These are the answers to the EQ questions that appeared in Circuit Cellar 258(January 2012).
Answer 1—Q1 and Q2 are wired as a differential amplifier, but in this digital application, it’s probably better to think of them as a current switch. Whenever VIN < VTH, the current through R1 is shunted to ground, and whenever VIN > VTH, the current passes instead through Q2 to the base of Q3, turning it on. As long as Q1 and Q2 are reasonably well matched, VTH is the input voltage at which the switchover occurs, and this is relatively stable with respect to temperature. For example, if VIN is being driven by standard TTL logic, you might set VTH to 1.5 V.
Answer 2—R1, combined with VTH, sets the amount of current flowing through Q2 to drive the base of Q3. This current should be sufficient to drive Q3 into saturation, given the expected load on it (R3 plus the external circuit). R2 serves to make sure that Q3 isn’t turned on by any leakage current through Q2 when it’s supposed to be off. For example, a value of 100 kΩ would bypass currents of up to 6 µA or so.
Answer 3—When VIN is high, there is virtually no current flowing through the input terminal—just the leakage current through Q1’s base-collector junction. When VIN is low, the driving circuit must sink the current set by R1 divided by the beta (current gain) of Q1. For example, if R1 is 4,300 Ω, giving a current through Q1 of about 1 mA, and the beta of Q1 is 50, then the driving circuit must sink about 20 µA.
Answer 4—Better: The component count is reduced by one transistor. Better: VTH is stabilized by the forward drop of a diode, making it less dependent on the exact value of the positive supply. Worse: The switching voltage is now determined by the combination of D1 and the base-emitter drop of Q2, both of which vary with temperature. Worse: Now the driving circuit must supply all of the base current for Q3 when it is in the high state.
Contributed by David Tweed (eq at circuitcellar.com)