Please look at the attached schematic. This is again from a beginning electronics book. And the schematic demonstrates how to turn off a transistor. Everything is clear here except the resistor on the base. The author further suggests that it can be between 1K and 1M, but why do we need a resistor there at all?

Again, I will appreciate simplicity and, if possible, fullness of explanation.

 

Please look at the attached schematic. This is again from a beginning electronics book. And the schematic demonstrates how to turn off a transistor. Everything is clear here except the resistor on the base. The author further suggests that it can be between 1K and 1M, but why do we need a resistor there at all?

Again, I will appreciate simplicity and, if possible, fullness of explanation.

The primary reason is turn-off speed. When you saturate a transistor, excess charge is stored in the base region. Before the transistor can turn off, the charge has to be removed. With no resistor, the charge has to drain off through the base-emitter junction, which becomes exponentially higher impedance as the charge drains off. A resistor to ground speeds up the turn-off. A lower value will result in faster turn-off.

 

Ron is quite correct. I would give a different answer to this as I see it as a transistor that is off and should remain off. To do so it needs R2.

There is essentially a reversed biased diode from collector to base, and it will (must) have some reverse leakage current. Without R2 this current has no path except into the base and out the emitter...so it can become base current.

Base current will allow more collector current, and more current means more power being dissipated, so the temperature goes up. Leakage currents also go up with temperature, so without R2 this circuit can literally turn itself on without any external base current applied.

Take a good ole 2N2222A. The I-CBO current is spec'ed as .01uA at 25°C and 10uA at 150°C, or a range of 1:1,000. If you put a 1 Meg resistor for R2 you will develop 1 E6 * .01 E-6 = .01 volts, which has the B-E diode well off.

 

Thanks, guys, I couldn't have imagined that being the reason. It's very nice to know why. Sorry I didn't thank you right away -- my communication preferences for the post were not set right.

 

A lower value will result in faster turn-off.

But no resistor would provide a really fast turn-off then, no?

f you put a 1 Meg resistor for R2 you will develop 1 E6 * .01 E-6 = .01 volts, which has the B-E diode well off.

Hmm, Ernie says a high resistor would work better and Ron says a low one, which one is right?

 

Think of the base as a capacitor. A low ohms resistor will discharge that quickly, and a high ohms resistor will discharge it more slowly. Using no resistor is equal to using a very high ohms resistor.

The time constant is calculated as RC, with R in ohms and C in farads. An RC tank charges or discharges nearly completely in 3 time constants, 3RC.

Unless you need high speed, the resistor value is not much concern and you can use almost any resistor to accomplish turning off the transistor. The faster you go, the smaller value you must use. I just borrowed a power transistor from an old monitor. It was used in the high frequency switching section and had a 200Ω from base to emitter, built right onto the little PCB it was on.

 

Using no resistor is equal to using a very high ohms resistor.

why?

The time constant is calculated as RC, with R in ohms and C in farads. An RC tank charges or discharges nearly completely in 3 time constants, 3RC.

So, ins't R=0 (no reistor) multiplied by any C is 0?

Also note, that in the book where the schematic is used there is no discussion on the transistor having been switched on before.

 

Careful here. No resistor means open circuit, R = ∞.

This is a common mistake.

Also note, no capacitor means C = 0, R = ∞.

(To be pedantic, no resistance means R = 0).

 

Careful here. No resistor means open circuit, R = ∞.

What I meant of course, there's no resistor but a direct connection from the base to the ground.

 

No resistor = infinite ohms, not zero ohms. Of course there is no such thing as true zero or true infinite. There will always be some path for a charge to dissipate.

With nothing to hold it down, the voltage on the base can "float" in an unpredictable way. Using the resistor is good form to add certainty. Can you get away without it? Can you ride a bike with no hands? Sometimes.

 

What I meant of course, there's no resistor but a direct connection from the base to the ground.

Not when the voltage is less than the diode voltage, about 0.65V. Well below that voltage, there is no "connection". Above that voltage, there is a good connection. Right around that voltage, things are complicated.

 

With nothing to hold it down, the voltage on the base can "float" in an unpredictable way. Using the resistor is good form to add certainty. Can you get away without it? Can you ride a bike with no hands? Sometimes.

That's exactly what I don't get: this floating issue, which is also why pull-down resistors are used, right? But I don't get how connecting directly to ground vs. connecting through a resistor helps certainty.

 

Extending this a further - A OFF transistor under base open condition operates under vceo ie lower collector Emitter Breakdown capability ,where in ICEO leakage is high. The collector -Emitter Breakdown voltage capabilty of the device can be increased to VCER ,by connecting resistor across emitter-Base ie the leackage current (at given voltage) Icer will be low.The Vcer capability depends upon the value of Base-Emmiter resistor. Lower the resistance value , higher is the C-E(Vcer) breakdown . As we increase Resistor value Vcer will tend towards Vceo.

 

But I don't get how connecting directly to ground vs. connecting through a resistor helps certainty.

? I don't get your question. You would never connect the base directly to ground, or more accurately, to the emitter. I gave you the example of the 200Ω resistor from base to emitter because I found that puzzlingly low at first, until I learned its function in high speed switching at high voltages. I've never seen a value less than 200Ω, although I'm sure someone here has.

 

? I don't get your question. You would never connect the base directly to ground, or more accurately, to the emitter. I gave you the example of the 200Ω resistor from base to emitter because I found that puzzlingly low at first, until I learned its function in high speed switching at high voltages. I've never seen a value less than 200Ω, although I'm sure someone here has.

For higher speed, active turn-off can work even better than a low-value resistor.

 

You would never connect the base directly to ground, or more accurately, to the emitter.

Why not? That would make them (base and emitter) at the same potential and prevent any base-emitter bias, thereby preventing any current in the Emitter-Collector path or the Base-Emitter junction.

I'm a newbie, so the no-no's of electronics are not known to me. If you see me asking something stupid, it means I don't know and I would like to know why.

 

Sorry, yes of course you could fully ground the base with a switch or such. I was thinking of only the "passive" arrangement, where a signal to the base has to overcome the base-emitter resistor. Obviously that cannot happen while the base is directly connected to the emitter at 0Ω.

 

So, where does it leave us? The resistor is between the base and the ground (as in the original picture) to prevent reverse COLLECTOR-BASE current?

 

So, where does it leave us? The resistor is between the base and the ground (as in the original picture) to prevent reverse COLLECTOR-BASE current?

Yes, and to speed up turn-off.
I have included a simulation to illustrate the effect of different methods of turning off a BJT.

 

Yes, and to speed up turn-off.
I have included a simulation to illustrate the effect of different methods of turning off a BJT.

Very nice! But I don't see the variant which is a modification of my original schematic: base and emitter at ground? What am I missing?