# Why would the transistor not switch?

I was reading an example from a text book. And for this circuit above the author claims when R3 is less than 100 ohm Q3 will not switch. I couldn't figure out the "reason" why. But I verified with LTSpice the author is right. He just doesn't explain the reason.

If lets say R3 is close to zero when Q2 on, why wouldn't Q3 also switch on?

• You seem to be at hard work,studying and analyzing this example.Nothing wrong with that.It just got my attention after seeing it twice.Good luck out there! – Daniel Tork Dec 18 '15 at 16:46

For Q3 to switch on, the voltage drop between its base and emitter must be about 0.6 V, which means that the same voltage must be dropped over R3, which means that the current flowing through R3 must be at least I3 = 0.6V / R3.

When there is less current flowing through R3, the voltage drop over R3 is smaller than Q3's minimal voltage drop, and Q3 will stay off.

For R3 = 100 Ω, the required current I3 would be 6 mA. However, in this circuit, the current through both R3 and Q3 is also limited by R2: a current of 6 mA would result in a voltage drop of 19.8 V over R2, which is not possible with a 15 V supply.
The largest possible voltage drop over R2 happens when Q2 is saturated, and is about 14 V, which results in a maximum possible current of about 14V/3.3kΩ = 4.2 mA.

• " which means that the same voltage must be dropped over R3," why same voltage has to drop? is that because kirchoff eq? – user16307 Dec 18 '15 at 14:55
• btw but when R3 is too small then the current can get greater and create a 0.7 volt to balance the emitter base voltage. im confused.. – user16307 Dec 18 '15 at 14:59
• @jjuserjr I think an easier way to roughly check whether Q3 should be on or not would be to see that with R3 ~ 0, Q3 would have similar voltage levels in it's emitter and base, but since it is pnp the emitter should be at a lower potential than the base for it to start conducting. If they are at a similar potential, Q3 would be off. – user13267 Dec 18 '15 at 16:41
• The ends of R3 and the base/emitter of Q3 are directly connected, so these points always have the same voltage. The current through R3 cannot get greater because R2 doesn't allow it. – CL. Dec 18 '15 at 20:11
• @user13267 When you wrote "since it is pnp the emitter should be at a lower potential than the base for it to start conducting.", I think you meant that the emitter should be at a higher potential than the base. – Deepak Dec 19 '15 at 4:06

PNP transistors turn on when $V_{EB}$ is large enough. When you make $R_3$ too small, there isn't enough voltage across the transistor's EB junction for it to turn on.

Intuitively, $V_{EB}$ is the same as the voltage across $R_3$. Since $R_2$ and $R_3$ are roughly a voltage divider (there's very little base current in $Q_3$), the voltage is $$V_{EB} \approx \frac{R_3}{R_2 + R_3} \cdot \text{15 V} \approx \frac{R_3}{R_2} \cdot \text{15 V}$$ if $R_3 << R_2$. Clearly, when the fraction $R_3 / R_2$ is too small, the transistor can't turn on.

• but when R3 is too small then the current can get greater and create a 0.7 volt to balance the emitter base voltage. im confused. – user16307 Dec 18 '15 at 14:58
• You should read en.wikipedia.org/wiki/Voltage_divider to understand why the increasing current will not cause an increased voltage. – Greg d'Eon Dec 18 '15 at 15:01
• no i meant that basically the pnp transistor should regulate the voltage drop across it right? so what ever the resistance it should regulate it. why cant it regulate? and if it regulates the current of R3 whatever it is small should increase. thats what i thought. – user16307 Dec 18 '15 at 15:02
• We're talking here about the current through the resistor (i.e. R3) not through the transistor, the latter of which (current) only is responsible for turning the transistor on. If R3 is too low, then there is not enough voltage on the base to turn the transistor on. The current through the transistor is given by R2, not R3. – user59864 Dec 18 '15 at 15:09
• And regarding Greg's answer: approximating R3/(R2+R3) as R3/R2 is not very useful here, especially when designing this divider so that Q3 actually goes into saturation. – Fizz Dec 18 '15 at 15:46

Since you're confused about the turn-on behavior of Q3 relative to R3 consider the equivalent circuit consisting only of the essential resistor divider (R3 and R2) and the base-emitter junction of Q3:

I'm varying here R3 over time from 0 to 1K. The BE diode turns at about 0.65V which corresponds to 150 ohms for R3. This is easily verified as 15V*150/(3300+150)=0.65V.

Since the current through a diode that's turned on has an exponential variation with the voltage across it (Shockley's equation), and since the current here is limited by R2, the BE voltage will be roughly constant once the diode is on. Once the junction is on, Vbe actually varies logarithmically with a diode current that has an upper bound (imposed by R2)... which is to say not much. Note that the V(BE) curve (red trace) has a sharper turn than the I(BE) current (magenta)... because of the logarithmic relationship it has with the diode current.

Before the diode turns on, the BE voltage is a linear function of R3 since it's just a resitive divider with R2. Also I(R2) doesn't vary a whole lot even before the diode turns on because the turn-on point is only at about R3=4.5% of the value of R2. But on a separate plot of I(R2) [in the lower pane] you can see that's "even more constant" past the turn-on point of the diode. So this verifies the usual assumption that Vbe is constant (and consequently so is I(R2) here) once the BE junction is actually on. Before that there's no restriction on what Vbe it can be as you can see; it only depends on the value of R3 when the diode is off.

Consider the voltage across a diode and the current that flows. Below are the curves for an old germanium diode (1N34A) and a silicon diode (1N914): -

Concentrate on the silicon diode (1N914). With 0.6 volts across it, the current is about 0.6mA. Now drop that voltage to 0.4 volts. The current falls to 10 uA and, with 0.2 volts across it the current is about 100 nA.

Now, the base-emitter junction in a BJT is a forward biased diode. The forward biasing comes from the voltage you place across it and this is usually via a biasing resistor. In your circuit, R2 and the power supply voltage define the current that can jointly flow into the base and into R3.

When R2 supplies a decent amount of current, most of it flows thru the base emitter junction because you are on that part of the diode curve and that part of the diode curve has a dynamic resistance that is much smaller than R3. As base-emitter voltage lowers, its dynamic resistance gets higher and R3 starts to become the "path" to which most of the current from R2 flows.

Dynamic resistance is the small change in applied voltage divided by the change in current. You could look at the diode graph above and pick some points: -

• At 0.60 volts the current is possibly 600 uA
• At 0.62 volts the current is about 1000 uA

Dynamic resistance would be 20mV/200uA = 100 ohms

• At 0.40 volts the current is about 10 uA
• At 0.42 volts the current is about 11 uA

Dynamic resistance would be 20mV/1uA = 20 kohms.

So, when R3 lowers it becomes more dominant that the base emitter junction and rapidly the junction current falls away. Given that we can approximate transistor action to a device with current gain, lowering R3 beyond a certain point means a rapidly falling collector current and, in effect, the transistor is regarded as turned-off.

A transistor needs about 0.7v VBE to start conducting. As you have the benefit of a simulator there, experiment with different R2/R3 values and look at the voltage developed across R3, and whether the transistor turns on.

As to why it's 0.7v, you need semiconductor physics!

• i thought i can understand on off beaviour by using aristotalian logic. "if this exceeds this turns on".. so on – user16307 Dec 18 '15 at 17:09

Well, I think all the complicated answers have been given, but for my two cents: anything below 150 ohms "shorts out" the base to emitter junction