# NPN Transistor Question - Reverse Voltage a concern?

#### Kurt_051

Joined Nov 11, 2015
6
In a lab I am working on, the following circuit is covered. V_in is an AC power source operating at a constant RMS voltage. It's important to note that the transistor will only be operated in saturation mode.

Explicitly mentioned in the introduction is that the 1n914 diode is used to prevent excessive reverse voltage. My question is, why would this even be a concern for a transistor operating in saturation mode? The voltage at $$V_{b}$$ is going to be greater than the voltages at $$V_c and V_e$$, so current will never be flowing backwards other than the typical AC direction switch. How is reverse voltage a concern here?

PS: This is my first time posting in All About Circuits and while I read the LaTeX sticky prior to posting this, it appears that my stuff isn't rendering, what am I doing wrong? The "plain" (command?) is present in my actual writing/editing but doesn't appear when posted and duly those sections are not in LaTeX.

#### dl324

Joined Mar 30, 2015
16,732
Explicitly mentioned in the introduction is that the 1n914 diode is used to prevent excessive reverse voltage. My question is, why would this even be a concern for a transistor operating in saturation mode?
What have you learned about the effect of breaking down the base emitter junction?

PS: This is my first time posting in All About Circuits and while I read the LaTeX sticky prior to posting this, it appears that my stuff isn't rendering, what am I doing wrong?
Leave out the "plain" directive.
The voltage at $$\small V_{b}$$ is going to be greater than the voltages at $$\small V_c$$ and $$\small V_e$$

#### Kurt_051

Joined Nov 11, 2015
6
What have you learned about the effect of breaking down the base emitter junction?

Wait, is this related to voltage breakdown since that path in the transistor is effectively a diode?

I'm guessing the reverse voltage consideration is if enough of a negative voltage is applied across the base-emitter path such that the diode begins to breakdown and is no longer able to block voltage/current in that direction, *then* it does become a concern. Is that right?

PS, how did you quote? Edit: And are we not able to edit our OP's in a thread? I don't see the edit button on my first post.

#### Jony130

Joined Feb 17, 2009
5,487

#### dl324

Joined Mar 30, 2015
16,732
Wait, is this related to voltage breakdown since that path in the transistor is effectively a diode?

I'm guessing the reverse voltage consideration is if enough of a negative voltage is applied across the base-emitter path such that the diode begins to breakdown and is no longer able to block voltage/current in that direction, *then* it does become a concern. Is that right?
If the base-emitter junction broke down, what effect would that have on the circuit?

Have you studied what breaking down the base-emitter junction could do to the transistor?
PS, how did you quote?
The default action to reply is to quote the message you're replying to. If you want to quote multiple messages, you use the "+Quote" button.
Edit: And are we not able to edit our OP's in a thread? I don't see the edit button on my first post.
You should have an "Edit" button available on any of your posts. I've only started one thread and it had an edit option.

#### Kurt_051

Joined Nov 11, 2015
6
If the base-emitter junction broke down, what effect would that have on the circuit?

Have you studied what breaking down the base-emitter junction could do to the transistor?
The default action to reply is to quote the message you're replying to. If you want to quote multiple messages, you use the "+Quote" button.
You should have an "Edit" button available on any of your posts. I've only started one thread and it had an edit option.
I'm not sure what breaking down the base-emitter junction would do other than change the direction of current heading through the transistor. I think it's fair to assume that if the base-emitter diode broke down that the voltage of the emitter is higher than the voltage of the base and thus current is headed back towards the power source.

Is, speaking really directly, the reason the 1n914 diode is there, to protect from reverse voltage resulting from a breakdown of the base-demitter diode, resulting from a voltage at the emitter higher than the voltage at the base that also was high enough to be the breakdown voltage? I'm not really sure based on your response whether I was right or not.

PS: It appears I only have an edit button on my posts for some time but then thereafter the option disappears (still in new forum format).

#### DickCappels

Joined Aug 21, 2008
10,128
I didn't see it mentioned, but the negative voltage shows up on the base when the input signal goes to ground, thus dumping the "speedup cap" across the base resistor into the base. If the 1N914 were not there, the collector could spike below ground.

In some circuits the base-emitter junction is intentionally driven into breakdown to get the fastest possible turn-off time.

#### Jony130

Joined Feb 17, 2009
5,487
I'm not sure what breaking down the base-emitter junction would do other than change the direction of current heading through the transistor.
At first glance nothing special will happen at negative half cycle when base-emitter junction is driven into breakdown. The collector current will not flow, so we can say that the BJT will be in cut-off. The only thing that is "not normal" is that the base current is now negative (current flow from emitter to base). But in real life BJT do not like this situation. Why? Because the operation of a transistors under emitter-base breakdown conditions will cause drastic reductions in current gain.
"But when you are all done with this transistor, you should (virtually) throw it away, as the Vbe and beta have been degraded by the zenering" Bob Pease.
And this is why we add a diode in parallel with base-emitter junction. This diode will start to conduct for Vin lower than -0.6V. And thanks to this base-emitter voltage will never reach breakdown conditions.

#### Jony130

Joined Feb 17, 2009
5,487
In some circuits the base-emitter junction is intentionally driven into breakdown to get the fastest possible turn-off time.
Are you sure ?? From what I know to speed-up turn-off time the reverse biased voltage is needed.

#### DickCappels

Joined Aug 21, 2008
10,128
Yes. Reverse breakdown, or is that the wrong term?

The only time I read about the degradation of beta as the result of driving the junction in to reverse breakdown, Motorola said that there was a degradation of low current beta but beta at higher currents was for all intents and purposes unaffected. That was in the days when we were replacing vacuum tubes with transistors.

#### crutschow

Joined Mar 14, 2008
34,078
Yes. Reverse breakdown, or is that the wrong term?

The only time I read about the degradation of beta as the result of driving the junction in to reverse breakdown, Motorola said that there was a degradation of low current beta but beta at higher currents was for all intents and purposes unaffected. That was in the days when we were replacing vacuum tubes with transistors.
The base-emitter junction can be driven into reverse bias to rapidly pull the excess charge carriers out of the base region to speed up the turn-off (minimize saturation delay).
BJTs optimized for switching are specially doped to rapidly kill the excess carrier charge to minimize this delay (a 2N2369 is an example).
But you normally don't want the reverse voltage high enough to cause reverse breakdown as that can degrade or damage the junction.