What's wrong with thermal runaway?

[Plecto]

Hello. I've long been having issues with biasing a class AB push-pull stage and I hope someone could help me get things sorted I've previously been biasing it using resistors only, but I've been told many times that this is a bad way of biasing because of the risk of thermal runaway. I'm advised to use diodes instead of resistors to prevent this, but using diodes for biasing introduces problems that I can't quite fix. Here's three ways of biasing and my thoughts about them:



This is the configuration I see most often, but I don't see how this can be done in practice. Assuming equal characteristics of all PN-junctions the biasing current will be ((Vs-1.4)/(R3+R4)*R3)/2*Hfe or half the current flowing through R3 and R4 times the gain. Unless very high bias current is needed, R3 and R4 has to be of some size, but that would severely limit the output voltage as the output voltage is limited by 0.7+IL/Hfe*R3.



I find this configuration more realistic. The bias current can now be tweaked by adjusting the values of R7 and R8, but the output voltage will still be limited by 1.4+IL/Hfe*R3 so R3 and R4 still have to have pretty low values to preserve the output voltage of the AB stage. An issue with this is that because of the high gain of the darlington pairs the bias current becomes very sensitive to the value of R7 and R8 and after some testing, I could not repeatedly set an accurate bias current. Values that lead to 20mA of bias current on one design does not lead to the same current on another. Another point to make about this circuit is that about half the transistor diode drop resides above R7 and R8 so the bias current will still rise with temperature, although not as much.



As for this tabu-circuit. The bias current can be accurately and predictably set each time (within 30% or so at least) and the output voltage is not limited by the values of R3 and R4 anymore, but rather the values of R7 and R8 which can be much much smaller without getting an excessive bias current. One issue is of course thermal runaway, but I've made this design many times and never had an issue with it. Dissipating about 20W into a 1.8C/W heat sink will make the bias current rise about 50% due to thermal runaway and then settle in a predictable fashion every single time.

So the question is, what's so wrong about thermal runaway if the increase in bias current can be taken into account? In my experience it has been a very predictable. Am I right to say that the gain vs. temperature graph isn't as steep as the temperature vs. watts cooling graph for heat sinks? Giving that the bias current always settles down I assume that there will always be a point where the cooling of the heat sink get's so big that the thermal runaway stops?

[Kleinstein]

It depends on the heat sink and how much power depends on temperature, whether there will be true thermal runaway or only a reduced stability. If the heat sink is very large, an increase in power will not give much increase in temperature and thus not much additional power - so nothing dramatic happens. However if the heat sink is relatively small, the additional power will result in an significant temperature rise and thus even more power and so on. In the simple linear model, once the system gets unstable, there is nothing holding it. So it might well heat up to destruction. Nonlinear effects might save the unit or not.

Besides this, distortion depends on the bias current. So one also wants the bias current of the output stage to be rather constant, independent of temperature. So even if you don't get a thermal runaway, a temperature dependent bias is a bad thing.

The practical solution is to use a transistor and an adjustable pot as a so called V_be multiplier. This gives an adjustable voltage drop of something like 2.2 times the base emitter voltage. Thus something like 0.2 V_BE drops over the two resistors at the output. Details might be more complicated, as not all parts have the same temperature.
Usually R3/R4 in the circuit are not just resistors but more often current sources or the driving transistor. So there is no problem creating the biasing voltage. The case with just resistors even relies on a relatively constant current. It the current depends on temperature in the right way, this still can be a valid solution.

[codeboy2k]

If you want (or need) accurate bias currents, then we turn to Jim Williams. He did an interesting design in his 100A active load, where he replaced the bottom diode (D2) with a VN2222L FET, whose gate was driven by an opamp.  This opamp has one input set to a reference voltage, and the other input comes from sensing the voltage across a resistor in the output stage.  So by sensing the voltage drop across a small resistor, closed a loop that controls the bias current, and he maintained the needed bias current over all output loads.

Figure 5 in this EDN article: http://www.linear.com/docs/40601

[calexanian]

In audio this is handled in two different ways. The most common with your first schematic would be to thermally bond the bias diode set to the transistors so they track reasonable close to each other. Over bias is limited (Hopefully) by the emitter resistors and under bias to a small degree is helped by having a very high slew rate in your feedback circuit. To get more precise control a transistor collector to emitter is substituted for the series diodes with a pot from collector to emitter with the wiper on the base and some other resistors, this in turn in thermal contact with the following transistors. This gives a degree of bias adjustment. Ideally you would watch the bias current of the whole stage cycle every few seconds to a minute or so by a small amount where if things start getting away, the bias element lowers its dropping voltage reducing later stage bias current. I recommend looking at the Elliot Sound Products pages. he goes over this in depth.

[T3sl4co1l]

Required reading:
http://en.wikipedia.org/wiki/Bipolar_junction_transistor#Ebers.E2.80.93Moll_model

Vbe is NOT "0.7V".  Obviously if it were fixed, resistors would work, and that would be the end of that.  So where's the tempco?  Duhh....    Right?  So, obviously, calling it 0.7V, full stop, isn't right.

If you take the exponential,
I ~= Is * exp(V/Vth)
(Vth = thermal voltage, also, give or take the emission coefficient; Is = saturation current)
then you will see, for some fixed terminal current (such as for the bias diodes), the terminal voltage must go as ln(I/Is).  The logarithm is a very slowly increasing function (over a factor of 10x up or down, Vf only varies about +/-0.1V), but it's nonetheless variable.

Vth is proportional to absolute temperature.  Is also varies with temperature, but it's usually the less significant factor (it's not inside an exponential!).

The BJT does the same, except it does Ic ~= Is * exp(Vbe/Vth).  Ib is similar, and the ratio between them is hFE, but as long as this ratio is large, we don't care about it, and in particular, we always want to design circuits such that they are as insensitive to hFE as possible.  hFE varies widely with manufacture, temperature, Ic and Vce.

The result is that, if we have a resistor-diode voltage divider (such as the bias network in the first example), we get a voltage offset that's characteristically proportional to this log-exp function, including its tempco.  Connect that B-E and you get the exact same current back, give or take the differences in doping and junction area between the diode and transistor.

To further enhance stability, emitter resistors are used, which either complement the resistor in the bias chain (if it's a +V - pullup - base - diode-diode-resistor - other base - pulldown - -V configuration), or simply buck the current mirror behavior altogether (linearizing the transistor's transfer function retains the diode's log character, making something like a Widlar current mirror, i.e., some current in, log current out).

I mentioned doping and area.  Saturation current is a device constant -- more junction area means more Is, straightforward as that.  If you had a power transistor that was made on the same process as a 2N3904, but which was rated for 2A instead of 0.2A, you should find Is measures 10x higher.  But you're more likely to get disparate transistors (say, a TIP31C with a 3A 100V rating, clearly using a different process than the 60V rated 2N3904), which means both Vbe and Is will be different.  You can very easily end up with a runaway tempco, even with correct diode compensation in place.  This is most typical with excessively high amperage transistors (and is doubly a problem, because big transistors can sink big currents if they want to!).

The concept of "ratio of areas = ratio of voltages" arises in the design of the balanced bandgap voltage reference, or the intentionally unbalanced temperature sensor.

Tim

[dom0]

I dunno how they do it in really, really large linear power stages, but I'd feel really uncomfortable with just a diode biasing network and some additional emitter resistance. I think I'd go straight for a closed-loop bias current control.

[Plecto]

Thanks for the replies. So how exactly could I tackle this problem, or at least come up with a solution that's better than the ones I've already mentioned without increasing the complexity too much? How about something like this:



If R26 and R27 are reasonably small, the ouput voltage of the device shouldn't be that limited. If Q10 is mounted on the same heatsink as the other two transistors, shouldn't that counter thermal runaway some what?

[mikerj]

That is a fairly classic biasing scheme for a Class AB output stage.  Good thermal coupling between the three transistors will give reasonably stable bias, though it never going to be perfect.  Usually there would be a potentiometer in the resistor chain so that bias can be set to a specific initial value.  Note that you have omitted the emitter degeneration resistors which are still required.

[Odysseus]

 I don't think you can do much better than something like this.  Runaway and biasing is a non-issue.  Resistors can be used in place of current sources. Q1-Q4 are small signal transistors. Q5,6 are your power transistors. Choose Ix around 1-10mA.  Choose Ra so that Ra*Ix to be about a volt or a few tenths of a volt.  Then choose Rb about 5-20 times smaller than Ra, depending on the minimum Hfe of your power transistors and the bias current you want.

[dom0]

Mmmh, the sziklai pairs in the output stage might need some additional Miller capacitance (depending on the actual devices used), they tend to oscillate. Or at least, they tend to do when I build them I think they still have some voltage gain left (in theory it should be very close to 1), probably by virtue of device mismatch.