6
\$\begingroup\$

I'm studying this transistor matching circuit provided initially by Bob Moog. I think I mostly understand how it works, the op amp, bjt not under test and resistors form a current source, which stays constant for each transistor Vbe is measured for. What I'm a bit unclear on is the purpose of the 10k between the two transistors in the NPN measuring circuit, and why it's not needed in the PNP measuring circuit.

\$\endgroup\$
  • 1
    \$\begingroup\$ Hmmm yes it does seem a bit odd - I'm thinking that they should both have one but I'm also thinking that the circuits will work equally well without them. \$\endgroup\$ – Andy aka Jan 8 '16 at 18:02
  • \$\begingroup\$ Good question, I would have to agree with @Andy aka that the 10k is not needed. \$\endgroup\$ – user1582568 Jan 8 '16 at 18:20
  • 1
    \$\begingroup\$ There is another anomaly in the circuits - why is one of the biasing 10kohms specified as 5% 1/4W when the other isn't? \$\endgroup\$ – Andy aka Jan 8 '16 at 18:26
  • \$\begingroup\$ Yeah that is a strange way of saying "all resistors are 5% 1/4 W" if that's the intent. \$\endgroup\$ – mhz Jan 9 '16 at 0:18
2
\$\begingroup\$

The purpose of the resistor is probably to deal with "stuff happens". It's only dropping 1 volt, so the current sink still has about 4 V to work with.

As you say though, it's not really needed, and can just as well be left off, as was done in the bottom schematic. This is one of those personal whims of the designer, and the fact that the same designer used it in one place but not in another shows how he was on the fence about it too.

Personally, I would just leave it off.

\$\endgroup\$
  • \$\begingroup\$ Would you mind elaborating a bit more on "still has about 4 V to work with"? I'm having a hard time seeing it. \$\endgroup\$ – mhz Jan 9 '16 at 0:20
  • \$\begingroup\$ @mhz: The current sink works by keeping 5 V across the sense resistor. The 10 kOhm resistor drops another 1 V with 100 uA thru it. The whole thing is run from a -10 V supply, so that leaves 4 V across the transistor. \$\endgroup\$ – Olin Lathrop Jan 9 '16 at 14:06
  • \$\begingroup\$ Got it. isn't it 14V though, since it's a bipolar supply (20V total dropped across Vce1 Vce2 and the two resistors? \$\endgroup\$ – mhz Jan 9 '16 at 14:53
  • \$\begingroup\$ @mhz: Actually it's a bit less than 4 V since the base of the transistor under test is grounded. The emitter would be at around -700 mV, so more like 3.6 V. \$\endgroup\$ – Olin Lathrop Jan 9 '16 at 17:06
  • \$\begingroup\$ ah right, good point, thanks for clearing that up for me! \$\endgroup\$ – mhz Jan 9 '16 at 17:39
5
\$\begingroup\$

OK I think I can shed some light. As soon as I saw the article I was reminded of the early Moog synth voltage controlled oscillators - they are "so-called" logarithmic response VCOs because, for each extra volt that you applied to the input of the VCO, the output frequency would double. They are simply described as 1V per octave VCOs.

Why is this important you may ask and the reason is that if two VCOs were used in tandem from the same control voltage but, with one VCO set at one-mucical "fifth" difference (1.5 times the frequency of the other to put it in engineering terms), you would want both to remain at this ratio as you changed the control voltage. So with a smooth change in 1V on the control line, you'd hear the fundamental and musical-fifth rise together to reach an octave higher and remain in relative tune to each other all the way up.

OK that's the back story and the important thing now is to recognize that in most analogue synth VCOs NPN transistors were used for the "so-called log" conversion of voltage to frequency.

Next is this website that confirmed the reason for the rogue 10k resistor - they show a dual NPN transistor tester circuit and this has a 10kohm inserted in series with each NPN under test. This then allows the use of a differential amplifier to measure the voltage differences and use LEDs to state "go"/"no-go": -

http://www.musicfromouterspace.com/analogsynth_new/TRANSISTORMATCHER/images/go_nogo_transistor_matcher.gif

So, my suspicion is that Moog probably had a similar tester and that the circuit in the OP's question is a derivative of it that retains a 10k in line with the NPN transistor and, of course, it loses its meaning as a single NPN tester.

\$\endgroup\$
  • \$\begingroup\$ I didn't realize that the source of the moog image would be obscured by uploading it to imgur, but I pulled it from the same page as the one this schematic is from =) \$\endgroup\$ – mhz Jan 9 '16 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.