL_max = √(2*Pout*RL) = √(20W * 4Ω) ≈ 8.944V ≈ 9V
IL_max = UL_max/RL = 2.25A

For good thermal stability I choose
R10 = R9 = Vbe/ILmax ≈ 0.8V/2.25A ≈ 0.33Ω /1W
The output stage quiescent current I set via a pot connected across VBE multiplier Q8.
And this current we set on the lab bench Iq = (2...5)%*IL_Max ≈ 50mA.
R7 = R8 = 100Ω
a typical value.
Now if we assume Hfe_min = 20 for Q5 and Q7 and Hfe = 100 for Q4 and Q6 we can select ICQ3 current.
ICQ3 ≥ (5.... 20) * Ilmax/(20 * 100) ≈ 20mA
R5a+R5b ≈ 0.5Vcc/ICQ3 = 10V/20mA = 500Ω
R5a = R5b = 500Ω/2 = 220Ω

ICQ1 = 10 * ICQ3/Hfe_min = 10 * 20mA/100 = 2mA

R3 = Vbe/ICQ1 = 0.7V/2mA ≈ 330Ω

R4 = (0.5Vcc - Vbe)/(2 * ICQ1) = (10V - 0.7V)/4mA ≈ 2.2kΩ

R1 = R2 = 0.5Vcc/(10 * IcQ1/Hfe_min) ≈ 10V/0.2mA ≈ 47kΩ

Also notice that R1||R2 determine the amplifier input resistance. Rin = R1||R2 = 47kΩ/2 = 23kΩ
And to reduce the "DC offset" RF1 should be equal to Rin
RF1 = 22kΩ
RF2 = RF1/voltage gain = 22k/20 ≈ 1KΩ
C2 = 0.16/(RF2 * Fc) = 0.16/(1kΩ * 10Hz) ≈ 22μF
C1 - 0.16/(Rin * Fc) = 0.16/(23kΩ * 10Hz) ≈ 1μF
Cout = 0.16/(RL * Fc) = 0.16/(4Ω*10Hz) ≈ 4700μF

will limit the amplifier bandwidth from high frequency side.

Cf = 0.16/(RF1 * F) = 0.16/(22kΩ * 30kHz) = 220pF
Cm = 10pF
but we select the correct size of this capacitor by testing it on the lab.

The average power dissipation in output stage transistor (in Q5 and Q7) is equal to
Ptot ≈ 0.1 * (0.5Vcc)^2/RL = 0.1 * 10V^2/4Ω = 0.1 * 100/4 = 2.5W plus a DC current looses (Iq * 0.5Vcc = 50mA * 10V= 0.5W)
And the peak power dissipation P = 10V^2/(4 *4Ω) = 6.2W
The power in Q4 and Q6 is Ptot/Hfe . Power dissipation in Q3 is 20mA * 10V = 0.2W.
And Q8 (Vbe multiplier transistor) should be thermally connected with Q4.

And again you are way too optimistic about your headroom voltage.
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Thread Starter


Joined Dec 18, 2014
thanks very much, that helps alot, ill have a play.
to increase headroom voltage i need to raise supply voltage whilst aiming for the same power at the output. Adjusting the resistor values accordingly?


Joined Nov 30, 2010
Not much. Increase your power supply by 3 or 4 volts and half your resistors will not have to be changed to accommodate that.

Thread Starter


Joined Dec 18, 2014
ah, didn't see the reply, thanks. so that would work, just design with a 24V supply and aim for the same 10W over 4 ohms assuming 10V peak out? seems too simple ^^


Joined Nov 30, 2010
NO! Not the feedback loop.
Minor changes might happen in the idle current settings. The feedback loop is about gain, is supposed to be centered, and will not change.