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PNP bipolar transistor

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hannover90

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Hello all,

when the base (B) and collector (C) nodes of a pnp bipolar transistor (with VD0=0.7 V) are tied together and their voltage (VB) is raised from 0 to 1 V and VE (voltage of the emitter) is set to 0 V, how operate the diodes D1 of the emitter and bias junction and D2 of the collector and bias junction for VB> 0.7 V?

I know only that the transistor operates as a diode in this condition. I can imagine that D1 is forward-biased but I don't know what happens with D2?


Thanks for any reply in advance
 

I know only that the transistor operates as a diode in this condition. I can imagine that D1 is forward-biased but I don't know what happens with D2?
Click to expand...

If the base and collector are tied together and apply to them voltage with respect to the emitter and is a PNP transistor, then clearly the transistor is OFF, meaning, both diodes are reverse biased.
 

CataM, thanks for your reply.
I think, when |VBE|<0.7 V, the transistor is off, but when it exceeds 0.7 V, it switches on.
The cadence simulation shows me, that for |VBE|>0.7, the emitter current IE is equal to | IB+IC |

In the case that the base is connected to the emitter, yes the transistor is off because |VBE|=0 V.
 
Last edited:

hannover90 said:
I think, when |VBE|<0.7 V is the transistor off, but when it exceeds 0.7 V, the transistor switches on.
.
Click to expand...

The transistor is more than a simple switch. Don`t forget - it can be used as a quasi-linear amplifier.
Hence, it is not "off" for |VBE|<0.7.
When the C-B junction is reversed biased there is an exponential collector current Ic=f(Vbe).
That means: There is, for example, a collector current also for VBE=0.1 Volt.
 

LvW thanks for the information.
Do you know how operate the diodes D1 (Emitter/Bias) and D2 (Collector/Bias) for |VBE|>0.7 V? As I wrote, I think D1 operates in forward-biased but what is about D2? The simulation shows that IE =| IB+IC |.
If a current flows through the collector, how operates D2 ?
 

Lets see again what you have said in post #1.

* PNP transistor
* Base and Collector tied together
* VB voltage rises from 0 to 1 V
* VE (voltage of the emitter) is set to 0 V

Again, if I do not miss something, the BE diode is reverse biased and the BC diode has across 0 V.

Now, if you say that |VBE|>0.7 V , I shall understand that VE > VB right ?

Then, the BE junction diode (PNP transistor) is foreword-biased, and the CB junction diode has across 0 V (i.e. 0 A through it)
 

VE is set to 0 V and the voltage of B and C rises from 0 V to 1.0 V, which means |VBE| rises from 0V to 1 V.
 

When the collector is connected to the base of a BJT it acts as a fairly ideal diode.
It will conduct in the forward direction with a diode's voltage-current logarithmic relation (giving about 0.7V at 10mA), and block current in the reverse direction (up to the base-emitter break-down voltage which is typically below 10V).

If you put 1V in the forward direction you will get rather high current, depending upon the transistor's current rating.
 

Thanks crutschow.
I am sorry, it is a mistake in my first post. The voltage of B & C, (which are tied together) varies from -1 V to 0 V.
 

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