converting V to I using a resistor and PNP

[dentaku]

I love the way this circuit is described. You can follow along with the schematic and see how it flows.
http://www.birthofasynth.com/Thomas_Henry/Pages/Bass_Plus.html
I pretty much get how all the parts work (I've been messing with synth circuits long enough now) but I'm not sure about the two PNPs with their bases connected to ground, Q3 and Q4.
In the description it says...
"This voltage feeds R18 and is converted to a current by Q3."
and
"the wiper of the "Sweep" control taps off a variable amount of the total voltage, and this is converted to a current by R23 and Q4"

What's the theory behind using a resistor and a PNP with the base connected to ground to do this?

[JPortici]

http://www.ti.com/lit/ds/symlink/lm13700.pdf
The output current of a LM13700 OTA is proportional to the current going inside the amp bias input. It's a current controlled current source, zero current in means zero current out. 

Q3 and Q4 are used in a common base configuration. Long story short this is also called "current buffer", meaning the current going in is about the current going out but why this was used? nobody forbids you to apply voltage and convert it to current via a resistor like so
IABC = (VIN-V_IABC) / R
but in this case you already have a current output from Q1 and the resistor node before Q4 emitter. instead of converting it to voltage then to current again a current buffer was used so the voltage between emitter and ground is totally independent of the voltage beween collector and ground. this simplifies the circuit a lot

[David Hess]

The transistor's emitter and collector currents are almost identical and limited only by the transistor's alpha which is just under 1 do to the loss of current through the base.  Alpha = beta/(1+beta) so higher current gain yields an alpha closer to 1.

With the base tied to ground, the emitter will be 1 Vbe (about 0.6 volts) lower or higher depending on whether an NPN or PNP is used so the voltage across the emitter resistor is the input voltage - Vbe.  Since the collector current is equal to the emitter current minus the base current , the current through the resistor shows up at the collector and it is insensitive to changes in collector voltage.

Vbe has a temperature coefficient of about 2mV/C which may add an undesirable error in some applications.  To fix this, a diode or another transistor can be used to offset the base from ground by the same Vbe so now the emitter is at ground potential and the temperature coefficient is compensated for.

The circuit uses this voltage to current circuit because the LM13700 OTA wants an input current to control transconductance.

[dentaku]

http://www.ti.com/lit/ds/symlink/lm13700.pdf
The output current of a LM13700 OTA is proportional to the current going inside the amp bias input. It's a current controlled current source, zero current in means zero current out. 

Q3 and Q4 are used in a common base configuration. Long story short this is also called "current buffer", meaning the current going in is about the current going out but why this was used? nobody forbids you to apply voltage and convert it to current via a resistor like so
IABC = (VIN-V_IABC) / R
but in this case you already have a current output from Q1 and the resistor node before Q4 emitter. instead of converting it to voltage then to current again a current buffer was used so the voltage between emitter and ground is totally independent of the voltage beween collector and ground. this simplifies the circuit a lot

Interesting stuff.
I'm going to have to read some more about this. Thomas explained his circuit very well but I guess getting into the specifics of current buffers was a bit beyond what that article was about.