The transistor in the vocoder has its base tied to ground, collector connected to the control pin of the LM13600 and the emitter to the output of an opamp (and a trim pot) through a resistor.
Unlike other converters I've seen, it does not have the transistor inside the feedback loop of the op amp.
I suspect that the converter does not rely on the \(\beta\) or Hfe of the transistor, it only has to be high enough. I believe that what is important is the relationship between \(V_b\), \(V_e\) and the output voltage of the op amp.
I've chosen the VCA in the internal excitation circuit as my reference when analysing the circuit.
R34 is a 10k resistor, you can see these in most all designs using the LM13600 OTA. It's most likely there to protect the lm13600 from self destructing - the maximum control current of an LM13600 is 2mA. See my post on the Xonik VCA for an explanation of its value.
So how do we calculate the collector current (which is what controls the OTA, Iabc)?
Here is a page that explains how a transistor may be used as a constant current source:
http://www.radio-electronics.com/info/circuits/transistor/active-constant-current-source.php
It says that
\(I_{load} = \frac{\beta \cdot V_e}{(\beta + 1) \cdot R_e}\)
If \(\beta\), the transistor gain, is large, \(\frac{\beta}{\beta + 1}\) is approximately 1 (and thus \(I_{load} = I_c = I_e\)). Also, \(V_e\) is always one diode drop below \(V_b\) when the transistor is on (\(V_e = V_b - 0.6V\)), which means that the formula above may be simplified to
\(I_{load} = \frac{V_b - 0.6V}{Re}\)
In our case, \(I_c = I_{load}\) and \(R_e = R32\) (=22k).
What the page above fails to mention is that the bottom of \(R_e\) must be connected to ground, or at very least that \(V_e\) is the voltage across \(R_e\).
This final detail is of importance to us, because in our case the voltage at the op amp end of \(R_e\) (R32) is what varies, not the voltage at the base. The base is stuck at 0V/GND. Thus, the voltage across the resistor is not \(V_e\), it is \(V_{out} - V_e\), where \(V_{out}\) is the output voltage of the opamp (IC8b).
We also have to take into account that the vocoder uses a PNP transistor, not an NPN as in the constant current source above. while the emitter of an NPN transistor is 0.6V below the base, the emitter of a PNP transistor is 0.6V above the base. Thus, for a PNP transistor:
\(V_e = V_b + 0.6V\)
Now we can calculate the emitter current = current through \(R_e\), which will also be approximately the collector current if \(\beta\) is large:
\(I_e = \frac{V_{out} - V_e}{R_e} = \frac{V_{out} - V_b -0.6V}{R_e} = \frac{V_{out} - 0V -0.6V}{22k}\)
\(I_e = \frac{V_{out} - 0.6V}{22k}\)
\(I_e = \frac{V_{out} - 0.6V}{22k}\)
What does this mean
Well, first of all, Ie is independent of the transistor \(\beta\), which means that we can replace the BC212L transistor with something else without having to look too hard for a perfect match.Second, we can calculate the maximum current through the base. Knowing that
\(I_e = (\beta + 1) \cdot I_b\)
we get that
\(I_b = \frac{\frac{V_{out} - 0.6V}{22k}}{\beta + 1}\)
The vocoder uses a +/-12V power supply. Thus, the maximum output value of the opamp is +/-12V (in practice it will be a bit lower than this, and the vocoder may even be designed to use an even lower control voltage, but the absolute theoretical maximum is +/-12V).
The transistor only conducts if the emitter is more positive than the base. Thus, only \(V_{out}\) values > 0.6V will turn on the transistor. As the maximum \(V_{out}\) is 12V, the maximum \(I_b\) can be found as
\(I_e = \frac{12V - 0.6V}{22k} = \frac{11.4V}{22k} = 0.52mA\)
\(I_b = \frac{0.52mA}{\beta + 1}\)
Thus, when selecting the transistor to use, we need to make sure that it has a maximum \(I_b\) higher than this when entering its \(\beta\) in the formula above.
Also, we can see that \(I_e = I_c = 0.52mA\) is well within the 2mA maximum for the LM13600.
Trimming
As for the trimmer potentiometer PR1 and its associated resistor R33, I assume they add a constant current that trims the 0-point of the VCA.The potentiometer acts as a voltage divider, so the maximum current through R33 will be:
positive:
\(\frac{12V - 0.6V}{470k} = 0.024mA\)
negative:
\(\frac{-12V -0.6V}{470k} = -0.027mA\)
A final note
I suspect that the diode D5 in the op amp feedback loop is there for this reason:When the opamp positive input is at 0V, the opamp output must be at 0.6V (one diode drop above to keep the negative output at 0V. This means that the opamp output always stays 0.6V above its positive input, which cancels out the -0.6V from \(V_e\) in the calculations above.
The voltage to current conversion formula will then be
\(I_e = \frac{V_{out} - 0.6V}{R_e}\)
\(I_e = \frac{V_{in} + 0.6V - 0.6V}{R_e}\)
\(I_e = \frac{V_{in}}{R_e}\)
where \(V_{in}\) is the input voltage at the positive opamp input terminal, i.e. the control voltage.\(I_e = \frac{V_{in} + 0.6V - 0.6V}{R_e}\)
\(I_e = \frac{V_{in}}{R_e}\)
I tried breadboarding the circuit to confirm my suspicions.
R31, D5 and IC8b form what is called a simple precision rectifier, see this wikipedia post: https://en.wikipedia.org/wiki/Precision_rectifier. As we tap the output at the opamp output instead of at the negative input, we should see the 0.6V offset.
My measurements clearly showed that this is indeed true. The positive input and the output of IC8b follow each other closely, with a difference of 0.6V.
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| Input at the bottom, output on top. The difference is almost exactly 0.6V |
Also, to confirm that this is indeed the precision rectifier from the wikipedia article, once the input goes below 0V, the output immediately drops to the negative rail - no surprises there. As the input is connected to the previous rectifier however, this doesn't really matter, the CV will never be negative.
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| The output saturates to the negative rail once the input is less than 0V |
As for R31, removing it makes the circuit act like a normal buffer. Increasing the value from 3.9k to 12k has no effect at all. Replacing the diode with a wire also changes the circuit back to a buffer.
Some transistor rules:
The load should always be on the collector side
\(V_b = V_e + 0.6V\) for NPN
\(V_b = V_e -0.6V\) for PNP
The \(V_b\) to \(V_e\) relationship stays constant, the currents are what change.
If \(\beta\) is large, \(V_c = V_e\). Good designs do not rely on \(\beta\) as is varies widely.
