# Does a PNP transistor pull up itself?

Blog schematics for PNP transistors sometimes connect +5V to the emitter and a voltage divider between the base and an open drain output, and sometimes only connect a base resistor to the low side switch. In the simpler configuration will the current flowing from the emitter to the base pull up the base and turn off the transistor? How long will this take?

• Sample schematic, please. Commented Jun 16, 2017 at 14:00
• Bloggers might not be representatives for good engineering practices. Commented Jun 16, 2017 at 14:01
• PNPs don't have drain terminals. Do you mean open collector? Commented Jun 16, 2017 at 14:21
• The low side switch is a fet. Commented Jun 16, 2017 at 14:28
• Schematic given Commented Jun 16, 2017 at 14:50

That's actually pretty easy. You have a schematic editor you could use here. I'll use it now. I've paired them, side by side, to match up with your pictures in your question:

simulate this circuit – Schematic created using CircuitLab

As you probably already know, $R_1$ is used to sink the needed base recombination current for $Q_1$, when it's activated as a switch by $SW_1$. Your question, as I read it, asks about $R_2$ in the right-hand case.

$R_2$ is used here as a "pull-up" of $Q_1$'s base. It's added security of a sort. For example, assume for a moment that there was an accumulated of dirt on the switch contacts which allowed a very slight current to flow across its contacts. Or that someone might be able to "brush by" and accidentally touch something that was connected to $R_1$ or directly to the base of $Q_1$ through dirt accumulation or other unintended means. Or also just realize that there may be very small capacitances which can hold a slight charge that might, for a moment, provide enough current at $Q_1$'s base to cause it to turn on momentarily.

To cheaply provide some added insurance against all of those cases, a "weak" pull-up resistor can be added to the circuit. This provides a way of pulling unintended charges away from the base of $Q_1$. Only when $SW_1$ is engaged fully and with definite intention will (then) enough current be supplied to overwhelm $R_2$'s weak action and turn the BJT on.

But both work when used in ideal situations. But adding the pull-up is cheap and it helps reduce the likelihood of unintended switching events when the situations are not ideal. (Which is most of the time.)

In normal situations (manual switch), there is no meaningful difference in speed between the two schematics. But even when a microcontroller is used to activate $Q_1$ as a switching device, there will usually be no meaningful difference in speed. When using a BJT as a highly saturated switch, like this, there is charge storage in the BJT that takes a certain amount of time to remove and a weak pull-up doesn't change it a lot. So for most uses, you can consider them of equivalent speed.

However, $R_2$ does improve the speed a tiny bit in practice. The stronger you make it (up to a point), the more improvement in speed you get. (This improvement will be in the turn-off time and NOT the turn-on time, which remains fast either way.) It still won't be that much of an improvement, though. In the end, you might normally see a few microseconds of delay in a turn-off and the improvement might be 10% or 20% of that time. (The actual percent of improvement will depend on details about the BJT device, itself.)

Here's a simulation to show you what's happening on the turn-off side:

There are both cases shown above. The green trace shows you the base current of the right-side PNP BJT and a dark blue trace shows you the base current of the left-side PNP BJT. The dark blue trace is almost invisible because it is everywhere exactly the same as the green trace, except for a tiny bit that you may be able to see near a "porch" that appears when both BJTs are being turned off. This porch takes less time for the green trace (faster circuit) than for the dark blue trace (slower, left-side circuit.)

The red trace is for the faster (right-side) schematic's output. As you can see, it still takes a while for the BJT to come out of saturation. The light blue trace is for the slower (left-side) schematic's output. And as you can see here, it also still takes a while for the BJT to come out of saturation. But as you can see it takes even longer because $R_2$ is not present to help. Either way, it's not a LOT of difference.

In most cases, this is just added insurance and has nothing to do with improving speed. In some rare circumstances using a BJT as a switch like this (and as a hobbyist life it has never been a consideration), you might need that extra few hundred nanoseconds of speed in the turn-off time. (Perhaps to slightly improve the accuracy of the output duty cycle?) But there are better methods then.

• I built the circuit to do high side switching on LEDs with a micro and it works great. Commented Jun 28, 2017 at 20:25
• @joeforker Glad to hear it!! (It's nice when things that are supposed to work, actually do work when you build them.) In the case where the high side voltage is higher than the MCU's I/O pin voltage, you'll need another BJT. But that's a different story.
– jonk
Commented Jun 28, 2017 at 20:27
• I happened to have a low side MOSFET on the board, but I'm sure the 5V tolerant I/O pin would have worked too. Commented Jun 30, 2017 at 18:31