Quick question: is this a voltage divider?

[greigoroth]

Hi!

After nearly a year of slowly collecting parts and what not I finally started building the Beginner's Project last night - wohoo! So far so good. But in all of my purchasing over the past year I managed to purchase pretty much every pot except for a 5k pot... so I thought of changing one of my existing pots into a 5k. I found this (awesome) thread which explains the process http://www.diystompboxes.com/smfforum/index.php?topic=62920.0 but Mark H mentions two different methods of adjusting a pot's value depending on if it is a variable resistor or a voltage divider. I have been reading up on electronics as much as I can and can recognise a voltage divider if it looks like this



However, I became confused when looking at the schematic for the Beginner's Project:



Does the 5k pot form a voltage divider with the 10k resistor on the other side of the transistor or is it simply a variable resistance? Does the input to a voltage divider always have to be connected to Z1 (as shown in the first diagram?) Can you even have a voltage divider with a component in the middle?

Thanks for any help guys!

[R.G.]

However, I became confused when looking at the schematic for the Beginner's Project:

That's because you're out in the deep water for a beginner with no background instruction. It can be confusing all right.

Does the 5k pot form a voltage divider with the 10k resistor on the other side of the transistor or is it simply a variable resistance?

In this particular circuit, the 5K pot is being used as a simple resistor for DC and a variable resistor for AC signal. (I did say the water is deep here...)

The DC part is that it is acting as an emitter resistor for the transistor. That does not change with pot position. But the capacitor attached from the wiper to ground effectively "short circuits" the portion of the 5K pot it covers. To an AC signal from the transistor's emitter, only the un-bypassed part of the 5K pot exists, the remainder being bypassed by the 47uF capacitor.

There is a set of judgement calls to be made for quickly deciding what affects what in circuits like this. They are largely based on impedance. The reason that you don't try to combine the 10K and the 5K pot into a voltage divider is that the transistor's emitter is connected to the top of the 5K pot, and the transistor emitter is much, much lower impedance to AC signals than the 5K pot, so the transistor emitter dominates what happens, not the pot. This is a big topic in itself, many pages of reading to cover it all.

Here's an explanation of what is happening.
1. There is a voltage divider there - it's the 100K/47K across the power supply and ground. That makes a voltage of (47/147) or about 1/3 of the DC power supply to bias the transistor input.
2. The 10K pulls the base up to this voltage. 10K is a lower impedance than the impedance of the 100k/47k divider, but the transistor base looks like a resistor of the transistor's gain times the 5K emitter resistor at DC. That's about 100*5K = 500K or more, so the 10K is effectively in series with a big resistor and does not load down the 100k/47k divider.
3. However at AC signal frequencies, the 100k/47k divider is pulled around by the 22uF cap to the transistor emitter. The signal is in phase with the input signal at the base and only slightly smaller. This makes the 10K resistor have essentially the same AC signal on both ends, so no AC current flows in it - it looks like a very big resistor. This setup is called "bootstrapping", and it raises the input impedance of the whole circuit a lot. Bootstrapping is an advanced technique all in itself.
4. The gain of the transistor to AC is determined by the ratio of the collector resistor to the resistance seen at the emitter plus the transistor's internal "emitter resistor". So the change of effectively resistance at AC by changing the 5K pot setting changes the gain of the transistor for AC signal.

Overall: At DC, the capacitors look like open circuits, and the resistors all act to bias the transistor at one DC condition. At signal frequencies, the capacitors act like short circuits, and make for a high input impedance and variable gain with the 5K pot.

Does the input to a voltage divider always have to be connected to Z1 (as shown in the first diagram?)

Can you even have a voltage divider with a component in the middle?

[greigoroth]

Wow, thanks for the great reply R.G.!

I still find it amazing that you and the other gurus around here take the time to answer beginner questions... thanks again!

/Matt