MOSFET Amp design

[tpaairman]

This is from my thread about op-amps that morphed into the MOSFET push-pull discussion. I figured I start a new thread that's on topic of the title.

I had to back burner the project for a bit. My computer bought the farm, though being 8 years old, it was time. But that had me occupied. Also, my plan is to build the cabinet in my Dad's workshop as he has about every type of wood working power tool Sears ever sold. Due to his age, he hasn't been able to work in the shop for several years, and it turns out most of the items in the shop need fixing up. Some have power cords that are dry rotted to where you can literally see the copper. I'm glad they were not left plugged in.

Anyway, as I'm making my back to this, what I have built on the breadboard is actually working quite well. It does need some dialing in and a bit of tweaking, but I don't think I'm that far from done.

One question I have on biasing. It was suggested to use a Vbe multiplier in place of the diodes, which would be resistors with a transistor. This might be a dumb question, but why could I not just use another resistor on it's own, possibly a variable resistor? As it sits now, the two resistors I have with the two diodes all form a voltage divider with the diodes providing a difference between the sources, which a resistor in their place should do the same.

Another question from observation - I'm currently powering this from my dual voltage power supply at 10V +/- with an 8Ω speaker. And it's actually a pretty good volume level at the speaker. So I connected a 4Ω speaker that I have, and the volume seems lower. But shouldn't that mean roughly twice the power and hence a louder volume?

[J M Fahey]

Anyway, as I'm making my back to this, what I have built on the breadboard is actually working quite well. It does need some dialing in and a bit of tweaking, but I don't think I'm that far from done.

Coolo , please post schematic of what you have done.

One question I have on biasing. It was suggested to use a Vbe multiplier in place of the diodes, which would be resistors with a transistor. This might be a dumb question, but why could I not just use another resistor on it's own, possibly a variable resistor? As it sits now, the two resistors I have with the two diodes all form a voltage divider with the diodes providing a difference between the sources, which a resistor in their place should do the same.

You can bias with resistors alone, even better fine tune with a cariable resistor.
Problem is that proper bias varies all the time with temperature, and temperature varies all the time depending on playing.
So either you hire a midget Tech to live inside the amplifier, all the time measuring temperature and constantly adjusting bias or you design some self adjusting circuit.
A Vbe multiplier transistor glued or clamped to power transistor heat sink will self adjust if properly designed.

Another question from observation - I'm currently powering this from my dual voltage power supply at 10V +/- with an 8Ω speaker. And it's actually a pretty good volume level at the speaker. So I connected a 4Ω speaker that I have, and the volume seems lower. But shouldn't that mean roughly twice the power and hence a louder volume?

1) simplest explanation:your 4 ohm speaker has less sensitivity/efficiency than the 8 ohm one, so it pulls out more power from the amplifier but sounds weaker.

2) your MosFets can not supply enough current, so the lower impedance actually means less power.
Most know but forget to consider that Power=V*V/Z but it's also I*I*Z .
So keeping I at a certain level will mean lower Z means lower power.

Tubes in principle are strongly current limited, because of their working principle, only "so many" electrons can boil off the cathode and that with great difficulty, we need hundreds of Volts to pull them, so in tube amps lowering load impedance does usually NOT increase power.

3) your supply can not supply enough current, and it limits or, worse, "folds back" under load.

Like the wise skiing Sumo fighter often says:"the amplifier is just that thingie between power supply and speaker" .... with which I strongly agree.

[R.G.]

To add just a bit to JMs accurate observations:

The MOSFETs are acting as a complementary source follower. To see what this involves, imagine that the "bottom" P-channel MOSFET is not there, and that the load "resistor" is connected from the source of the MOSFET to the negative supply. The opamp then moves the gate around, and the source follows the gate.

Well, it follows it for AC signals. It follows the DC level, but the source must be lower than the gate by the MOSFET's threshold voltage (Vth) plus an enhancement voltage of the output current divided by the transconductance of the MOSFET, generally about one volt per amp. In this simple follower, the opamp will sense the output voltage on the source and resistor load, and adjust its own output voltage up by enough voltage to make the MOSFET produce enough current. That turns out to be Vth plus about one volt per amp.

This is much like the idea that bipolar transistors don't turn on (not very much at least) until their base-emitter voltage gets higher than 0.5 to 0.6V.

The single MOSFET follower is horribly inefficient, so we add a second follower, a P-channel follower, to pull the output "resistor" down, and now we can hook the load resistor to the midpoint of the power supply. The N-channel MOSFET pulls the output load up, the P-channel pulls it down. But now both N and P followers have to have a drive voltage that's higher (for the N-channel) and lower (for the P-channel) by their own Vth (which is slightly different for every device, and varies with temperature). So the N-device needs its gate pulled above the output voltage (which is attached to its source); the P-device needs its gate pulled below its source by its own Vth. You want the gates to wobble up and down together, but to be offset by the sum of the N-type Vth plus the P-type Vth.

People have used resistors for this. It works fine as long as temperature change doesn't wobble the Vths of the output MOSFETs. But this is a POWER amp, and power implies heat, so temperature is not constant. You can use strings of diodes and such to make a temperature sensitive DC voltage, but it is simpler to use ONE bipolar transistor set up to produce a voltage across its collector and emitter that is both enough voltage to make the positive and negative Vth biases, plus a little bit to defeat crossover distortion. The temperature of the bipolar naturally modifies its Vbe, and so a Vbe multiplier naturally produces temperature-changing voltage that tracks temperature.

As JM points out, you don't have much voltage. Vth is generally 3V to 7V, depending on the MOSFET. Let's assume for the purposes of understanding that the N Vth is 5V and the P type Vth is 6V (i.e. the gate has to be 6V more negative than the P-source to get it to conduct at all). So the bias voltage for the two devices is then 11V to get them to both just barely conduct and have the output sit in the middle of the +/- 10V power supply. To get the N-channel to pull the output voltage to +10V, you have to pull its gate up to +15V. To get the P-channel to pull its source down to -10V, you have to pull the P-gate to -16V. And the opamp has to have enough power supply voltage to do this pulling.

Bipolar transistor output stages have this same problem, but their cut-in voltage, the base emitter voltage needed to start conduction, is only about 0.5 to 0.6V, so it's a much smaller problem.

Ask questions.

[tpaairman]

One other thing that has come to mind. Per the original circuit, I have the op-amp to take care of crossover distortion, but since the diodes are supposed to help take care of that, and presumably the Vbe multiplier, do I even need the op-amp?

[R.G.]

In my estimation, the original circuit may have been somewhat mis-described. The opamp doesn't take care of the crossover distortion, it can only reduce it somewhat, and how much it can reduce it depends on the details of the crossover distortion. The opamp will do pretty well reducing crossover at bass frequencies, but its effective gain gets smaller with frequency, so the higher frequency components of the crossover distortion will get through to some extent.

How much of the remaining crossover distortion you can withstand and hence whether you can do without the opamp or not depends on what your objectives are with the amp, and what signals you expect to drive it.

Presumably you want some amount of audio power out. Speakers are rated in sound pressure level above the average auditory threshold with a drive of 1W and a distance of 1m. Common values for speakers are 80-120db spl for 1W of drive depending very much on the speaker. Guitar speakers are often 94-110 db in this setup. So you can get pretty loud with 1W. One watt into 8 ohms is calculated as:

P = Vsquared/R, where V is the AC rms value. So for 1W and 8 ohms, 2.828Vrms, or about 4V peak.

So no biggie, right? You have 10v to swing (at least if you can arrange the opamp to drive the gates to 5V above and below the 10V supplies).

Wrong. Music has a crest factor, the ratio of loudest signal to average signal. The crest factor for quite a lot of un-compressed music is about 20db. So for an average music level of 1W, you need to process peaks of 10W to keep the peaks from being distorted in a big (and objectionable) way. So the 1W which is easy gets to being 10W if you want to use it for actual music. And that needs a swing of 8.9Vrms, 12.6Vguipeak, and your +/-10V supply won't be enough voltage, even if it can supply peaks of 1.25A of current.

The guitar world recognizes amps of about 10-20W as being OK-ish for bedroom practice stuff only. They're not usable in band situations.

A whole lot of this story is explained at more length and depth than I can reproduce in one of several books on audio power amp design. I heartily recommend books on audio power amp design by Doug Self and/or Bob Cordell. You might be able to get one of these used for a small sum, though amazon, abebooks, or Powell's. Some reading will save you a whole lot of time and parts.

[Gingertube]

Have you considered one of these chips to handle your bias (US$6.50):
https://www.analog.com/media/en/tech...ets/1166fa.pdf

Cheers,
Ian

[R.G.]

I'm guessing not, GT.

Looks like a clever chip, but we're trying to get him to go for something other than a resistor or diode, and to understand what bias is for, and how the voltage amplifier stage can't eliminate crossover. And to fill in the rest of the existing power amp design methodology.