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Here is the schematic of the circuit I'm trying to build (source) but I only have an IRF9530N Mosfet; I don't have any NPN transistors at hand.

1 watt audio amplifier

What would the schematic look like with a MOSFET instead of the NPN?

Dave Tweed
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Tamás
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  • Put component designators on your schematic! It is a hassle talking about a circuit without designators. – Olin Lathrop Jun 22 '14 at 12:22
  • I don't know what that means. I thought this Q&A site wasn't only for professionals. – Tamás Jun 22 '14 at 12:27
  • Anyone, amateur or professional, still has a obligation to communicate circuits effectively. That means clear schematics that include component designators. http://electronics.stackexchange.com/a/28255/4512 – Olin Lathrop Jun 22 '14 at 12:33
  • I don't know what the above schematic is missing. I seriously don't. – Tamás Jun 22 '14 at 12:37
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    What's missing? Hint : it's easier to refer to "R3" than "the 100kohm resistor near the left : no not that one, the other one." –  Jun 22 '14 at 12:48
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    DC through a speaker? Ouch. – Kaz Jun 22 '14 at 17:18

2 Answers2

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Bipolar transistors are cheap and available. Get some.

This is a pretty crappy circuit in the first place. It is in no way capable of providing 1 W of AC power into the speaker. Even ignoring the inevitable off-center biasing of the transistor and any voltage drop accross it, at best this circuit is a 4.5 Vpp AC source with 50 Ω output impedance. 4.5 Vpp is 3.2 V RMS. The highest power transfer happens when the load impedance matches the source impedance, in other words when the speaker is also 50 Ω. That means the load (the speaker) will see half the open circuit RMS voltage, which is 1.6 V. (1.6 V)2 / 50 Ω = 51 mW. Or about 1/20 of the advertised output power. With a typical 8 Ω speaker, the output power will be considerably lower than that.

Another problem with this circuit is that it puts DC current thru the speaker. That's generally not a good idea, but at these low power levels won't cause any harm to a normal size speaker. This circuit is so poor that the additional distortion caused by DC thru the speaker is the least of the problems.

This circuit could work with a N channel MOSFET replacing the darlington, but the same output power calculations still apply. MOSFETs don't have as well a defined gate voltage threshold where they quickly change between off and on. Because of that, the bias point produced by the two 100 kΩ resistors can't be computed up front. If this is just for experimenting, use a 100-500 kΩ pot for the two bias resistors and adjust it so that the drain sits at about half the supply voltage (4.5 V) when there is no input signal.

Olin Lathrop
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  • I understand that it's a very basic, inefficient amplifier, but as you mentioned too it is just for experimenting. What legs ona mosfet correspond to the NPN's emitter-base-collector legs? – Tamás Jun 22 '14 at 12:25
  • Base => Gate; Collector => Source; Emitter => Drain; ? Am I correct? – Tamás Jun 22 '14 at 12:31
  • @Tamas: No, you have source and drain swapped. – Olin Lathrop Jun 22 '14 at 12:35
  • @OlinLathrop, I've seen this many times: " .. adjust it so that the drain sits at about half the supply voltage .. ". Why is this? Each time I have seen this remark it has been in the context of amplifier circuits. Does you bias the drain of the FET to be `Vcc/2` in order to provide for maximum potential voltage swing due to the amplified signal? – sherrellbc Jun 22 '14 at 16:24
  • @sherr: Yes, you bias the output in the middle of its range so that it has the maximum symmetric +- swing available. – Olin Lathrop Jun 22 '14 at 20:43
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In answer to your question, this is how your circuit would look if you replaced the bipolar Darlington with a MOSFET:

enter image description here

Notice that an IRF 9530 is not N-channel, it's a P-channel MOSFET, so the supply polarity has been reversed.

This circuit is fraught with problems and is barely suitable for experimentation, but for starters I'd cap-couple the source and the load to the circuit and then work with the biasing and feedback.

EM Fields
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  • So that's why it didn't work. I also had to replace the 100K resistors with 22K ones. I wonder if the MOSFET still works or I broke it with my wrong wiring. – Tamás Jun 22 '14 at 17:31
  • With two 22k resistors you are now using about 5x more power in that portion of your circuit - although still very little. – sherrellbc Jun 22 '14 at 20:48