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I’m designing a Class AB amplifier that needs to deliver up to 10 A RMS at frequencies as high as 100 kHz. The output stage uses Darlington transistors (TIP142 for NPN and TIP147 for PNP), powered from ±15 V rails, with an op amp driving the transistor bases. The load includes a 2 µH inductor and a 1 Ω resistor.

My main concern is determining how to calculate (or at least estimate) the power dissipation in each Darlington. So far, I’ve tried to estimate the power dissipation by calculating the instantaneous VCE (or VEC) for each transistor, multiplying it by the current flowing through each transistor. I’m assuming each Darlington only conducts on its respective half-cycle (positive or negative, not both), and I’ve been treating voltage and current as roughly in phase (since the load’s inductance is small).

My questions:

  • Does this approach (multiplying instantaneous voltage by instantaneous current and integrating/summing over the conduction period) make sense for a Class AB stage at these current/frequency levels?
  • What are the best practices (analytical or SPICE-based) for accurately gauging how much heat each transistor will have to dissipate over a cycle?
  • Are there particular considerations for Darlingtons (e.g., higher saturation voltage, extra diode drop, etc.) that might significantly affect the power dissipation calculation?

Any guidance, references, or rules of thumb are greatly appreciated. Thank you in advance!

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  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$ Commented Mar 15 at 0:30
  • \$\begingroup\$ Sebastian, You are talking about dissipating 100 W in the driven load. Your circuit is moving into class-C, from class-B. With a true class-B the best you can hope for will be something around 100 W dissipation in your driver circuit. With class-AB it would be higher. With class-C but near class-B, perhaps a little less. But power dissipation seriously complicates a design. The design you show is already likely wrong. But without more details about exactly what's being driven, I can't be sure. You are holding cards back. So I can only guess. \$\endgroup\$ Commented Mar 15 at 8:21

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I’m designing a Class AB amplifier that needs to deliver up to 10 A RMS at frequencies as high as 100 kHz. The output stage uses Darlington transistors (TIP142 for NPN and TIP147 for PNP), powered from ±15 V rails, with an op amp driving the transistor bases. The load includes a 2 µH inductor and a 1 Ω resistor.

The datasheets for the TIP142 & TIP147 darlington devices rarely specify frequency performance. The only OEM datasheet that I have seen that shows a plot of gain vs frequency is from Onsemi, and it suggests that at 1MHz gain has fallen to less than 50 (refer image below).
enter image description here

That might seem OK, until you realise that DC (0Hz) gain is over 4,000 for the NPN, and over 5,000 for the PNP device (@25C, refer image below). This means that gain falls-off very rapidly as frequency increases. You may be better off choosing different devices for the output stage.

enter image description here

The other problem with the circuit as it now stands is that the output stage is not correctly biased for class AB, it is biased as purely class B (one could even argue it is class C), which will further reduce the ability of this circuit to provide a good output at a frequency of 100kHz. You need to add a bias network, something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The string of diodes D1 to D4 provide a voltage that keeps both output darlingtons partially turned on when output voltage is 0V. There are better ways to do this, for example a "VBE-multiplier" that thermally tracks the output transistors to keep bias currents constant as temperature changes.

Note that the feedback network shown the schematic (R4, R5, R6, C1, C2) is a suggestion only, it has not been designed to suit your application. Also note that all other components are nominal only, not to be taken as final selections - for example, the output devices are not to be construed as shown, the schematic editor did not have a the TIP14x series devices in the library.

There are other answers on this site that may help you, I will add these to this answer at a later time.

Responding to your specific questions:

Does this approach (multiplying instantaneous voltage by instantaneous current and integrating/summing over the conduction period) make sense for a Class AB stage at these current/frequency levels?

Yes, provided you are multiplying instantaneous values, and not some convenient way of expressing time-varying values with a single number, eg: RMS or average. Use a simulator for this task.

What are the best practices (analytical or SPICE-based) for accurately gauging how much heat each transistor will have to dissipate over a cycle?

Use spice. LTspice is free (seems you already have it).

Are there particular considerations for Darlingtons (e.g., higher saturation voltage, extra diode drop, etc.) that might significantly affect the power dissipation calculation?

The calculation of the power dissipation will be accurate, that is, will not be affected by any of the parameters you mentioned, provided that:
(a) you use a good simulator (eg: LTspice)
and
(b) have a good model for the Darlingtons.

You will find that the power dissipation in the output stage will be almost independent of the type of output devices used (Darlingtons, discrete BJTs, or MOSFETs). That is by the very nature of linear amplifier circuits.

UPDATE 17-Mar-2025
Your specifications seem to indicate an output power of about 100W (10A into 1Ω). A normal audio amplifier may not suit, as these are designed to deliver rated power into loads of 8Ω or 4Ω. 10A into 8Ω is 800W ! However, in general the designs are similar, but you may have to increase the number of output devices to get the higher current you need.

Also, your load is very inductive - at 100kHz, the 2uH inductor has an impedance of 1.26Ω which is larger than the 1Ω resistor. This will have consequences for the power dissipation in the output stage - power dissipation will be much higher than for a purely resistive load, which you will have to take into consideration. The simulations will reveal all this to you.

Here are some links to help you find a good amplifier design.

Introduction to audio amplifier design, with a decent 20W design in the answers:
Designing a 20W audio power amplifier

60-100W Audio Amplifier:
https://sound-au.com/project3a.htm

300-500W Audio Ampilfier:
https://sound-au.com/project68.htm

1500W Audio Ampilfier:
https://sound-au.com/project117.htm

Consider using class-G or class H to reduce power dissipation:
https://sound-au.com/articles/class-g.htm

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    \$\begingroup\$ ("[small signal gain @1 MHz 1% of DC gain] means that gain falls-off very rapidly as frequency increases" Or starts decreasing early, or a combination thereof.) \$\endgroup\$ Commented Mar 15 at 10:56
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Push pull driven bjt’s power dissipation can be derived as follows: From the data you provided the load impedance ZL=RL +jXL = 1 + j2*10^-6

The instantenious power dissipated by the top transistor Q1 is equivalent to the product of the collector-emitter voltage Vce_Q1 and collector current Ic_Q1 that is

P_Q1(t)= Vce_Q1(t) - Ic_Q1(t) assuming that the emmitter resistance Re of transistor Q1 is far far less than the load resistance RL, so you will need to derive intantaneous Vce and Ic

Vce_Q1(t) is nothing but the supply voltage Vcc - the output voltage Vo that is to say

Vce_Q1(t) = Vcc(t)- Vo(t), since you are using a sinusoidal input voltage the load voltage will be also sinusoidal in the form Vmsin(ωt)

so after substitutiion we get

Vce_Q1(t) = Vcc(t)- Vmsin(ωt) where Vm is peak voltage

instantaneous collector current flowing through transitor Q1 is equal to the quiscent current Iquiscent_Q1 plus Imsin(ωt-θ) (where Im is peak current) the quiscent current being the current flowing through the transitor for biasing and the sinusoidal current induced by the sinusoidal input, notice the current lag at an angle θ due inductive load. Lastly derive the phase using :

arctan(XL/RL) or

arctan(ωL/RL). also peak current Im can be found using :

Im = Vo/XL = Vm/ |ZL| where peak voltage Vm= sqrt(2)xVrms and Vrms = sqrt( average powerx RL) where average power is the power you desire to drive the load normally selected in the beginning of design.

With the derived equations, substiting yields

P_Q1(t) = (15-Vmsin(2π100000t))(Iquiscent_Q1+sin(2π100000t - arctan(2π100000210^-6/1) substitute Vm and Im after deriving them, generate a waveplot of the power dissipated in matlab or any simulation software. with the waveplot you can move the cursor to find instantaneous power at any time(t)

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