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I'm trying to learn electronics and electrical engineering through building small practical projects (toys mostly). I've been learning a ton from adafruit and spark fun, but I'm trying to branch out a bit and learn outside of their tutorials from time to time.

I have a couple of projects that I want to add sound to and provide volume control for. I had problems achieving this on my last project (the potentiometer "worked", but the volume only changed along a small portion of the pot's range and the change was very drastic).

I did some searching around and found a really rough video on youtube where the maker built out an amplifier for his phone. I adapted the circuit diagram to work with tone signals I'm sending out from my adafruit micro controller and it works.

The fact that it functions is great, but I don't completely understand what's going on. This is the first time I've worked with a mosfet (which I know is a transistor and operates like the NPN transistors I've used previously) and a capacitor.

I get that the mosfet is allowing me to use the smaller 3v signal coming out of my controller (p1) to control the higher 9v power source effectively amplifying the signal but I'm a little hazy on some of the other details:

  • Does this circuit look correct? (it physically works, but I don't know if I drew the circuit correctly :/ )
  • What does the capacitor actually do in this circuit?
  • Why do I need a resistor across the gate and drain legs?

I definitely appreciate any insight you can give.

schematic

simulate this circuit – Schematic created using CircuitLab

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Does this circuit look correct? (it physically works, but I don't know if I drew the circuit correctly :/ )

Yup. Looks reasonable. At least, as reasonable as the circuit is, which isn't very.

What does the capacitor actually do in this circuit?

It allows the FET to ignore the DC level on the input. So, you can present it with 1 VAC around ground, or 1 VAC riding on 5 volts, or whatever. As I'll get to in a moment, this can be important.

Why do I need a resistor across the gate and drain legs?

Now, this is where it gets interesting, and not entirely in a good way.

The FET works like this: above some threshold (called Vgs(th) on the data sheet), increasing the GS (gate-source) voltage, causes the apparent resistance between DS (drain-source) to drop. In this circuit, this causes the current in the speaker and the FET to increase, which causes the drain (and hence the gate) voltage to drop. At some point, the two effects will stabilize, and this arrangement is called negative feedback.

Here's where this circuit shows its problems. Remember the threshold I mentioned? For your part it's in the range of 2 to 4 volts. Let's say 4. So, if the gate voltage is less than 4 volts the FET will be off and a full 9 volts available at the gate. This will cause the resistance of the FET to drop, and at some point (about 4 volts) the FET will be turned on enough to give 4 volts at the gate. Problem is, this means that the speaker is dropping 5 volts. With an 8 ohm speaker, this means that, with no input, the circuit will draw 5 volts/8 ohms, or about 0.6 amps. This will quickly drain any battery you use for power. Not only that, it will produce about 3 watts of heating in the speaker. Since your maximum possible music output is 0.7 (2.5 ^2/8), or about 0.54 watts, this is a grossly inefficient system.

Now, as to R2. It more or less isolates the AC part of the gate signal from the drain. Without it, the signal passing through the capacitor would effectively have to drive 8 ohms, and this would produce a very small signal at the gate, since you would have a voltage divider of 8 ohms/10,000.

ETA - Oh yes, and I forgot. Remember when I mentioned the utility of a blocking capacitor (C1)? While this circuit is not a good choice for a power amplifier, something like it is quite useful as a preamplifier to provide an input for a power amplifier which IS a good choice. Such a preamplifier circuit will (unless it also incorporates a blocking capacitor on the output), have a large DC level on its output. About 4 volts in this case. Or it may be centered around zero if a more sophisticated circuit is used. Since you never know what sort of weird circuit may be hooked up to an amplifier, putting a DC blocking capacitor on the input is a very good idea.

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  • \$\begingroup\$ Awesome. Thanks for the detailed answer. I read through it, get some of it, and will be re-reading it as I look up some of the terms and concepts I'm not familiar with yet. As far as this not being a great circuit because it will drain a battery fast, do you know of a tutorial you could point me to that would be a better option for a battery powered project? Thanks again! \$\endgroup\$ – Chris Schmitz Aug 2 '18 at 14:11
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This circuit used a diode-biased FET: the gate and the drain are biased at the same voltage; your linear region is very small, hence the "drastic change" in volume as the pot is changed. The resistor ensures the gate and drain are at the same (DC) average voltage.

The current will be high, assuming this is a power MOSFET. Any battery will not last.

The capacitor provides a flexibility in voltages at the gate versus the pot wiper; this behavior is classically named "DC blocking".

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  • \$\begingroup\$ Thanks for your answer. Between your and WhatRoughBeast's answer I know I need to go back to the drawing board to figure out a more efficient solution for my project. I definitely appreciate you taking the time to answer and give me some starting points for further researching and understanding. Thank you! \$\endgroup\$ – Chris Schmitz Aug 2 '18 at 14:12
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this is not an answer, but a comment

You can simplify the look of the circuit by eliminating the ground bus from the circuit, thereby making the signal path through the circuit clearer.

Also make it obvious that R1 is a voltage divider by making it vertical.

Untangle the speaker connections and make that part of the circuit vertical, so that it shows more clearly that M1 switches 9V through the speaker

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Ah nice! That definitely makes the diagram easier to read/understand. I've seen other diagrams drawn this way, but seeing one that I've drawn re-drawn in this style makes the advantage super clear. Thanks for posting this! \$\endgroup\$ – Chris Schmitz Aug 2 '18 at 17:09

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