PNP circuit - The Art of Electronics

Question

I was reading the BJT section of The Art of Electronics on page 77 Figure 2.10.B.

Shouldn't the divider voltage sit at 14.4 V rather than 11.6 V? the opposite what the author mentioned. Since if 𝑄₃ is in saturation, the base voltage of 𝑄₃ should be ~0.6 V lower than the collector which is 14.4 V.

The “divider” formed by 𝑅₂ 𝑅₃ may be confusing: 𝑅₃’s job is to keep 𝑄₃ off when 𝑄₂ is off; and when 𝑄₂ pulls its collector low, most of its collector current comes from 𝑄₃’s base (because only ∼0.6 mA of the 4.4 mA collector current comes from 𝑅₃ - make sure you understand why). That is, 𝑅₃ does not have much effect on 𝑄₃’s saturation. Another way to say it is that the divider would sit at about +11.6 V (rather than +14.4 V), were it not for 𝑄₃’s base-emitter diode, which consequently gets most of 𝑄₂’s collector current. In any case, the value of 𝑅₃ is not critical and could be made larger; the tradeoff is slower turn-off of 𝑄₃, owing to capacitive effects.¹²

Figure 2.10. Switching the high side of a load returned to ground.

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^{12) But don’t make it too small: 𝑄₃ would not switch at all if 𝑅₃ were reduced to 100 Ω (why?). We were surprised to see this basic error in an instrument, the rest of which displayed circuit design of the highest sophistication.}

Answer 1

The text clearly states that the base-emitter junction of Q3 dominates the voltage across R3. If Vcc supply is +15 volts, this 'locks' the voltage at the base of Q3 at +15-.6, or about 14.4 volts.

The 0.6 volt B-E drop is an average, as bjt transistors (NPN and PNP) have a B-E drop of about .55volts to .65 volts, depending on its beta and the manufactures needs.

Notice the comment that R3 could be increased a great deal, to 10K perhaps. The only minor penalty would be a slower response due to transistor capacitance. R1, R2 and R3 could also be lower by 50%, increasing the response speed of the circuit, but at the small penalty of a higher 'ON' current.