Why does signal output come from collector and emitter goes to ground?

MBFX

Well-known member
Hi everyone! I have graduated from putting together purchased PCBs to assembling my own circuits on breadboards, and I am trying to learn how signal flow works.

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Above is the schematic for a booster I found online just now. I also built an LPB-1 and a Fuzz Face last night. I noticed that all three effects have the signal output flowing from the collector of the transistors, instead of from the emitter like I would expect. Why is that? It seems wrong - the signal should come from the emitter, right? That's the way the arrow points! I realize I am wrong, I just don't understand why.

If someone is feeling particularly kind, I have a few more questions too:

The AC signal from the pickups goes from IN to MPC POT to ground via lugs 1 and 3 of the pot. Lug 2 of the pot is the wiper, and that goes to a 120K resistor. Why is that - doesn't this make the already tiny input signal even smaller?

From the 120K resistor, we connect to a capacitor. I can't really read the value, .1uf maybe? Kinda looks like a 2, but that would be a weird value. Anyway, is this input capacitor used for keeping 9VDC power from flowing back to the input?

The circuit branches off here, with a lead to the base of the transistor, and a lead to a 2.2M resistor. The input goes to the Base to be amplified and sent out of the Emitter, is that correct? If so, that links to my initial question.

The 2.2M resistor connects to a 4.7K resistor and the collector of Q1, both of which connect to +9VDC. So, the collector gets a lot more of the +9VDC current than the base, which is how a transistor is supposed to amplify out of the emitter. Is that correct?

The emitter goes straight to ground through a 470R resistor. By now, kind reader, you've already told me why. Thank you!

The output capacitor, .1uf, is also for DC-blocking. Is that correct?
 
  • The input and output caps pass AC signal but block DC from getting out of the circuit back to the guitar or out to the amplifier.
  • The 120k resistor provides some signal attenuation before amplification. That's purely a a design choice here.
  • The signal comes out of the collector because this transistor is being used as an amplifier, which is its primary function (phase inverted). Transistors can also be used as simple signal buffers (phase neutral), in which case the input signal passes to the emitter. The collector does not require any special bias voltage in that case.
  • The 2M2 provides DC current to the base to bias the input of the transistor for maximum output. You need to pull up the weaker guitar output signal to a high enough DC voltage to achieve maximum clean amplification from the transistor without unwanted gating or distortion. A high value is chosen in order to create a high input impedance to the transistor, as well. Necessary for the first gain stage in most guitar pedals (though not always).
  • The 470R resistor on the emitter is for gain attenuation. IOW, a higher value limits the transistor amplification. However, if you put a DC bypass cap in parallel with it, it becomes a gain multiplier (IOW, it squares up the signal into distortion, which we like).
If you really want to get into discrete semiconductors and all their various properties and intricacies then look up transistors as amplifiers on youTube. They get much more detailed. The above is put more in terms of understanding guitar pedals circuits rather than being a true technical description.

This is a great channel: https://www.youtube.com/@Lantertronics
 
It's also worth noting that signal can come out of the emitter. What you've posted is a "common emitter" setup, which is used as an inverting gainstage. The same transistor can be set up as an "emitter follower" which is non-inverting but typically serves as a buffer as the gain level is generally right around 1.
 
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