Transistor clipping

[frets]

Hi Guys😸😸😸
Happy Boxing Day!!

I am working on designing some more guitar pedals and I am trying to get transistors to clip but the diagram is confusing me.

Here it is.





Okay, we have BC and E on a transistor.  But this makes it look like there are 4 leads.  The connection at the rear of the transistor has me confused.

Will someone clarify this for me?
As usual, thanks!😸

[newey]

These are MOSFETs. They have 4 terminals. There are a number of different ways of designating these in a schematic, so here's a tutorial:

www.allaboutcircuits.com/technical-articles/schematic-symbols-for-electronic-components-transistors/

[thetragichero]

mosfets can be used for clipping, like any other diode. will not amplify setup like that though

[MattB]

That diagram doesn't show the bit that's doing the clipping.

MOSFETs have what's called a "body diode" between drain and source. It's part of the basic structure- a diode is a silicon junction and a MOSFET is made of layers of silicon, so the diode is kind of an unavoidable side effect. Some schematic symbols show the body diode, like this:

The MOSFETs in this setup aren't really doing anything- the gate for each MOSFET can't turn on because the body diode in the other MOSFET conducts before the gate can reach a high enough voltage.

 The body diodes are what clips the signal, and they behave like a regular diode. Functionally this circuit works the same as a simple pair of diodes connected to ground.

Some people like the sound of MOSFETs used like this more than  regular silicon diodes- I've never tried it myself so I can't say.

[frets]

Thanks guys,
I have never seen one. I am going to buy some to try them out.

[Yogi B]

Dec 26, 2022 13:06:02 GMT -5 frets said:

Okay, we have BC and E on a transistor. But this makes it look like there are 4 leads. The connection at the rear of the transistor has me confused.

Dec 26, 2022 13:28:27 GMT -5 newey said:

These are MOSFETs. They have 4 terminals.

To clarify, though the devices themselves internally have four terminals (Drain, Gate, Source, Body/Bulk), but in most cases a discrete device will have the source & body connected internally, this is the connection which is shown within the right hand side of the symbol's envelope (the enclosing circle). And thus resulting in an actual part resulting in only three external terminals — note how there are only three wires which actually cross the envelope.

Without going into too much semiconductor theory: a diode is formed via a PN junction which allows current to flow only in one direction, from P(ositive) to N(egative), following the direction of the arrow that forms the standard diode symbol. Within a MOSFET there are also p-type & n-type regions and thus PN junctions. In the presented diagram you have N-channel MOSFETs, the channel being the dashed line adjacent to the gate terminal on the left (dashed in this case because this is an enhancement mode device, meaning without a voltage difference between gate & source there should be no conduction between drain & source) and we know the channel (or more correctly, with an enhancement mode device, it's end points: the source & drain terminals) is n-type — as, just like a diode, the arrow points from P to N. This therefore also means the body is p-type.

The internal connection of the source to body means we have an intrinsic "body diode" pointing from source to drain (this direction is reversed in a p-channel MOSFET), and this is the basis of the clipping in the shown arrangement.

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What about the external connection between gate & drain?

In this case, with two MOSFETs in alternating directions I don't think this matters much. What it allows is for the MOSFET to turn on and conduct in the direction opposite to the body diode, however the threshold voltage for the MOSFET will most likely be larger (e.g. 1.5V) than the conduction threshold of the (other) MOSFET's body diode (e.g. 0.6V).

This means with a single MOSFET you can get asymmetrical clipping (e.g. will clip outside the region -0.6V to 1.5V), but with a parallel pair you'll just be using the body diodes.

(Anyone else: is there a technical reason why you would actively prefer the arrangement shown rather than connecting the gate to source?)

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Dec 26, 2022 13:06:02 GMT -5 frets said:

I am working on designing some more guitar pedals and I am trying to get transistors to clip but the diagram is confusing me.

Finally another important point that's perhaps just off the sides of the given diagram are (de)coupling capacitors. Since the "diodes" are connected to ground they'll clip at ±0.6V with respect to ground, so for symmetrical clipping you'll want the signal to be biased at ground. If instead, for example, the signal was biased at half the supply voltage (4.5V, e.g. the output of an op amp or other 'clean boost' stage) Q2 would basically always conduct thereby reducing your output to a near constant 0.6V, i.e. not varying in response to the input signal.

Alternatively you could theoretically connect the 'diodes' to the relevant reference voltage instead (and only AC couple them to ground, via a cap), but I don't recall ever seeing this done in practice.

[ashcatlt]

Dec 26, 2022 20:40:03 GMT -5 Yogi B said:

Alternatively you could theoretically connect the 'diodes' to the relevant reference voltage instead (and only AC couple them to ground, via a cap), but I don't recall ever seeing this done in practice.

There are some schemes out there that do it. Can’t think of one off the top of my head, but I know I’ve seen it. It’s actually better to do it this way when you’re shooting for asymmetrical clipping because an asymmetrical square wave looks very much like a square wave with a DC bias. Coupling caps won’t let that stand, and will tend to float toward the actual center point.

[JohnH]

I'm interested in the outcome, but TBH Im not very hopeful of a great tone when used like this, effectively as a passive diode clipper. Whereas active circuits with Mosfets or jfets can sound really good!

[Yogi B]

Dec 27, 2022 12:03:05 GMT -5 ashcatlt said:

Dec 26, 2022 20:40:03 GMT -5 Yogi B said:

Alternatively you could theoretically connect the 'diodes' to the relevant reference voltage instead (and only AC couple them to ground, via a cap), but I don't recall ever seeing this done in practice.

There are some schemes out there that do it. Can’t think of one off the top of my head, but I know I’ve seen it.

Having slept on it, I've since realised there's a family of circuits that often do this, and to boot I've been looking at them pretty recently: bajaman's amp-to-pedal translations over at FSB (for a specific example see the string of three 1N4148s after IC2b in the Mars Hall 1959).

I had to go digging through schematics to find a more mainstream example: the Fulltone OCD.