Minimoogs do it discretely

How do Minimoogs do it? With discrete transistors. I've highlighted the keyboard resistor string and current source. Here are the differences between the Minimoog from the Prodigy:

• The current source uses two BJTs instead of an op-amp.
• There are 43 resistors instead of 31 because there are 44 keys instead of 32.
• The resistors are 10 ohm instead of 100 ohm. This is because we are using discrete BJTs for the current source, which will give a smaller, less ideal, current source output resistance. This is overcome by making the load smaller.
• The current needs to be 10x that of the Prodigy because of the smaller resistors.

The current shown on the schematic is 8.33 mA, which gives the 1/12 V for each key. Let's solve for this current using some quick and dirty approximations. We will assume the transistor base current is negligible and for both transistors Vbe = 0.7 V.

1. R20 and R22 form a voltage divider between 10 V and ground. Vb of Q11 is 5 V.
2. If Q11 Vbe is 0.7 V, then Q9 Vb is 4.3 V.
3. If Q9 Vbe is 0.7 V, then Q9 Ve is 5 V.
4. Q9 Ie is (10-5)/590 = 8.5 mA.
5. The current source comes from Q9 Ic, which will be almost equal to Ie = 8.5 mA.

This is pretty close to 8.33 mA. Where did I go wrong? I assumed the Vbe of the transistors was the same. First of all, they aren't match and one is NPN, the other PNP. Second of all, The current through them is not equal, so the Vbe values will be different. The current through Q11 is about 4.3/1k = 4.3 mA. We found the current through Q9 is almost twice this. The Q9 Vbe will be about 18 mV greater than Q11 Vbe based on the Ebers-Moll BJT model (26 mV*ln(2)). Q9 Ve will be about 5.018 V. Q9 Ie will be (10-5.018)/590 = 8.44 mA. Now assume Q9 has a Beta of 160. Our current source will be 8.45*159/160 = 8.39 mA. Now we're less than 1% off from the expected 8.33 mA. That's close enough considering we have 1% resistors.