2N3904 Maximum Allowable Ratings

Thread Starter

elec_eng_55

Joined May 13, 2018
214
Hi:

I am not sure how Vebo rating works. Is the voltage drop across the emitter-base
junction no always approximately 0.6, 0.65 or 0.7 volts?

If the quiescent voltages on a 2N3904 are VCQ = 12.00, VBQ = 7.05 and VEQ = 6.35
then VCEO = 6.05, VCBO = 4.95 and VEBO = 0.70. This would indicate that none of the ratings are exceeded?

I think that I am missing something here.

upload_2019-1-4_10-43-22.png

Thanks,

David
 
Last edited:

MrChips

Joined Oct 2, 2009
21,877
The voltage drop across the base-emitter junction is not 0.6, 0.65, or 0.7V.

Here is the V-I characteristics of a hypothetical p-n junction.



This graph shows that the voltage across the junction can be almost any value.
The 0.6-0.7V that you are familiar with is the threshold, turn on,or knee voltage as shown on the characteristic curve. It means that if you can somehow limit the current through the junction to a safe operating value, the p-n junction voltage will settle to around 0.7V.

VEBO at 6.0V is the maximum you should apply to the junction.
Of course, if you apply 6V to the junction without limiting the current you are sure to destroy the junction.
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
The voltage drop across the base-emitter junction is not 0.6, 0.65, or 0.7V.

Here is the V-I characteristics of a hypothetical p-n junction.



This graph shows that the voltage across the junction can be almost any value.
The 0.6-0.7V that you are familiar with is the threshold, turn on,or knee voltage as shown on the characteristic curve. It means that if you can somehow limit the current through the junction to a safe operating value, the p-n junction voltage will settle to around 0.7V.

VEBO at 6.0V is the maximum you should apply to the junction.
If you are looking at a circuit, how would you know if the VEBO exceeds the 6 volt max when
VB - VE = 0.7 volts when looking at the quiescent voltage?
 

dl324

Joined Mar 30, 2015
11,543
I am not sure how Vebo rating works. Is the voltage drop across the emitter-base
junction no always approximately 0.6, 0.65 or 0.7 volts?
No. It depends on current:
upload_2019-1-4_8-1-3.png
upload_2019-1-4_8-1-17.png

If the quiescent voltages on a 2N3904 are VCQ = 12.00, VBQ = 7.05 and VEQ = 6.35
then VCEO = 6.05, VCBO = 4.95 and VEBO = 0.70. This would indicate that none of the ratings are exceeded?
The O in those parameters mean that the unspecified terminal is open circuit.

VEBO is the voltage from the emitter to base, in that order. In your example, you have the polarity wrong.
 

WBahn

Joined Mar 31, 2012
26,156
You need to consider the very important distinction between Vbe and Veb.

The forward voltage when in the active region is the base voltage relative to the emitter voltage, or Vbe.

The spec sheet is talking about the maximum Veb, or the emitter voltage relative to the base voltage.

You can't take the base more than 6 V BELOW the emitter and expect the component to survive.
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
You need to consider the very important distinction between Vbe and Veb.

The forward voltage when in the active region is the base voltage relative to the emitter voltage, or Vbe.

The spec sheet is talking about the maximum Veb, or the emitter voltage relative to the base voltage.

You can't take the base more than 6 V BELOW the emitter and expect the component to survive.

So, Vebo = Ve - Vb = 6.35 - 7.05 = -0.70 volts. Since the emitter is at 6.35 volts (6.35 - 6.00 = .35 volts)
that means that if the base goes lower than 0.35 volts, the transistor will go poof?
 

dl324

Joined Mar 30, 2015
11,543
So, Vebo = Ve - Vb = 6.35 - 7.05 = -0.70 volts. Since the emitter is at 6.35 volts (6.35 - 6.00 = .35 volts)
that means that if the base goes lower than 0.35 volts, the transistor will go poof?
No.

Lowering the voltage at the base would also decrease emitter current and the emitter voltage would also decrease.
 

WBahn

Joined Mar 31, 2012
26,156
So, Vebo = Ve - Vb = 6.35 - 7.05 = -0.70 volts. Since the emitter is at 6.35 volts (6.35 - 6.00 = .35 volts)
that means that if the base goes lower than 0.35 volts, the transistor will go poof?
IF the emitter remains at 6.35 V when you take the base down to 0.35 V -- in the circuit you are envisioning, why would the emitter stay at 6.35 V when the transistor is turned very hard off? -- then you are at the limit of exceeding the absolute ratings. That does not mean that the transistor will suddenly die -- the spec sheet is essentially saying that, under the specified conditions (which in this case it specifies a temperature of 25 °C), the device will NOT die BEFORE that limit is reached. Due to manufacturing tolerances, if nothing else, it will likely survive at least somewhat beyond that. At a different temperature, however, it may fail before that.

There is also the issue of HOW it dies. It could die spectacularly and never work again, it could continue to work but with degraded performance, or it could appear to work normally for quite some time but fail without warning long before reaching its normal life expectancy.
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
IF the emitter remains at 6.35 V when you take the base down to 0.35 V -- in the circuit you are envisioning, why would the emitter stay at 6.35 V when the transistor is turned very hard off? -- then you are at the limit of exceeding the absolute ratings. That does not mean that the transistor will suddenly die -- the spec sheet is essentially saying that, under the specified conditions (which in this case it specifies a temperature of 25 °C), the device will NOT die BEFORE that limit is reached. Due to manufacturing tolerances, if nothing else, it will likely survive at least somewhat beyond that. At a different temperature, however, it may fail before that.

There is also the issue of HOW it dies. It could die spectacularly and never work again, it could continue to work but with degraded performance, or it could appear to work normally for quite some time but fail without warning long before reaching its normal life expectancy.

If you take the base down to 0.35 volts does the emitter not drop to -0.35 volts? Then
the Vebo limit would never be reached?
 

Jony130

Joined Feb 17, 2009
5,183
Here you can read/see how Vebo can be measured in real life.



So you can use BJT as a Zener diode.

https://forum.allaboutcircuits.com/threads/astable-multivibrator-how-does-it-work.32164/#post-200215

What is more interesting is Veco


Because now BJT's behaviors just like a poor's man tunnel diode (Esaki effect). And tunnel diodes will have "negative resistor" region.
And this negative resistance region occurs only for NPN BJT's.
http://www.cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html
 
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