# BJT transistors: can one see them as voltage dividers or should they be analyzed differently?

I am currently reading The Art of Electronics 3rd edition and I am struggling to understand basic transistors.

On page 77 there is an example circuit with a PNP transistor driven by an NPN transistor. The second paragraph asks a question and in parenthesis it says "make sure you understand why", which I don't!

I am struggling to understand the voltages and currents present in this circuit. Can I treat the resistors as a simple voltage divider and perform ohms law and KVL/KCL as a normal circuit or do I have to analyze the circuit differently because of the transistor?

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.

• silicon bjts need how much base-emitter voltage to turn ON? Commented Jan 5, 2017 at 16:26
• @bufo333 You can use impedance ratios for everything keeping in mind power limits and slew rate reduces with current. Normally data sheets give switch saturation for Vce(sat) which * I = Pd and times Rja ( gives temp rise. including sink if added) They standardize on saturation for Ic:Ib=10 ( regardless of hFE ) then report the Vce(sat for that current level and ratio. Try to remember this current ratio for BJT's and it can be relaxed to 50:1 for very low current or very high hFE devices >500 or in between ( my rule of thumb is 10% of hFE ) Commented Jan 5, 2017 at 17:23
• The footnote refers to R ratios on pull-up with 3k3/(100+3k3 being too high a ratio ~1 vs 3k3/(1k+3.3k) *15V drop from 15V with 0.75k =Req . But the R3 is important necessary for faster turn off. Thus Load R is derived from Ic:Ib ratio of 10:1 Commented Jan 5, 2017 at 17:30
• I really want to thank everyone for responding to this question, I was banging my head over and over this question, trying to puzzle out the currents and voltages and the cause and effects. I still feel confused but I am slowly getting it. I must have read each comment a dozen times over the last hour or so. I want to thank Tony Stewart, Olin Lathrop, petEEY, and Peter Smith. It really took me reading all of your comments for the answer to start to make sense for me. I guess if you say it enough times it eventually sinks in. Commented Jan 5, 2017 at 20:56
• Other questions about this circuit: PNP Circuit-The Art of Electronics, Why would the transistor not switch?
– CL.
Commented Jan 6, 2017 at 9:57

You seem to be asking multiple questions.

The first thing to understand is what the NPN in front of the PNP does for you. As the page correctly points out, the left circuit needs to hold the input at nearly 15 V for the transistor to be off. Understand why before proceeding.

The reason is that the B-E junction of these transistors looks like a diode to the driving circuit. This diode drops only 500-750 mV for normal currents. Therefore when current is drawn from the base of the PNP, that base is only about 700 mV below the emitter, so a bit above 14 V. As the book says, that is inconvenient when you want to control something with a 0 to 3.3 V digital logic signal.

The solution is the right circuit. The NPN has the same diode-looking characteristics, but flipped around in polarity. Now the base will be about 700 mV above the emitter, so 700 mV above ground, when the transistor is being turned on. That's a level that digital logic can do.

The two resistors R2 and R3 are to interface the easy to control NPN switch with the PNP switch you actually want to control.

R2 limits the current out the base of Q3 and into the collector of Q2. If it wasn't there, that current could get high enough to damage both transistors.

R3 isn't strictly necessary, but makes sure Q3 is really off when Q2 is off. Note that when Q2 is off, it's just not driving current thru the base of Q3. It doesn't actively hold the base of Q3 close to its emitter to make sure Q3 is really off and not randomly turning on from stray noise. That's what R3 does.

As the text says, there is wide latitude in picking the value of R3. However, it can't be so low that the current limited by R2 can't turn on Q3 anymore.

The book wants you to understand why most of the current through Q2 comes from Q3 and not from R3.

The voltage across R3 is the same as the voltage across the PNP's emitter and base. This emitter and base junction functions a bit like a diode because any voltage increase past ~0.6V increases current exponentially. ~0.6V across a 1K resistor is only ~0.6mA (the current mentioned in the book).

simulate this circuit – Schematic created using CircuitLab

Ic:Ib=10 minimum to 30:1 or 50:1 in extreme cases>500 since hFE drops rapidly when Vce < 1V.

There are several ways to quickly analyze this. From R Ratios or KVL-KCL as long as Vbe is forward biased and Vbe is included in KVL and treated as a voltage source and forward biased unlike when Rb pullup = 100 ohms in example of question.

(neglecting L1 and C1 and input frequency for this case but C has a non-zero ESR)

When Q2 is turned on, it will pull Q3s base to about 14.4V (a diode drop below its emitter).

Q2 therefore will have a collector current of 14.4V / 3.3k = 4.4mA.

The current through R3 will be 0.6V (as the resistor sees the same voltage as the base - emitter drop) / 1k = 0.6mA

As there is a total of 4.4mA in R2 is 4.4mA and the current in R3 is 0.6mA, the remaining current (3.8mA) must be flowing through the base of Q3

• You might want to add the calculation if R3 is reduced to 100R (see note 12 at bottom of page) Commented Jan 5, 2017 at 16:50
• @peter This finally clicked for me, let me see if I am understanding this correctly. The voltage source is 15v, the voltage through Q3 (Veb) has a 0.6v diode drop, which gives you 14.4v. That means that the current through R2 is 14.4v/3300 gives you ~4.4ma while the current through r2 is the difference in voltage 0.6 volts / 1000R3 = 0.6ma. Maybe I came to the right answer for the wrong reasons. I am confuses why the 0.6ma is not added to the 4.4ma, instead it is a part of the 4.4ma? Commented Jan 5, 2017 at 21:22