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Context: output stage of Class-AB audio power amplifier.

enter image description here

In the example schematic above, the output stage is biased in Class-AB via a bias voltage source. In use, output transistors will heat up depending on signal but the bias voltage generator will only track these temperature changes after a while, since it is usually implemented as a Vbe multiplier using a BJT mounted on the heat sink.

So how much does transient heating of the output transistors upset their Vbe, which upsets bias setting and how much distortion does this generate ? This is complicated to measure, it is simpler to measure the transient thermal impedance of each transistor separately and make a thermal model that can then be used in simulations.

So I want to measure how much the Vbe of these output transistors moves around due to transient thermal effects. The way to do this is usually to make the transistor dissipate a pulse of power then switch to a lower power and record Vbe as it cools down. I've seen setups like that:

enter image description here

However this one has two constant current sources, which begs the question: what is the settling time and transient behavior of these current sources and how will that interfere?

So instead I went for simplicity:

enter image description here

The setup is powered by a bench supply. Q1 (the DUT) is wired as 2 Amp constant current source, its base grounded through a resistor. When MOSFET M1 is OFF, there is about the same voltage drop in R1 and R2 so Q1's Vce is a few volts. With M1 switched ON then Q1's Vce is 32.7V. So Q1's dissipation can be switched between low and high power, and the scope can measure Vbe and also hFe (via voltage on R3).

Current draw from the power supply is constant, so I don't need to worry about its transient behavior (and I can check if the supply voltage stays flat). It seems to give results:

enter image description here

So, is this test setup sane and can it be trusted? The other setups I've found on the net are much more complicated, so it feels like I've overlooked something...

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  • \$\begingroup\$ So you want to measure the "thermal" distortion? \$\endgroup\$
    – G36
    Jan 4, 2020 at 15:48
  • \$\begingroup\$ @G36 This subject usually generates generates lots of talk on audio forums, much hot air, and very little experimental results... I'd rather measure to know if there's something interesting or not. Also a thermal model would be useful to calculate transient Tj to make sure the protection circuit actually protects the devices... \$\endgroup\$
    – bobflux
    Jan 4, 2020 at 16:02
  • \$\begingroup\$ Been lurking on this thread for a while...I'm going to be picky here. The phrase "transient thermal impedance" really doesn't make much sense. I'm familiar with the term "thermal resistance" and "transient thermal response" or "thermal rise" from our thermal analysis folks, but the the OP's title really doesn't explain things.. \$\endgroup\$
    – SteveSh
    Jan 8, 2020 at 23:30
  • \$\begingroup\$ @SteveSh It is usually called "transient thermal impedance" in datasheets. I agree it makes little sense. "Transient thermal response" would make more sense (and that's what I want to measure). \$\endgroup\$
    – bobflux
    Jan 9, 2020 at 19:59
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    \$\begingroup\$ Learn something new everyday! Thanks, peufeu. \$\endgroup\$
    – SteveSh
    Jan 9, 2020 at 20:28

2 Answers 2

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You can see in your graphs that, on a time scale of about 50ms, your Hfe doesn't change much more than 3-4%. In your first schematic, however, Vbe is your parameter which will contribute more to the distortion, and that's under 2% variation. These represent the maximum slopes of those curves, starting at t0.

50ms is a complete cycle at 20Hz, typically the lowest frequency of interest in audio design...so there will be very little distortion from an audio standpoint. Over the longer term, the gain may change slightly, but this won't cause harmonic distortion. If it did, the distortion that comes from your sub-$10000 speaker system will easily overwhelm it.

If you're a real perfectionist, drive the finals with an op amp and take the feedback from the output.

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  • \$\begingroup\$ Yes indeed. But I'm not asking how to interpret the results... I'm asking if the measurement setup is likely to give valid results. \$\endgroup\$
    – bobflux
    Jan 7, 2020 at 1:14
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    \$\begingroup\$ I'd call them valid...not too precise, but enough for audio. You're not quite getting 2A through there, probably more like 1.86 due to the drop across R3 and Q1be. If you set the negative supply to -32.1 it would be closer. Also, if you want to be that choosy, set the positive to +28V to get closer to 30V on Q1ce. (Your current configuration has those two compensating so you're still close to 60W total on Q1.) Also, you might want to check out how R2 varies over temperature. \$\endgroup\$ Jan 7, 2020 at 13:51
  • \$\begingroup\$ Thanks! I did check R2's tempco, it's alright. It is actually a huge cemented wirewound resistor which is about 8 inches long, so it has a bit of thermal mass too. \$\endgroup\$
    – bobflux
    Jan 9, 2020 at 20:14
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I would say that the short answer is yes, it is a decent measurement setup. The longer, and slightly more annoying answer is, as it often is in engineering, it depends on what accuracy you are after

Regardless however of the accuracy that we are after we need a model in order to say something about what we are measuring, and how well.

Let's first create a formal mathematical description of the quantity that we are trying to measure, if I have understood the question correctly, we are trying to find the transfer function between the power dissipation and the base emitter voltage or;

$$G(s)=\frac{V_{BE}}{P}$$

The following is an approximation to the Ebers–Moll model for a BJT, it is sufficiently accurate for now;

$$I_E=I_{ES}(e^{\frac{V_{BE}}{V_T}}-1)$$

We don't care about the emitter current but rather the base emitter voltage, and its dependence on V_T (the thermal voltage), so let's rearrange;

$$V_{BE}=V_T\ln{(\frac{I_E}{I_{ES}}+1)}$$

Since I_ES is assumed to be constant, and since I_E is kept constant, we see that indeed V_BE is proportional to V_T.

For V_T we have the equation;

$$V_T=\frac{kT}{q}$$

Where k is boltzmann's constant, q is the elemental charge (both are constants), and T is the temperature in Kelvin.

For now let's assume that the transfer function between the power dissipation and the junction temperature can be modelled as a single pole low pass filter as follows;

$$\frac{T}{P}=\frac{R_{th}}{1+\frac{s}{\omega_c}}$$

Where R_th is the thermal resistance and omega_c is the corner frequency

Combining all of the above we have the transfer function;

$$G(s)=\frac{V_{BE}}{P}=\frac{kR_{th}}{q(1+\frac{s}{\omega_c})}\ln{(\frac{I_E}{I_{ES}}+1)}$$

So if we assume that the thermal resistance is the same between your test setup and the final application then the right hand side of the equation above is just a constant and your measurements hold.

Now to tear it appart

Let's re-evaluate the approximations that we made before;

  • We assumed that the simplified Ebers-Moll model was sufficiently accurate, is it?

Well the full Ebers-Moll model for the emitter current in a BJT is as follows;

$$I_E=I_S((e^{\frac{V_{BE}}{V_T}}-e^{\frac{V_{BC}}{V_T}})+\frac{1}{\beta_F}(e^{\frac{V_{BE}}{V_T}}-1))$$

Let's again solve for V_BE to get;

$$V_{BE}=V_T\ln{(\frac{\frac{I_E}{I_S}+e^{\frac{V_{BC}}{V_T}}+\frac{1}{\beta_F}}{\frac{1}{\beta_F}+1})}$$

We now see that the base emitter voltage is not just proportional to the thermal voltage V_T, but is also dependent on the base collector voltage V_BC, which in the case of your test setup is not a constant.

  • We also assumed that the emitter current I_E was a constant, is it?

Well it's probably close enough, but in reality the voltage across the bottom resistor R_2, and hence the emitter current, is going to depend on the base emitter voltage itself, as well as R_3 and the base current. I am not going to calculate by how much right now as it's a harry bit of math work and because I think the error it introduces is fairly negligible, but it is worth noting that this causes the factor on V_T to be non constant.

  • We assumed that the thermal resistance R_th was the same between your test setup and the final application, is this the case?

I don't know your test setup well enough to draw any conclusions in this regard, so I will leave that as a question for you.

  • We didn't considder any parasitics such as collector-emitter and base-emitter capacitances, is this reasonable?

Yes, since we are dealing with audio frequencies and low impedances it is completely reasonable.

  • Did you make any mistakes in the execution?

Well one thing that I will note is that I would definitely not, ever, have connected the gnd of my oscilloscope to anything other than gnd, Vcc or eq. (low impedance point) of my circuit. This is because inside the osc. its gnd is capacitively coupled to earth, or worse, if not earthed it can even be capacitively coupled to line, through a mains input filter.

Another thing is related to the circuit diagram that you show in your question; The FET used to switch the collector current is an NMOS, and it has its source connected to the supply, I assume that this is an error, and that you meant for it to be a PMOS?

Also, I did not address the point of how you drive the FET, because you did not show any gate-drive circuit, and I have just assumed that you have measured the collector voltage, and that it does indeed switch fully on and off.

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  • \$\begingroup\$ Thanks for the detailed answer! My comments: "We assumed that the thermal resistance R_th was the same between your test setup and the final application" -- I will try measuring with no heat sink, a very low thermal inertia heat sink (vapor chamber) and a high thermal inertia one (big chunk of aluminium) to separate thermal response of the transistor vs thermal response of the heat sink. \$\endgroup\$
    – bobflux
    Jan 9, 2020 at 20:05
  • \$\begingroup\$ "We now see that the base emitter voltage is not just proportional to the thermal voltage V_T, but is also dependent on the base collector voltage V_BC, which in the case of your test setup is not a constant." --> Yes. I only take the Vbe waveform after the MOSFET has switched, since before the FET switches, Vce and Vbe will be different. I'm making a note to check if Vce/Vcb settles quickly and cleanly. I'll also recheck power supply voltages for flatness. \$\endgroup\$
    – bobflux
    Jan 9, 2020 at 20:08
  • \$\begingroup\$ Good catch with the capacitive coupling between scope ground and mains. I'll measure Vbe and hFe separately, which will allow connecting scope ground in a more "orthodox" manner and see if there is a difference. // FET is indeed wrong on schematic, although it is correct in the setup, sorry about that. I omitted the driver for simplicity, but yes it does switch properly. \$\endgroup\$
    – bobflux
    Jan 9, 2020 at 20:11
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    \$\begingroup\$ "Yes. I only take the Vbe waveform after the MOSFET has switched, since before the FET switches, Vce and Vbe will be different. I'm making a note to check if Vce/Vcb settles quickly and cleanly. I'll also recheck power supply voltages for flatness" You not only need to check that the voltages settle "quickly and cleanly" but you also have to make sure that the voltage across the collector-emittor changes quickly enough (high enough dV/dt) as what you wan't to do is convert a step respons to a transfer function, so ideally the change in power (not voltage) has to be instantaneous. \$\endgroup\$
    – Vinzent
    Jan 9, 2020 at 20:53
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    \$\begingroup\$ Seems like you've got it under control then, I would say that, to answer your original question; your settup IS sane, mostly it depends on the amount of understanding excersised by the person using the setup. I hope that I could be helpful in eleborating on the pitfalls of measuring like this (: \$\endgroup\$
    – Vinzent
    Jan 10, 2020 at 19:15

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