Understanding the XR VCO

This page takes a look at parts of Thomas Henry's XR VCO in an attempt to understand as much as we can from it's various subcircuits. The module is designed around the XR-2206 Function Generator chip.

The schematics for it can be found here:

1 Volt / Octave Control Signal

The first sub-circuit of interest is the input area on schematic page 1, where the pitch control voltages are brought in. A linear signal is brought in by the 1V/OCT input and then put through an exponential conversion in the matched LM394 transistor pair.

The simplified circuit without J2, Coarse and Fine adjustments, is shown at right. I've also omitted the HF trim pot R42, R23 and diode D1. These are present to improve the range of the VCO but are otherwise unimportant to its overall function.

The simulation run varies the input control voltage (V3) from 0 volts up to 7.75 volts. The graph (below) plots this in mauve as V(1v/oct). 1V/Octave response

Internal to IC1 (XR-2206) is a +3 Volt supply that appears on pins 7 and 8. By drawing current from pin 7, the VCO frequency is increased.

The absolute maximum current that can be safely drawn is 3 mA. In this simulation graph it comes close to reaching that at a peak of just under 2.2 mA, when the input control voltage reaches 7.75 volts.

Note the yellow plot shows the current flowing through Q2 out of pin 7 of the XR-2206. It rises exponentially with the input control voltage. Q2's emitter is connected to the output of the opamp IC2b, which acts as a virtual ground.

Exponential Conversion

So where does the exponential conversion come from?

The opamp IC2b is using Q1 in it's feedback loop. The transistor is not a linear device, so the feedback loop will have an exponential change in response to Q1's base voltage changes. See the article Common Emitter Linearity if you need to review the transistor's exponential response.

Assume initially that the input control voltage is constant at 0 volts. This holds the base of Q1 at a steady voltage level, at nearly ground potential.

The opamp IC2b is going to struggle with it's output to make the opamp's (-) and (+) inputs to arrive at the same voltage. Initially the voltage divider R43 and R29 (and R35), is going to put the voltage of the (-) at about 1 volt. So the output of the opamp IC2b is going to go negative, with current flowing through Q1 to subtract voltage away from that (-) point until it becomes 0 volts, like it's grounded (+) input. At that point, the opamp reaches a steady state.

The Input Opamp

Before we get too far into the internals, let's now examine the gain of the Input opamp IC2a. The inverting gain of it is:

G = \frac { R7 } { R26 } = \frac { 2k } { 100k } = 0.02

The gain has been scaled down from a range of about 0 to 8 volts down to the range that can be handled by the base of Q1. So a large but linear gain reduction (of 50) has been done in Q1's base circuit. So aside from acting as an input buffer, IC2a also scales the control voltage to something that Q1 can handle.

Now consider what happens when Q1's base voltage changes with the input pitch control voltage. As the base of Q1's voltage changes, the Q1's emitter is going to change with it. The emitter will maintain a voltage of about 0.6 volts lower than that of the base (the Vbe difference for silicon). So as the base changes linearly, we find that the emitter does also.

As a result of Q1's emitter voltage changing, the opamp IC2a will be doing some adjustments as a side-effect. But as far as Q1 and Q2 are concerned, there is no further reason to care about the opamp here.


As the emitter voltage of Q1 changes, so does Q2's of course because they are tied together. So a linear change at Q2's emitter with it's base held constant at ground potential, is going to cause an exponential change in current flow at Q2's collector.

Linear FM

In the preceding section, you might have wondered why all the bother with the transistor Q1 if it really didn't contribute to the exponential response (Q2 did, but not Q1). Changing the pitch control voltage upset the balance in IC2b, and the opamp corrected it's output for that but otherwise did little else to contribute. The critical success factor was that the pitch input CV linearly changes the emitter voltage for Q1 and Q2.

Logarithmic opamp output and linear Q2 current Q1's big role occurs when you modify the voltage at the Linear FM input. The Linear FM input will add to or subtract from the opamp's input summing point. This forces it's output to try to cancel that change.

Because the voltage at the base of Q1 is held steady (no pitch changes) while the Linear FM input is changing, we can see that any change at the emitter of Q1 will result an exponential change in it's current flow feeding back to the (-) input. As that (-) input changes in a linear way, the opamp needs to affect a linear cancellation.

But this circuit has an exponential function in it's feedback loop. So to create a linear response through that feedback loop, requires that output to follow a logarithmic function.

How does a logarithmic change at the output of the opamp affect Q2? If you change Q2's emitter voltage in a logarithmic way, the change get's amplified in an exponential way such that the result in Q2 is a linear one! The logarithm and exponentiation cancel each other out to make the change linear.

Obviously this is a complicated way to affect a linear change if this were the only thing being done. But the design is this way to allow both an exponential input control (CV) and a linear control (Linear FM) with a minimum of parts.

In the plot at right, V(n010) is the voltage at the emitter of Q1 and Q2, as the opamp attempts to correct the voltage appearing at it's (-) input. Not shown is the the Linear FM signal increasing from 0 volts to about 7 volts from the start to end. The V(n010) plot is an inverted logarithmic response to the Linear FM input signal.

The blue trace Ic(Q2) is a straight line (unlike when the pitch CV is changed). This is due to the logarithmic emitter voltage change being canceled by Q2's exponential response.

Matched Q1 & Q2

What? You don't have an LM394 matched pair? Depending upon your requirements, it may not be so critical to have them exactly matched. Mind you, this statement is based upon LTspice simulation runs, not actual measurements with real parts.

Based upon the LTspice experiments with mismatched Q1 and Q2 (using different transistors), the following simulation observations were made with mismatched Q1 and Q2:

  • Mismatch did not seem to affect the Linear FM operation (it was still linear, though the current in Q2 may have differed slightly).
  • Exponential response was still exponential, though a different current level was reached in Q2.
  • Beta differences will also result in different currents in Q2.

Any time a different current in Q2 is reached (from the expected) means that you have a pitch change side-effect in the XR-2206. So all of these “changes” will affect the oscillator from a pitch perspective.

Large changes in Beta in Q1 vs Q2, are likely to upset the spacing between octaves due to a shift in the exponential current response. To counteract that you might need to change the gain of IC2a or simply get a better Beta match between Q1 and Q2.

Note that even the LM394 pair can vary Beta between 150 (1 uA) to 700 (1 mA) depending upon the part you're using. The pair might be super balanced, but the Beta can still vary.

Since the Linear FM is not tied to absolute values like the 1V/OCT input, you can probably live with any change that a different Beta or mismatch may bring about.

If you were to substitute ordinary transistors in place of the LM394, choose parts that can handle a collector current of 2 uA to 3 mA, with a Beta near 300 or more. Measure the Beta and use closely matched pairs.

synth_xrvco_inside.txt · Last modified: 2012/01/16 02:46 by ve3wwg
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