I need a simple single supply amplifier that draws a lot of current when signal is applied. Without going into why, let's just say this is supposed to be part of a compressor. Distortion performance is not important. What is important is that it has a high ratio of current drawn in relation to the signal. So if the signal is low, the current drawn is low. If the signal is high, the current drawn is 100X as much.

So I have modeled and breadboarded the following circuit:

not so good push-pull

[Note: Right click and select "Open Image in New Tab" for larger version]

This circuit works fine (with a 100pF cap across the PI outputs) but it doesn't draw nearly enough current. The transformer is a standard Radio Shack style output transformer (actually an Eagle LT700) with a 1K2 primary and 3R4 secondary.

Ultimately the problem is that I cannot pull more than 2mA through the power transistors. Can someone explain why?

My first guess is that the impedance relation between the phase inverter and power section stinks. But it's not crystal clear how to resolve that.

Ultimately I would like a current gain of 100:1. If I could make a 1W power amp out of this circuit, that would be ideal.

Any ideas?

  • \$\begingroup\$ I did not initially notice the inductors were part of a transformer. Where did you get this circuit/schematic from originally? \$\endgroup\$ – HL-SDK Sep 4 '13 at 18:55
  • \$\begingroup\$ Where you thinking spaghetti when you were drawing the circuit diagram? You should really limit unnecessary crossings in the diagram and draw an differential amplifier as differential amplifier. That way people see more easily how the circuit works. It took me a few moments to realize you drew an differential amplifier (the right two transistors) which is fed from a phase shift stage (leftmost transistor). The circuit is pretty much symmetric. \$\endgroup\$ – jippie Sep 4 '13 at 19:00
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    \$\begingroup\$ @jippie Although it de facto looks like a diff amp, I suspect that the intent is to make a non-complementary push pull output stage, whereby the output transistors take turns being activated on opposite half cycles, drawing current through two taps of an output transformer, much like a tube output stage. \$\endgroup\$ – Kaz Sep 4 '13 at 19:18
  • \$\begingroup\$ What is your supply voltage? What impedance do you believe the transistors "see" ? \$\endgroup\$ – Chris Stratton Sep 4 '13 at 19:58
  • \$\begingroup\$ @Chris Stratton - It's 9V in the schematic but I'm not limited to that voltage. Note that if you open the image in a new tab it will be legible. I think the transistors see something like 4k7+10k. But the transistor input impedance is going to be really low I think. I wonder if perhaps mosfets would work better here. I have 2n7000 on hand. \$\endgroup\$ – squarewav Sep 4 '13 at 21:24

Here's my thoughts. You want a power amplifier to drive a speaker right? And you also want to take progressively more current from its power supply as the signal gets bigger. Why don't you use a regular 1W amplifer IC and design, on the side, a small circuit that rectifies the peak signal amplitude and uses this to drive a current source that takes the required current from the PSU.

What does this solve? Firstly you can get 1W IC amplifiers from lots of places and they are not complex to build or get working. Secondly, a peak signal capture circuit (given that you can drive it from the amplifier output) is just a diode, a capacitor and a resistor. Thirdly, a circuit that takes current from the supply based on a demand signal (the output from the diode, capacitor and resistor) is easy too; it's an op-amp, a power transistor on a heatsink and a 1W, 1 ohm resistor.

  • \$\begingroup\$ But then it won't be a 1960's era design with a non-complementary transistor pair driving an output transformer. If this is for music production gear, it's an important aspect. \$\endgroup\$ – Kaz Sep 4 '13 at 20:26
  • \$\begingroup\$ As Kaz points out this is in fact a 60's era design. I am ultimately starting with a speaker driver. But then I'm going to deliberately limit the power supply to explore audio compression effects. But first I need a power amplifier with a decent current range. Unfornately nobody seems to know what I'm even talking about (except for Kaz apparently). Oh well. \$\endgroup\$ – squarewav Sep 4 '13 at 21:19
  • \$\begingroup\$ @ioplex Which part of your question mentioned this? Why does your power supply need a decent current range - this will prevent it from doing what you apparently want I would have thought? \$\endgroup\$ – Andy aka Sep 4 '13 at 22:01
  • \$\begingroup\$ @Andy aka - The fact that it's a 60's era design is not necessary knowledge to answer the question. But Kaz noticed that the topology is important wrt your answer because the IC might not react in the same way. The IC will almost certainly use a complementary push pull pair in which case limiting the supply current would not "compress" the output of both signal swings equally. Your answer is appreciated. But I need a non-complimentary push-pull topology. So I would like to just see if I can get this circuit to work. \$\endgroup\$ – squarewav Sep 4 '13 at 23:54
  • \$\begingroup\$ @ioplex What is the idea behind this then. I'm a guitarist and it sounds like it could be of interest. \$\endgroup\$ – Andy aka Sep 5 '13 at 15:59

Use an IC amplifier such as an LM386, this is if you want compression effects, albeit you need to know how a compressor works, first.


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