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I'm trying to build a simple amplifier for my guitar. I created the exact same circuit as a common-source amplifier, I biased it on the necessary Q-point, and AC signals are decoupled.

It worked nicely; I get the amplified guitar output on my osilloscope. The output signal was really satisfying, I get 2 to 10 V Vmax, depending on which guitar string is oscillating.

But the problem is that the output signal disappears when I connect a speaker to the circuit. By "disappear" I mean that the signal gets very low and the speaker isn't being driven. When I put my ear on the speaker and play something on the guitar I get a distorted guitar sound which is pretty low-volume, and I can barely hear it.

I connected the speaker right after the coupling capacitor on the drain point. I also tried to connect a 1 MΩ resistor in series with the capacitor and then connected the speaker in parallel with the resistor, but still could't get any results.

What should I do to get my speaker to be driven?

My circuit is like this (when I add a source resistor, the signal also disappears. It doesn't matter if I add the capacitor between source and ground):

enter image description here

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    \$\begingroup\$ Could you add a schematic of the amplifier you built? \$\endgroup\$
    – ocrdu
    Dec 22, 2021 at 22:20
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    \$\begingroup\$ What is the impedance of your speaker - 8ohms? And what is the output impedance of your amplifier (mostly dominated by the resistor you probably have between the drain of your output FET and the positive supply)? How do they compare to each other ... .? \$\endgroup\$
    – brhans
    Dec 22, 2021 at 22:22
  • \$\begingroup\$ Realize that a simple common source amplifier isn't enough to properly drive a low impedance (8 Ohm) speaker. Look at other audio amplifier designs and notice how basically all of them have a push pull output stage. \$\endgroup\$ Dec 22, 2021 at 22:24
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    \$\begingroup\$ A simple common-source (or common-emitter for a BJT) amplifier is rarely (if ever) a suitable design for driving a low-impedance transducer like a normal 8-ohm speaker. The output impedance of such a design is far to high to be able to deliver any significant amount of power to the speaker. \$\endgroup\$
    – brhans
    Dec 22, 2021 at 22:38
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    \$\begingroup\$ @jwdomain You need a "push pull" or 2-quadrant driver. You could go with class-A or class-AB or do like many ICs do and go class-D, class-E, class-F... (whatever the current labeling of these things are now.) (You can find, technically, some ugly designs on youtube. And these will reproduce sound. See Single BJT, 1 W and Adventures in a one transistor audio amplifier. But they are very very inefficient. And there is no possible way I'd help you consider one of them.) \$\endgroup\$
    – jonk
    Dec 22, 2021 at 22:38

2 Answers 2

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What you want is an audio power amplifier, but what you have built is more like a preamp. Let's look at the current path for the positive peaks of a 400 Hz audio waveform. It starts at +9 V and ends at GND. The path is through R3, C4, and the speaker.

R3 = 1500 ohms

At 400 Hz, C4 has an impedance of almost exactly 400 ohms

The speaker is 8 ohms

This is a voltage divider, with R3 and C4 as the series leg and the speaker as the shunt leg. If Vin is 9 V, then

Vout = Vin x 8 / (1500 + 400 + 8) = 9 x 8 / 1908 = 72 / 1908 = 0.0042 V.

That is an output voltage of 4.2 millivolts. That is why the volume is so low.

There are other things going on, such as phase shift caused by C4, but this is a rough guess at how much signal is lost when the load impedance is way lower than the source impedance.

There is such a thing as a 1-transistor power amplifier that can drive a speaker, called a class A amp. In fact that approach is the favorite of a small community of audiophiles. If you really want one, then R3 must be much lower and M1 will need a significant heatsink. A search for

class a amplifier circuit

or

class a amplifier schematic

will bring up tons of schematics and project pages.

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You aren't providing any current to the speaker with this design. The current has to go through a 1k5 resistor. The most current that can pass through that resistor at 9.8V is about 6.5 milliamps and you'll actually be getting less than that. You need a low impedance output stage.

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  • \$\begingroup\$ How can i achieve a low impedance output stage ? I actually tried connecting resistors lower than 1.5k but results didn't change much either. \$\endgroup\$
    – jwdomain
    Dec 22, 2021 at 22:31
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    \$\begingroup\$ @jwdomain Actually, there are a number of problems. And even if you managed to solve Ian's thoughts (which are incomplete), you'd still have a problem. You need a driver that can actively both sink and source through a low impedance transducer. And that means 2-quadrants. Resistors won't cut it either on the high or low side. And that's all you have. \$\endgroup\$
    – jonk
    Dec 22, 2021 at 22:35
  • \$\begingroup\$ If you use a push-pull output stage then your very low power supply voltage of only 9.8V will limit the output power in an 8 ohm speaker to only 0.6W like a little LM386 amplifier IC. If you use a modern bridged amplifier IC with a 12V supply then the output in an 8 ohm speaker will be 8W. \$\endgroup\$
    – Audioguru
    Dec 23, 2021 at 1:01

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