Don't Understand Amplifier Circuit Results

Thread Starter

elec_dude

Joined Mar 31, 2017
20
I built this amplifier circuit. The voltage readings on the power transistors start dropping after a few seconds and the current draw on the power supply starts climbing.

The power transistors start to heat up. I know that the circuit is built correctly per the schematic. The components are all new and I have replaced the transistors with the same results. I have not injected a signal into the circuit.

See attached schematic.

Any advice would be appreciated.

Would the transistors be damaged from heating up?

John

upload_2018-1-27_11-32-55.png
 

ColinPatra

Joined Jan 27, 2018
28
Hi John,
I've just one a quick DC calculation. I get the voltage across R4 is 12V * (15k)/(115k) = 1.56 V. Allow 0.56 V for the Vbe of TR3 and there is 1V across R4 so a current of 10 mA. That means a drop of 10V across R3. (ignoring the base currents of TR1 and TR2) . This looks strange!
 

BobTPH

Joined Jun 5, 2013
2,047
With no signal the only current path through the output transistors is through both of them. With two diode drops between the two bases, one of the two must be off, or the may both be slightly on, which should not cause any heating.

Either some component is bad, or you have wired it wrong.

Please
Report the votage at the two bases.

Bob
 

Thread Starter

elec_dude

Joined Mar 31, 2017
20
Voltage Readings are as follows:

TR1 Base 7
TR1 Collector 6

TR2 Case 1.2
TR2 Collector 5.6
TR2 Emitter 0.5

Circuit is wired correctly.

John
 

ColinPatra

Joined Jan 27, 2018
28
Hi John
If these are readings wrt 0V, I don't understand. The collector of Tr1 (TIP31A) goes to the 12V supply. and the collector of Tr2 (TIP32A) goes to 0V. Please confirm!
Clin
 

Thread Starter

elec_dude

Joined Mar 31, 2017
20
The voltage readings are as follows:

TR1 Collector 12.0V
Base 6.7V
Emitter 6.0V

TR2 Collector 0.0V
Base 5.3V
Emitter 6.0V

TR3 Collector 5.3V
Base 1.2V
Emitter 0.5V

These values start to fall as the transistors warm up and the power supply current
drain steadily increases. Reminds me of thermal runaway with the old geranium
transistors. The circuit is built according to the schematic. Is there a problem with
the schematic and/or the component values?

John
 

crutschow

Joined Mar 14, 2008
23,487
My LTspice simulation of the circuit showed excessive bias current (0.26A), which is causing the heating.
Below is the circuit modified to reduce the output bias current.
I added 1 ohm emitter resistors, R5 and R6, to stabilize the bias current at about 75mA.
I changed the values of R1, R2 and R3 to improve the bias stability and move the output bias point to the center of the supply voltage.

upload_2018-1-27_11-48-18.png
 

Attachments

ColinPatra

Joined Jan 27, 2018
28
Hi John

The voltages across R3 is less than I had expected. But taking it as correct, the other voltages agree with a 1.2 V drop across D1 and D2 . BUT that gives a Vbe of 0.7 V for both the output transistors - which I think means they are conducting quite heavily even with no signal. I would have expected a small resistor in series with each emitter to provide a little dc feedback.
Colin
 

WBahn

Joined Mar 31, 2012
24,850
I built this amplifier circuit. The voltage readings on the power transistors start dropping after a few seconds and the current draw on the power supply starts climbing.

The power transistors start to heat up. I know that the circuit is built correctly per the schematic. The components are all new and I have replaced the transistors with the same results. I have not injected a signal into the circuit.

See attached schematic.

Any advice would be appreciated.

Would the transistors be damaged from heating up?

John

View attachment 144500
There is nothing magical or engraved in stone about the forward voltage drop across a diode or the Vbe of a transistor being exactly 0.7 V.

What I suspect is happening here is that initially you have enough bias current to result in a combined voltage drop across the diodes to be high enough that both transistors are being turned on at the same time. The result is "shoot-through" current that dissipated power in the transistors heating them up. But as they heat up, the Vbe needed to sustain a given collector current drops the current increases, which results in further heating, which results in a further reduction in Vbe. This is classic thermal runaway.

You either need to decrease the voltage across the two diodes so that it is not large enough to allow both transistors to be on in any significant way at the same time, or you need to increase the voltage required between the bases of the two output transistors so that the voltage across two diodes is not large enough to allow both transistors to be on in any significant way at the same time.


There are a few ways to accomplish this.

First, as you reduce the current in the bias circuit, the voltage across the two diodes will decrease. If you can decrease the bias current enough, then the two transistors will no longer conduct at the same time. This is a twitchy approach and may not work reliably over the full range of operation, especially as the temperature of everything changes.

Second, you can accomplish this same thing in a more reliable way by using a circuit known as a Vbe multiplier. This will allow you to adjust the voltage between the bases and set it to something that is (hopefully) small enough to prevent shoot through put large enough to still keep crossover distortion acceptably low.

While the second approach is more reliable, it is still a largely manual fix that is incapable to responding to changing conditions. What you really want is a change that will automatically adjust to changing temperature and keep the runaway from happening.

One way to accomplish this is to put a small resistor between the two emitters. For symmetry, put two equal resistors there so that the output can be taken from the junction.

Now, as Vbe drops across the transistors due to heating, a slight increase in shoot through current will result in an increase in voltage drop across the emitter resistors that will compensate for it. It doesn't take much resistance, either. The optimal choice depends on a number of factors, but you will likely find that just throwing in a couple of 1 Ω resistors is close enough. From there, you can adjust them up or down until the no-signal quiescent current through them is a perhaps few percent of the max current you want them to conduct.
 

Thread Starter

elec_dude

Joined Mar 31, 2017
20
I built this circuit and had the same result, overheating and runaway current draw.
When I fed a 1Khz 100mV sine into the circuit, it amplified for a few seconds and
then quit.

The circuit is built correctly and the voltages are as follows:
Q1 Collector 11.8V
Q1 Base 8.4V
Q1 Emitter 7.7V

Q2 Collector 0.0V
Q2 Base 7.0V
Q2 Emitter 7.6V

Q3 Collector 7.0V
Q3 Base 1.4V
Q3 Emitter 0.7V

The current drawn starts at 34mA and just keeps on rising.

John
 

sghioto

Joined Dec 31, 2017
1,038
You can also replace one of the diodes with a 500 ohm potentiometer to adjust the bias. Usually adjusted to just eliminate the crossover distortion but somewhere in the 20 to 30 ma range may work.

SG
 
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WBahn

Joined Mar 31, 2012
24,850
To WBahn:

Do you mean I should put a 1 ohm resistor across each of the two diodes?

John
I assumed that the circuit you built is the one that crutschow presented in Post #9. This circuit embodies the recommendation I made in my post (even used the same starting value of 1 Ω).

It would be nice if you reported your results to three sig figs (two places to the right of the decimal point, in this case).

If you have two 1 Ω resistors between the emitters and are seeing a drop of 0.1 V across them with no signal, then that means that the quiescent shoot through current is about 50 mA and that you are dumping about 300 mW of power in each one. That will warm them up pretty good, particularly if you don't have descent heat sinks on them.

The thermal coefficient for Vbe is generally taken to be -2 mV/K. If you are operating without a heatsink and using a TO-220 package, then the rule of thumb is a thermal conduction of about 70 K/W, so at 300 mW you would be have an increase of about 20 K, which would depress the Vbe by about 40 mV. Since, near room temperature, it only takes about 20 mV to double the collector current, this is significant as it would quadruple the current. But with a 1 Ω resistance (in each emitter), the full 40 mV increase would be consumed by an increase in collector current of only 40 mV, or less than a doubling. But it would still result in a noticeable increase in current and, hence power. But it should still limit itself fairly quickly.

If you increase the emitter resistors to 10 Ω (which will probably result in undesirable performance in other respects, but that's not the concern right now), then it would only take an increase of 4 mA to offset the lowering of the Vbe.

EDIT: Fixed 300 mA -> 300 mW. Thanks, sghioto.
 
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