# Push-pull power amplifier on complementary darlingtons

I decided to make a power DC-amplifier. I used complementary darlington pair TIP142 and TIP147. Also TL082 used for negative feedback. The schematic is as below:

I gave 100 kHz sine signal and under oscilloscope I noticed the crossover distortion. It seems opamp is too slow to compensate ±1.4V "dead zone" (distorion on oscilloscope above).

Then I decided to bias darlingtons bases using four 1n4148 diodes as follow:

Unfortunately after 5 seconds TIP147 blew up although transistors were mounted on a heat sink and the fun was over. :D

I was reflecting what actually gone wrong. I suppose that:

1. I didn't put diodes on the same heat sink as darlingtons. Consequently BE voltage dropped under the diodes bias.
2. In Horowitz's "The Art of Electronics" I have read if I we are using emiter resistors, then four diodes biasing are insufficient.

I would to know are my conclusions correct and also how to effectively get rid of crossover distortion.

• Are you stuck on using those TIP147's? Could someone propose a radically different way to do this if it "effectively got rid of cross-over distortion?" What are your constraints here?? Seems very, very wide open to me, right now. And yes, if the diodes are kept cold and the TIP147's heat up substantially, their own internal Vbe drops will radically shrink and your class AB quiescent current will rise like a rocket. (Even assuming you got it right to start.) – jonk Mar 21 '17 at 22:09
• By the way, your rails (consistent with a $1\:\Omega$ load) are about what to expect for a $100\:\textrm{W}$ (rms voltages) amplifier! Are you serious? – jonk Mar 21 '17 at 22:16
• My constraints are to keep schematic not much complicated (I am new into electronics), so I decided to use those TIPs. It provides high amplification. Next time I will decrease voltage on rails. I am interested how to reduce distortion and stay with this TIP pair. – Madras Mar 21 '17 at 22:27
• What loads? Are you serious about $100\:\textrm{kHz}$ (or more) for a frequency of operation? What peak voltage must be delivered into the load? What heat sinks are you using? (The TIP147 by itself can only deliver about 3 W into air. I know you are using heat sinks. But that tells me NOTHING about what those TIP147 can REALLY handle.) Does it need to work at DC? Or am I allowed to show you something that works up to 100 Hz and delivers 1 mW into a load of my choosing? In short..... write more!! – jonk Mar 21 '17 at 22:33
• (You are breadboarding the wiring for a 100 W power output stage? Wow! Also, I can't help you with the power estimate from a picture of a heatsink. So you will just have to live and learn with it, I suppose.) The diode issue is simple. The TIP147 BE junctions get hot and their Vbe (for same Ic) diminishes by as much as $-2.4\:\frac{\textrm{mV}}{^\circ C}$. But the diodes don't change -- they are cold. Say you have 2.8 V across the diodes. That doesn't change. But the TIP147s now need only 2.3 V as they are hot. Do you see the implications for the quiescent currents? – jonk Mar 21 '17 at 23:27

Four diodes provide too strong bias. Together with positive temperature coefficient of output transistors and lack of thermal coupling, the result is sadly expected.

An option is to use three diodes, but the circuit is still thermally unstable until diodes are on the heatsink, and it has smaller but still very high crossover distortion.

simulate this circuit – Schematic created using CircuitLab

Better solution is to use transistor. It must be mounted on the same heatsink, preferably on top of one darlington. Quiescent current should be adjusted to about 20mA (start with maximum resistence, measure mV on emitter resistor).

On 1 you are correct. The diodes are really temperature sensors.

On 2 the answer depends on your definition of "sufficient". Generally you want to have the same number of pn junctions in the bias section. As the idle current is small typically in the ma range the emitter resistor isn't really a factor.

In your case I think issue 1 is a far bigger driver . insufficient biasing would have saved you, everything else being equal.

Btw, oscillation many times kills amps too. So keep an eye on that.

Edit as your rail is single ended you will need a capacitor on the output to block dc.

• I use symmetrical +/- 17V rails as on the attached schematics. I do not want to block DC. It should amplify DC voltage too. Did you mean oscillations due to negative feedback? – Madras Mar 21 '17 at 22:35
• yes. check to see if your opamp is unity stable. – dannyf Mar 21 '17 at 23:19
• I would also short one of the diodes (anyone) and measure the idle current. if you want to keep four diodes, increase the emitter resistors to something bigger, like .47 or even 1R. – dannyf Mar 21 '17 at 23:20
• doesn't really matter where it is: the opamp will output the right dc offset to make sure that the output is center around 0v. – dannyf Mar 21 '17 at 23:43
• > A class... take a look at the datasheet and specifically the soa / derating curves and see what you can get out of a pair of devices like that in class A. BTW, the way you constructed your vbe multiplier is very dangerous: what if the wiper goes open? – dannyf Mar 23 '17 at 0:20

Place a 1 microfarad ceramic cap between the darlington bases and look for waveform improvement on the scope .This improvement should be more visible at say 20KHz. This is because loop gain decreases with frequency paticularly for opamps.Increase the value of the emitter resistors .A rule of thumb is up to 10% of the load resistance .It is a trade off between output swing and thermal stability .Your diode circuit is adequate but the thermal coupling should be adressed .What I have done is make my own diodes out of TO126 transistors like BD139 or what ever is handy .Bolting this to the same heatsink has made for reliable Audio amplifiers operating well into class A .I have tied the base to the collector to make a diode .Also I have jacked it up with BE and BC resistors making a VBE multiplier. I have seen this VBE multiplier called a Rubber Diode .The 1 mic cap is still just as useful . The key here is that the thermal resistance of the TO126 package to the heatsink is really low compared to your 4148 or other peoples BC547 .There is little point in designing fancy bias circuits before your thermal impedance is dealt to .

• Thank you for the answer, but I am going to replace all the diodes with Vbe multiplier. Advantage of this solution are only one additional transistor (for example BD139) on a common heatsink and gives possibility to adjust a quiescent current, too. Are your tips still compatible with this modification? If so, then could you explain how 1 uF ceramic cap may improve parameters of the amplifier? Capacitor between bases seems to be something another than bootstrapping. – Madras Mar 26 '17 at 15:11

try to avoid BJT transistors because BJT has large collector emitter resistor (CE(on)). in high current applications they loss more power and badly heat up. Try MOSFETs because its (RDS(on)) is very low. Transistors are current controlled device and MOSFET is voltage controlled device. Try this circuit