I have a problem I can't quite figure out.

I essentially want to "buffer" a 50mVpp audio signal to drive a 16 ohm load.

I preferably want to use primitive components so I thought that maybe a Darlington emitter follower might be used.

The problem is that AC coupling, the way I do it, makes the Darlington amplifiers output distort the waveform and massively shrinks the voltage amplitude.

Like this:

enter image description here

But there is no distortion and no shrinking of amplitude when disabling the AC coupling:

enter image description here

Altough AC coupling works for with a bigger load:

enter image description here

But since I want to drive smaller loads, I tried to see if a Darlington quadruple would work. And the waveform and the amplitude actually looks a little bit better, but it's still far from perfect:

enter image description here

I wonder why the AC coupling creates this problem for me?

Correction: Actually, the Darlington quadruple was not better in any way. The better wave form was only a result of a higher bias voltage (before the transistors).

  • \$\begingroup\$ I like the question and the effort you made researching this. However I think you are actually asking two things here: 1. What is the cause of the (change in) distortion and 2. What would be a better way to buffer the signal. Which of the two questions do you want an answer for? \$\endgroup\$ – jippie Oct 15 '15 at 18:59
  • \$\begingroup\$ Thanks! You're probably right. Why "the problem" is created, and actually I am also interested in one (or several) "work-arounds" if there's something relatively simple. \$\endgroup\$ – fredrik.hjarner Oct 15 '15 at 19:47

Your capacitor is making a resistor divider!


simulate this circuit – Schematic created using CircuitLab

Look at it like this schematic (same thing as your first circuit, darlington transistor as a voltage source, which should be (or at least you want to be) equivalent to your audio signal.

Now, the capacitor is 100uF to DC, but to AC it looks like a resistor of value 40 ohms to a 40 Hz input signal.

40 isn't a lot compared to 1k (your third example), but it is a lot compared to 16 (your first example). Of course, if you short it (second example), then it is zero, and the divider effect goes away.

  • \$\begingroup\$ Doh. Thank you very much! That makes a lot of sense. \$\endgroup\$ – fredrik.hjarner Oct 15 '15 at 19:23

In addition to Ken's answer about the impedance of C1 (the output capacitor), there is a second problem. You're using a 10 kΩ pulldown resistor on the emitter of the Darlington pair, which is a poor match for the load.

When the voltage is rising, the output impedance of the emitter is very low — you can think of it as a voltage source. However, when the voltage is falling, the transistor is cut off, and the only path to discharge the output capacitor is through the 10 kΩ resistor.

This huge asymmetry in the output impedance driving the left side of the capacitor explains the distortion you see in the first waveform.

So, bottom line, you want that pulldown resistor to be about the same order of magnitude as your load impedance, and you need to make sure that the impedance of the capacitor is small relative to the load impedance at the frequencies of interest.

If you get this right, then the Darlington configuration should not be necessary at all — a single transistor should suffice.

  • \$\begingroup\$ Thank you! Very clear addition! Yes, when trying the circuit with a pull-down resistor closer in value to the small load, then it's better (but it draws current like a beast). \$\endgroup\$ – fredrik.hjarner Oct 15 '15 at 19:56

Boiling down what Ken said R1 needs to be low .Infact R1 should be in the ballpark of the speaker impedence for a credible output voltage swing that isnt too distorted .You can use a current sink for R1 and you will get a nice class A amplifier that would be 25% efficient if the darlington was ideal.You could use a SRPP scheme to prospectively get the standard theoretical class A efficiency of 50%.Whats generaly done these days is a complementry emitter follower which can give class B efficiency .What was done in yesteryear was a choke or output transformer which should give normal class A efficiency.Chokes have got more expensive and transistors are much much cheaper.Speaking from experience the SRPP sounds better than a complementry emitter follower.

  • \$\begingroup\$ You might want to define SRPP and link to an example of it. \$\endgroup\$ – Brian Drummond Oct 15 '15 at 19:57
  • \$\begingroup\$ Brian Drummond .I will one day get with the program and learn to do a link.My screenname is not entirely innacurate!.The SRPP dates back to the valve days ,in 1958 it was used on the phillips HiZ Amplifiers .Its used in valve OTL amps .I have built these with N channel mosfets . Its better cousin is the loftan white follower .You can use darlingtons .The phillips electronics set in the 1960s for kids did a 9V SRPP with 2 AC126s driving a 80 ohm speaker. Orthodox SRPP runs class A giving big single ended sound without the output transformer. \$\endgroup\$ – Autistic Oct 16 '15 at 12:12

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