Please need help to calculate gain from circuit.

[FriedMule]

Can anyone help calculating this amplifier circuit's gain?
I do not know if this task are a insane large one or even manageable. For me it is totally impossible.
You have properly seen this circuit before, that Is what I constantly are bothering you with! :-)

[DuPe]

roughly ×5?
(100k//100k) / (10k+2k).
anything right of level shifter are emitter followers, having gain of about 1

[FriedMule]

not more then 5?! wow that was not much!!

Does that mean that a i.e. CD-player with a line out at 1Vpp only end in 5Vpp to the speaker?!

[orbanp]

roughly ×5?
(100k//100k) / (10k+2k).
anything right of level shifter are emitter followers, having gain of about 1

It is more than that!

The 10k resistor is bypassed by the 100uF // 100nF caps!
So the gain is (100k//100k) / 2k = 25

Peter

[FriedMule]

Thanks a lot, so it is about 25Vpp out pr channel?
Does that not sound as about 36W out?

EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why?

[Zero999]

Yes, 25 sounds about right.

Thanks a lot, so it is about 25Vpp out pr channel?
Does that not sound as about 36W out?

EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why?

What's the load impedance?

For a sine wave, the RMS power is equal to half the peak power, so with a 4Ohm load: P = V²/(2R) = 25²/(2*4) = 625/8 = 78W.

A 26V transformer will output 36V, after losses in the rectifier, so 25V peak out seems reasonable.

I doubt it'll be able to output full power continuously, as the smoothing capacitors aren't big enough, but the output transistors will probably overheat too, at that power level. It's also not a problem, since peak power occurs typically less than 10% of the time, even at full volume.

[b_force]

Yes, 25 sounds about right.

Thanks a lot, so it is about 25Vpp out pr channel?
Does that not sound as about 36W out?

EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why?

What's the load impedance?

For a sine wave, the RMS power is equal to half the peak power, so with a 4Ohm load: P = V²/(2R) = 25²/(2*4) = 625/4 = 78W.

A 26V transformer will output 36V, after losses in the rectifier, so 25V peak out seems reasonable.

I doubt it'll be able to output full power continuously, as the smoothing capacitors aren't big enough, but the output transistors will probably overheat too, at that power level. It's also not a problem, since peak power occurs typically less than 10% of the time, even at full volume.

Ehm, well better is the use to divide the rail voltage (which is the peak voltage of the sine wave) by the square root of two (= 1.41).
Which is obviously the same as the Rms value of the AC voltage (minus some losses), so in this case 26Vrms.
(26^2)/8 = 85W @ 8ohm or 170W @ 4ohm (the rail voltage is 26 * 1.41 = 36.7V)

However, with Class-AB amplifiers it's not uncommon to give a lot more headroom in the rail voltage.
The efficiency of a AB stage is about 60-70% at most, so that means that transistors will dissipate around 50W @ 4ohm.
That's quite a lot, so that makes me think, including the low amount of capacitive storage, this amp is suitable for 8ohm or higher.

[DuPe]

you are completely right. I missed the u and did read 2x100n instead.
quite likly I need new glasses

[FriedMule]

You are all so great, thanks!

Wikipedia tells me that 25Vpp = Nominal level, VRMS 0.316 x 25 = 7.9VRMS.
That does not sound as a lot: https://en.wikipedia.org/wiki/Line_level

That leads me to think that 36.7V is a lot more then needed with headroom and so on.


Is it possible to make it happy with 4Ohm speakers also?
How many volts shall the transformator have?

[David Hess]

The gain is 25 at high frequencies and 4 at low frequencies with a high pass cutoff of about 160Hz as shown but one of the capacitors bypassing the 10k resistor is not shown.  My guess is that it is suppose to be 1uF making the high pass cutoff 16Hz.

Noise for this design is high because of emitter degeneration of the differential input stage which was necessary for high full power bandwidth; this is common with discrete power amplifiers but it could be avoided if an operational amplifier was used in place of the input differential pair.  Distortion performance is poor because feedback does not come from the output.

[Audioguru]

I think it will sound awful with a lot of odd harmonics distortion since the output emitter-followers are not inside the negative feedback loop. Some crossover distortion also might be noticeable.
Why not make a "normal high fidelity amplifier circuit" instead, or use an LM3886 IC?

[Zero999]

Yes, 25 sounds about right.

Thanks a lot, so it is about 25Vpp out pr channel?
Does that not sound as about 36W out?

EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why?

What's the load impedance?

For a sine wave, the RMS power is equal to half the peak power, so with a 4Ohm load: P = V²/(2R) = 25²/(2*4) = 625/4 = 78W.

A 26V transformer will output 36V, after losses in the rectifier, so 25V peak out seems reasonable.

I doubt it'll be able to output full power continuously, as the smoothing capacitors aren't big enough, but the output transistors will probably overheat too, at that power level. It's also not a problem, since peak power occurs typically less than 10% of the time, even at full volume.

Ehm, well better is the use to divide the rail voltage (which is the peak voltage of the sine wave) by the square root of two (= 1.41).
Which is obviously the same as the Rms value of the AC voltage (minus some losses), so in this case 26Vrms.
(26^2)/8 = 85W @ 8ohm or 170W @ 4ohm (the rail voltage is 26 * 1.41 = 36.7V)

However, with Class-AB amplifiers it's not uncommon to give a lot more headroom in the rail voltage.
The efficiency of a AB stage is about 60-70% at most, so that means that transistors will dissipate around 50W @ 4ohm.
That's quite a lot, so that makes me think, including the low amount of capacitive storage, this amp is suitable for 8ohm or higher.

The original poster said the output voltage is 25V peak, so the output power will be 78W into a 4Ω load or half that into an 8Ω load, irrespective of the supply voltage.

[b_force]

Yes, 25 sounds about right.

Thanks a lot, so it is about 25Vpp out pr channel?
Does that not sound as about 36W out?

EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why?

What's the load impedance?

For a sine wave, the RMS power is equal to half the peak power, so with a 4Ohm load: P = V²/(2R) = 25²/(2*4) = 625/4 = 78W.

A 26V transformer will output 36V, after losses in the rectifier, so 25V peak out seems reasonable.

I doubt it'll be able to output full power continuously, as the smoothing capacitors aren't big enough, but the output transistors will probably overheat too, at that power level. It's also not a problem, since peak power occurs typically less than 10% of the time, even at full volume.

Ehm, well better is the use to divide the rail voltage (which is the peak voltage of the sine wave) by the square root of two (= 1.41).
Which is obviously the same as the Rms value of the AC voltage (minus some losses), so in this case 26Vrms.
(26^2)/8 = 85W @ 8ohm or 170W @ 4ohm (the rail voltage is 26 * 1.41 = 36.7V)

However, with Class-AB amplifiers it's not uncommon to give a lot more headroom in the rail voltage.
The efficiency of a AB stage is about 60-70% at most, so that means that transistors will dissipate around 50W @ 4ohm.
That's quite a lot, so that makes me think, including the low amount of capacitive storage, this amp is suitable for 8ohm or higher.

The original poster said the output voltage is 25V peak, so the output power will be 78W into a 4Ω load or half that into an 8Ω load, irrespective of the supply voltage.

Well, he said a 26V transformer.
But anyway, it doesn't matter.

Nothing is wrong here, I only wanted to point out that it helps to calculate the voltage, instead of dividing the power by two.
Both methods obviously work, but if someone is going to bridge an amplifier for example, it shows better that de power goes with a factor of two.