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**Amplifier Discussion / Peak dissipation in class B output stage**

« **on:**September 04, 2020, 06:27:15 AM »

Hi!

I'm trying to understand how to design a class B power amp, and I am a little confused with the method to define the peak of power dissipated by the output transistors.

All the mathematical methods I have seen are based on sinewaves.

But it look to me that with an assymetrical heavilly clipped signal, like you could have from a fuzz, things could be really worse. With a big phase shift, you could have all the voltage (Vcc - Vee) accross a transistor in the same it would pass full curent. So Pd(peak) could be I(peak) * (Vcc - Vee). Is this really pessimistic?

Here, https://sound-au.com/soa.htm, the result look the same, but the method look strange to me :

"Having discounted the idea of any 'rules-of-thumb', I'm going to give you one anyway . Let's assume that you want to deliver 100W into 8 ohms, so you need a power supply with ±42V rails (I'm going to ignore losses here). The amp must also be able to drive nominal 4 ohm loads, so expect the minimum impedance to be 3 ohms. Worst case (resistive load) dissipation is therefore ...

I = V / 2 / R = 21 / 3 = 7 Amps

P = V / 2 * I = 21 * 8 = 168 Watts (peak)

This accounts for the resistive part of the load, and as we saw above, the reactive part of the load causes dissipation to double. Just like second breakdown, we aren't interested in the average dissipation - this influences the size of heatsink needed, but not the transistor's safe area. Therefore, Ppeak will be ...

Ppeak = P * 2 = 168 * 2 = 336 Watts"

How could the transistor pass full current with the output being at Vcc/2 with a resistive load?

What would you assumed to be the whorse impedance and phase shift from a 8ohms guitar speaker please?

I am right assuming at the lower impedance, the load would look only resistive, so there won't be phase shift, so the the worse cases couldn't come hand in hand? I have not found anything about this yet.

Thanks a lot for your help!

Please apologise my poor English.

I'm trying to understand how to design a class B power amp, and I am a little confused with the method to define the peak of power dissipated by the output transistors.

All the mathematical methods I have seen are based on sinewaves.

But it look to me that with an assymetrical heavilly clipped signal, like you could have from a fuzz, things could be really worse. With a big phase shift, you could have all the voltage (Vcc - Vee) accross a transistor in the same it would pass full curent. So Pd(peak) could be I(peak) * (Vcc - Vee). Is this really pessimistic?

Here, https://sound-au.com/soa.htm, the result look the same, but the method look strange to me :

"Having discounted the idea of any 'rules-of-thumb', I'm going to give you one anyway . Let's assume that you want to deliver 100W into 8 ohms, so you need a power supply with ±42V rails (I'm going to ignore losses here). The amp must also be able to drive nominal 4 ohm loads, so expect the minimum impedance to be 3 ohms. Worst case (resistive load) dissipation is therefore ...

I = V / 2 / R = 21 / 3 = 7 Amps

P = V / 2 * I = 21 * 8 = 168 Watts (peak)

This accounts for the resistive part of the load, and as we saw above, the reactive part of the load causes dissipation to double. Just like second breakdown, we aren't interested in the average dissipation - this influences the size of heatsink needed, but not the transistor's safe area. Therefore, Ppeak will be ...

Ppeak = P * 2 = 168 * 2 = 336 Watts"

How could the transistor pass full current with the output being at Vcc/2 with a resistive load?

What would you assumed to be the whorse impedance and phase shift from a 8ohms guitar speaker please?

I am right assuming at the lower impedance, the load would look only resistive, so there won't be phase shift, so the the worse cases couldn't come hand in hand? I have not found anything about this yet.

Thanks a lot for your help!

Please apologise my poor English.