I am going through The Art of Electronics Third Edition. There is a note 12 under Figure 2.10.

The β€œdivider” formed by 𝑅2𝑅3 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 ∼0.6β€―mA of the 4.4β€―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 +11.6β€―V (rather than +14.4β€―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
Switching the high side of a load returned to ground.

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

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

Why 𝑄3 would not switch if 𝑅3 = 100β€―Ohm?

  • 2
    \$\begingroup\$ You need sufficient voltage drop across \$R_3\$. \$15\:\text{V}\cdot \frac{100\:\Omega}{100\:\Omega+3.3\:\text{k}\Omega}=440\:\text{mV}\$. Not enough to turn on \$Q_3\$. And +1 for the question! \$\endgroup\$
    – jonk
    Aug 11, 2021 at 20:04
  • \$\begingroup\$ What Vbe is required for Q3 to switch? What voltage do you get across R3 if it’s 100 ohm? \$\endgroup\$
    – winny
    Aug 11, 2021 at 20:04
  • 2
    \$\begingroup\$ @jonk you can't put the answer in comments ;-) \$\endgroup\$
    – Mitu Raj
    Aug 11, 2021 at 20:24
  • \$\begingroup\$ @MituRaj It's not an answer. ;) It's just a hint in the right direction. \$\endgroup\$
    – jonk
    Aug 11, 2021 at 20:44
  • \$\begingroup\$ Cahir, You haven't responded . Hope things are going okay for you. Best wishes. \$\endgroup\$
    – jonk
    Aug 24, 2021 at 3:58

1 Answer 1


Okay. I'll write an answer. Sheesh!

Why Q3 would not switch if R3 = 100 Ohm?

Assuming you already agree with the writing in AofE v3 that \$Q_2\$ can act as a reasonable and controllable switch, then we can take it from there and say the remaining circuit looks like this (when \$R_3=100\:\Omega\$):


simulate this circuit – Schematic created using CircuitLab

With the \$Q_2\$ switch engaged, the two resistors form a voltage divider:

\$V_{_\text{TH}}=15\:\text{V}\cdot \frac{3.3\:\text{k}\Omega}{3.3\:\text{k}\Omega+100\:\Omega}\approx 14.56\:\text{V}\$ and \$R_{_\text{TH}}=\frac{100\:\Omega\,\cdot\, 3.3\:\text{k}\Omega}{3.3\:\text{k}\Omega+100\:\Omega}\approx 97.06\:\Omega\$

So, for analysis, it now looks like this:


simulate this circuit

With \$V_{_\text{CC}}=15\:\text{V}\$, I think you can see that \$V_{_\text{CC}}-V_{_\text{TH}}=440\:\text{mV}\$. You could get buried in the weeds, trying to figure out just how much base current would result, but I can assure you it is relatively small. (I'll show you a way to estimate it, shortly.) Not enough by a long stretch to allow \$Q_3\$ to supply adequate source current into the load.

(You can perform the above analysis using \$R_3=1\:\text{k}\Omega\$ and see where that takes you. I'm sure you'll easily see the better news from that choice.)

As a general rule, the collector current of an active-mode BJT scales by a factor of 10 for every change of about \$60\:\text{mV}\$ in its base-emitter voltage. I mentally tend to use \$660\:\text{mV}\$ as my baseline for \$1\:\text{mA}\$ of collector current. Here, the difference is \$\Delta V = 440\:\text{mV}-660\:\text{mV}=-220\:\text{mV}\$. That calculates a power of \$\frac{-220\:\text{mV}}{60\:\text{mV}}=-3\frac23\$, so I'd work out that the collector current would be about \$10^{-3\frac23}=0.02\%\$ of \$1\:\text{mA}\$, or about \$200\:\text{nA}\$. That's just not going to be useful for most loads.

There are more details to all of this.

  1. Verify that \$Q_3\$ really is in active mode in this case. Well, with only \$200\:\text{nA}\$ of collector current, any reasonable load won't drop enough voltage to notice. So \$Q_3\$'s collector will be near ground. Therefore, it is in active mode. So the analysis holds.
  2. What about the voltage drop across the Thevenin base resistance? Well, again, with what's likely to be under \$2\:\text{nA}\$ passing through \$R_{_\text{TH}}\approx 97\:\Omega\$, there's going to be no voltage drop to speak of. Hence, all of \$V_{_\text{TH}}\$ appears across \$Q_3\$'s base-emitter junction. And once again, the analysis about \$\Delta V\$ also still holds.

At \$R_3=1\:\text{k}\Omega\$:

\$V_{_\text{TH}}=15\:\text{V}\cdot \frac{3.3\:\text{k}\Omega}{3.3\:\text{k}\Omega+1\:\text{k}\Omega}\approx 11.51\:\text{V}\$ and \$R_{_\text{TH}}=\frac{100\:\Omega\,\cdot\, 3.3\:\text{k}\Omega}{3.3\:\text{k}\Omega+100\:\Omega}\approx 767\:\Omega\$

\$V_{_\text{CC}}-V_{_\text{TH}}=3.5\:\text{V}\$ and assuming \$V_{_\text{BE}}=660\:\text{mV}\$ then \$I_{_{\text{B}_3}}=\frac{3.5\:\text{V}-660\:\text{mV}}{767\:\Omega}\approx 3.7\:\text{mA}\$. That's better. If you allow the BJT to go active, it could handle 100 times that in the load. If you wanted the BJT to be a saturated switch, then maybe 10-20 times that much for the load. But at least it's something usable for some loads.


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