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##### Amplifier Discussion / Re: Peak dissipation in class B output stage

« Last post by**shasam**on

*»*

**Today**at 08:20:37 AMI am going to use a 2*15V transformer, and NJW0281G (NPN) / NJW0302G (PNP) output transistors, in CFP configuration, for 8Ω output.

Datasheet can be found here : https://docs.rs-online.com/41ca/0900766b813e9eb0.pdf

For now, my calculations are :

Total DC power supply = 2 * (Veff *√2 - 2Vdiode) = 2 * (15 * √2 – 2) = 38,5V

Output power = (power supply - loss)² / (8*Rload) = (38,5 – 5)² / (8*8 ) = 17W (I found similar result in LTSpice)

Output peak current = (power supply - loss) / (2*Rload) = (38,5 – 5) / (2*8 ) = 2,1A

Output peak power = (power supply - loss) * Output peak current = (38,5 – 5) * 2,1 = 70,35W

With Rod Eliott rules-of-thumb previously quoted :

That give me : I = (Vrail / 2) / R = (19.25 / 2) / 8 = 1.203125A

Ppeak_resistive = Vrail / 2 * I = (19.25 / 2) * 1.203125 = 11.58...W(peak)

Ppeak_reactive = Ppeak_resistive * 2 = 23.16...W(peak)

Max dissipation for NJW0281G / NJW0302G is 150W at 25°, 0W at 150°

Derating = 150W / (150° - 25°) = 1.2W/°C

Maximum_case_temperature = (Pmax_at_25° - Ppeak_reactive) / derating + 25°

= (150 - 23.17) / 1.2 +25 = 130°C

I have use the formula given here for average worse dissipation : https://www.updatemydynaco.com/documents/Class_B_Amplifier_Dissipation_Calculations.pdf

Pdiss_worst_case = Vps² / (19.75 * RL) = 38,5² / (19,75 * 8 ) = 9,38...W

I don't really understand the calculation here. I have use it because it give me the worse result of all the calculation I have read online. What do you think about it?

Heatsink_max_thermal_resistance = (Maximum_case_temperature - Ambiant_temperature) / Pdiss_worst_case = (130 - 50) / 9.5 = 8.42... °C/W

I don't really know what to use for ambiant temperature, I have use 50°C here. I will use bigger heatsink to limit the max temperature lower than 130°C too.

I tried to make it as clear as possible. What do you think about it, please? Could you spot some mistakes? Should I take bigger margin?

Thanks a lot!

Datasheet can be found here : https://docs.rs-online.com/41ca/0900766b813e9eb0.pdf

For now, my calculations are :

Total DC power supply = 2 * (Veff *√2 - 2Vdiode) = 2 * (15 * √2 – 2) = 38,5V

Output power = (power supply - loss)² / (8*Rload) = (38,5 – 5)² / (8*8 ) = 17W (I found similar result in LTSpice)

Output peak current = (power supply - loss) / (2*Rload) = (38,5 – 5) / (2*8 ) = 2,1A

Output peak power = (power supply - loss) * Output peak current = (38,5 – 5) * 2,1 = 70,35W

With Rod Eliott rules-of-thumb previously quoted :

Quote

"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 ...(from https://sound-au.com/soa.htm)

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"

That give me : I = (Vrail / 2) / R = (19.25 / 2) / 8 = 1.203125A

Ppeak_resistive = Vrail / 2 * I = (19.25 / 2) * 1.203125 = 11.58...W(peak)

Ppeak_reactive = Ppeak_resistive * 2 = 23.16...W(peak)

Max dissipation for NJW0281G / NJW0302G is 150W at 25°, 0W at 150°

Derating = 150W / (150° - 25°) = 1.2W/°C

Maximum_case_temperature = (Pmax_at_25° - Ppeak_reactive) / derating + 25°

= (150 - 23.17) / 1.2 +25 = 130°C

I have use the formula given here for average worse dissipation : https://www.updatemydynaco.com/documents/Class_B_Amplifier_Dissipation_Calculations.pdf

Pdiss_worst_case = Vps² / (19.75 * RL) = 38,5² / (19,75 * 8 ) = 9,38...W

I don't really understand the calculation here. I have use it because it give me the worse result of all the calculation I have read online. What do you think about it?

Heatsink_max_thermal_resistance = (Maximum_case_temperature - Ambiant_temperature) / Pdiss_worst_case = (130 - 50) / 9.5 = 8.42... °C/W

I don't really know what to use for ambiant temperature, I have use 50°C here. I will use bigger heatsink to limit the max temperature lower than 130°C too.

I tried to make it as clear as possible. What do you think about it, please? Could you spot some mistakes? Should I take bigger margin?

Thanks a lot!