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The following transistor circuit appears in the schematic for the Apple II disk controller card.

Transistor subcircuit

I don't understand the purpose of this subcircuit. I think it must have something to do with increasing the current into the VCC input of the IC at the bottom of the figure, but I'm not sure. Also, why are the two transistors and resistors connected the way they are? Why are 2 transistors used instead of just one?

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    \$\begingroup\$ Google "Darlington transistors" \$\endgroup\$
    – John D
    May 5, 2018 at 17:20
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    \$\begingroup\$ I thought it might be a Darlington transistor, but I've never seen them with resistors configured like they are in this circuit. \$\endgroup\$
    – Davis
    May 5, 2018 at 17:22
  • \$\begingroup\$ Since the emitter of the upper transistor is right on the 5V rail, and the collector of the lower is on ground, you need the resistors for current limiting and to avoid excessive Vbe on Q2. So it's not necessarily a conventional Darlington, Q1 provides a level shift for the input signal as well to reference it to ground. Today we would probably just use a P-channel FET. \$\endgroup\$
    – John D
    May 5, 2018 at 17:28
  • \$\begingroup\$ Its a power-switch. With low-value resistors to ensure low Von despite transistor parameter variations. \$\endgroup\$
    – analogsystemsrf
    May 6, 2018 at 2:47

1 Answer 1

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When "I/O select" is low, Q1 drives Q2 'on'.

With I/O select at 0.4V, the emitter of Q1 will be at 1.1V and the base of Q2 at about 4.3V so the current through Q1 will be about 3.2V/150\$\Omega\$ = 21mA. About 10mA goes through R2 which leaves about 11mA for the base of Q2.

Q2 will drive about 100mA to 500mA depending the acceptable voltage drop.

R2 prevents leakage in Q1 from turning Q2 partially on, and greatly speeds the turn-off of Q2 when I/O select goes high. R2 determines the base current of Q2.

Two transistors are used rather than one to get more gain, so the loading on the input is lower- it will be in the 100's of uA rather than tens of mA.

I looked up and found the chip- which appears to have been an ancient 256x8 bipolar OTP (One Time Programmable) PROM (Programmable Read-Only Memory).. and they are interrupting the Vcc (an unconventional strategy from an unconventional Engineer- Woz) . It would have had a voltage specification and maximum Vcc current spec. A typical similar part is the DM74S471, which had a maximum Vcc current of 150mA, well within the range I mentioned- and commensurate with the limited working voltage range of the chip- it is not guaranteed to work if the voltage drop of Q2 is too large.

Edit: To follow up on question by @HerrderElektronik, below is the simulated switching performance (voltage across a 50 ohm load) with R2 = 68 ohms (pink) and with R2 = 68K (green). Transistor types 2N4403 and 2SAR533, similar to original circuit.

enter image description here

As you can see, R2 greatly reduces the turn-off time of Q2.

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  • \$\begingroup\$ Thank you for the answer. How did the engineer decide how much current gain was necessary? Was it based somehow on the IC's requirements? \$\endgroup\$
    – Davis
    May 5, 2018 at 17:39
  • \$\begingroup\$ Yes, see the datasheet linked above of a similar part to that used in the Apple II design. \$\endgroup\$
    – Spehro Pefhany
    May 5, 2018 at 17:48
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    \$\begingroup\$ Note to the OP to add to your comment that \$R_2\$ determines the base current into \$Q_2\$... When active, there is a roughly fixed voltage across \$R_1\$ (\$V_\text{CC}\$, less two [different] \$V_\text{BE}\$'s) leading to a fixed current through it. \$R_2\$ will also have a fixed voltage across it (the \$V_\text{BE}\$ of \$Q_2\$), so it's value determines how much of the current in \$R_1\$ is siphoned away before getting to the base of \$Q_2\$. \$\endgroup\$
    – jonk
    May 5, 2018 at 18:09
  • \$\begingroup\$ @SpehroPefhany "R2 prevents leakage in Q1 from turning Q2 partially on, and greatly speeds the turn-off of Q2 when I/O select goes high" ..... Could you please explain a bit more how R2 prevents leakage in Q1 ? and is it that Miller capacitor sees less impedance with presence of R2 speeding up Q2 turn off ? \$\endgroup\$
    – HerrderElektronik
    May 7, 2018 at 11:30
  • \$\begingroup\$ The collector leakage from Q1 is uA (at most) and won't result in enough voltage across R2 to result in significant current (which could be amplified by the gain of Q2). Yes, Miller + B-E capacitance + capacitance of Q1 + stray. \$\endgroup\$
    – Spehro Pefhany
    May 7, 2018 at 11:53

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