Exponential converter notes

At the bottom of this page (xonik.no) is the ultimate best series of articles I found about exponential converters (parts: 1, 2, 3, 4). This is a collection of my notes so that I remember the key points about how I understood the whole topic.


The whole idea of exponential conversion is based on the Ebers-Moll equation:

$${Ic = Is \cdot e^{\frac{V_{BE}}{V_{T}}} }$$

Here ${I_{S}}$ and ${V_{T}}$ are constants, so in a certain range around 0.7V base voltage, the collector current is exponential to the voltage difference between the base and the emitter (${V_{BE}}$) of the transistor.

As ${I_{S}} is highly temperature dependent, we need two (matched) transistors in the exponential converter circuit for temperature compensation. One of them works as a reference, the other one serves as the source for the output current (${I_{Out}}$).

Any of the two transistor's base can be used as the input for the external control voltage, the other one will have to be connected to ground. If the base of the reference is used as the input then the output current will be proportional to the input voltage. If the base of the source is used as an input, then the output current will be inversely proportional to the input voltage. This can be derived from the following equations:

For PNP transistors: ${I_{Out} = I_{Ref} \cdot e^{\frac{V_{B1} - V_{B2}}{V_{T}}}}$

For NPN transistors: ${I_{Out} = I_{Ref} \cdot e^{\frac{V_{B2} - V_{B1}}{V_{T}}}}$

At the input there's a voltage divider that converts 1V change on ${V_{VC}}$ to ~17.5mV. That is ${V_{VC} = 2V}$ the voltage divider outputs 2*17.5mV and, if ${V_{CV}}$ = -4V then it outpus -4*17.5mV and so on. The input is linear. The detailed behaviors how that works is excellently explained in the 2nd article by Xonik.

The reference transistor has a constant collector current ${I_{Ref}}$ which is ensured by a current source. Usually an small op-amp circuitry is used to provide this current.

The 4th article details 4 types of exponential converters. They all can be used for different purposes, depending on what's needed in the circuit after the exponential converter:

  1. A: uses PNP transistors, the base of the transistor in the feedback loop is connected to ground. This has positive output current which is inversely proportional with ${V_{CV}}$. This is great e.g. as a bias current of an OTA where the CV signal is mixed with a unity gain inverting op amp. A great example of this circuit is Thomas Henry's VCF-1.
  2. B: uses PNP transistors, the base of the transistor in the feedback loop is connected to ${V_{CV}}$. As this version also uses PNP transistors, it also has positive output current, but the output current is proportional with ${V_{CV}}$. Again, cool for VCFs or VCAs that use OTAs but the CV mixer is not buffered, like this one here.
  3. C: uses NPN transistors, the base of the transistor in the feedback loop is connected to ground. Has negative output current. It can be used for e.g. in a VCO circuit with a transimpedance amplifier.
  4. D: uses NPN transistors, the base of the transistor in the feedback loop is connected to ${V_{CV}}$. Output current is negative.
Source: https://www.xonik.no/theory/expo_converter/expo_converter.html
  A B C D
Feedback transistor Vb 0V VCV 0V VCV
Transistor PNP PNP NPN NPN
Output current Positive Positive Negative Negative
VCV relation Inverse Proportional Inverse Proportional

The voltage difference between the base and the emitter of the reference transistor is constant. This does not necessarily mean that the base and/or the emitter voltage is constant. The difference is.

$${V_{BE} = V_{T}[ln(I_{C}) - ln(I_{S})]}$$

This is universally true no matter which transistor we talk about. If ${I_{C}}$ is constant, then ${V_{BE}}$ is constant as all other tags in the equation are constant.

If ${V_{CV}}$ is connected to the base of the reference transistor (B and D) then that transistor's emitter voltage will change exactly the same as its base voltage which is proportional to the control voltage ${V_{CV}}$. Since the emitter voltage is changing with the control voltage and the base of the source transistor is connected to ground, ${V_{BE}}$ of the source transistor will change which will eventually change the amount of output current (see Ebers-Moll equation).

When the base of the reference transistor is grounded (A and C) ${V_{E}}$ will remain constant no matter what ${V_{CV}}$ is, as that's connected to the other transistor and the collector current of the reference transistor is constant. This means that any change in the control voltage (ie. the base of the source transistor) will also change ${V_{BE}}$ of the source transistor and result in changing the output current.

Calculating maximum output current

The maximum output current depends on the collector voltage of the source (${V_{Cout}}$) and the circuitry that comes after the converter. Let's consider the same example as in the article:

  • we use a PNP transistor setup
  • the reference current is constant ${I_{Ref} = 10µA}$

The output of the source PNP collector is connected to a 18kΩ resistor which is then connected to -15V. In this case the output current can be written as ${I_{Out} = \frac{V_{Cout} - 15V}{18kΩ}}$, from which:

$${V_{Cout} = I_{Out} \cdot 18kΩ - 15V}$$

So if we increase the output current, the source transistor's collector voltage also has to increase proportionally – i.e. it will change exponentially with input voltage. The collector voltage can only increase until there's at least 0.25V difference between the emitter and the collector voltage (below that, there's no conductance).

So the maximum output current will be exactly when the difference between the emitter and the collector voltage is 0.25V. How can we calculate this?

Since ${V_{E}}$ is costant ~0.7V, the maximum output current for a PNP setup is:

$${I_{Out Max} = \frac{(0.7V - 0.25V) - (-15)V}{18kΩ} ≈ 858µA}$$

And from ${I_{Out} = I_{Ref} \cdot e^{\frac{V_{B1} - V_{B2}}{V_{T}}}}$ where ${V_{B1} = 0}$ and ${V_{B2} = V_{CV}}$:

$${V_{CV Max} = -V_{T}[ln(\frac{I_{Out Max}}{I_{Ref}})] = -112.5mV}$$

From this if we consider the 1V/oct standard, the actual control voltage where the converter sources the maximum current is ${-112.5 / 17.5 = -6.43V}$ which looks correct, considering that as we increase the control voltage the output current is decreasing in this setup.

Note: this calculation is only valid for when ${V_{E}}$ is fixed so for setup A and C. For B and D it's approximately the same.