# Switching the high side of a load - The Art of Electronics

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

The βdividerβ formed by π2π3 may be confusing: π3βs job is to keep π3 off when π2 is off; and when π2 pulls its collector low, most of its collector current comes from π3βs base (because only βΌ0.6β―mA of the 4.4β―mA collector current comes from π3 - make sure you understand why). That is, π3 does not have much effect on π3β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 π3βs base-emitter diode, which consequently gets most of π2βs collector current. In any case, the value of π3 is not critical and could be made larger; the tradeoff is slower turn-off of π3, owing to capacitive effects.12

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

12) But donβt make it too small: π3 would not switch at all if π3 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 π3 would not switch if π3 = 100β―Ohm?

• 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!
– jonk
Aug 11, 2021 at 20:04
• What Vbe is required for Q3 to switch? What voltage do you get across R3 if itβs 100 ohm? Aug 11, 2021 at 20:04
• @jonk you can't put the answer in comments ;-) Aug 11, 2021 at 20:24
• @MituRaj It's not an answer. ;) It's just a hint in the right direction.
– jonk
Aug 11, 2021 at 20:44
• Cahir, You haven't responded . Hope things are going okay for you. Best wishes.
– jonk
Aug 24, 2021 at 3:58

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.