What voltage at the base emitter does a transistor conduct?

[avinsinanan]

Hello,

I was doing another questions and the questions is -

At what voltage does the Base- Emitter Juction of a silicon transistor have to be to conduct NORMALLY?

The keyword is NORMALLY. Because if the Base Emitter is 0 v the transistor is off, If the Base-Emitter is 0.6 to 0.7 it just begins to conduct and is termed PARTIALLY ON. If it is 1.4 the transistor is fully on.

SO what is the answer? 0.6 v or 1.4 v?

These questions are so vague its driving me crazy.

Thanks a million.

Yours Respectfully
Avin SInanan

 

[zachtheterrible]

It depends a lot on what transistor you are using. Just look at the transistor datasheet. go to google and type in the transistor number and you will find the datasheets, usually in .pdf form.

 

[Pyroandrew]

I'd answer

"depends on what tranistor it is!"

That is the only safe answer!

Is it bi-polar or uni-polar, is it of npn or pnp construction?
Also, for WHAT to conduct normally??? f*ck knows, the question doesn't tell you!

It is indeed a vague question, it is even more misleading though. What this specific question is actually asking is, "what is the di-electric strength of the BE junction in forward bias?". NOT what you think the question is asking. Where is this question from? it is increadibly poorly worded!

 

[David Bridgen]

When a silicon small-signal transistor is fully conducting the base-emitter voltage will be around 0.7.
The figure of 1.4 would apply to either a Darlington-connected pair or to a power transistor with a very high base current.

 

[audioguru]

Hi Avin,
Just look at the minimum and maximum base-emitter voltage spec on the transistor's datasheet. You can also look at the curve of typical base-emitter voltage vs collector or base current which also usually shows it changing with temperature.

A low-power small transistor such as a 2N3904 conducts a little at room temperature when its base-emitter voltage is about 0.6V. It conducts a lot (its maximum rating of 100mA collector current) when its base-emitter voltage is about 0.8V. A 2N3055 power transistor could have a base-emitter voltage of 2.5V with a 5A base current. On another 2N3055 it could be only 0.6V at a lower current.

The actual voltage is vague because it changes with current, temperature and luck. For two transistors having the same part number, the spec allows a different base-emitter voltage between them of 0.65V to 0.85V at the same current and temperature.

You shouldn't worry about the "normal" base-emitter voltage anyway, the base of a transistor is driven by current, not voltage.

 

[Nigel Goodwin]

When a silicon small-signal transistor is fully conducting the base-emitter voltage will be around 0.7.
The figure of 1.4 would apply to either a Darlington-connected pair or to a power transistor with a very high base current.

I agree, 0.7V is the value I usually use for any calculations - it can't go much higher than that unless the transistors has blown.

A darlington transistor (as it's two transistors together) will be double the Vbe drop, so would be 1.4V drop.

 

[JimB]

For a silicon transistor, Vbe will be 0.6 to 0.7 volts when the transistor is conducting.

If Vbe is as high as 1.4 volts, either,
you have a darlington transistor,
or
you have blown the base-emitter junction open circuit.

JimB

 

[audioguru]

Hey Jim,
A 2N3055 is a silicon transistor with a 15A max collector current. At only 10A, its Vbe is typically 1.8V and could be much higher with a 15A collector current:

 

[JimB]

Audio

OK point taken.

But for the average small signal transistor.....

1.4 Volts Vbe = Busted!

JimB

 

[audioguru]

Hi Jim,
A base-emitter has zener breakdown at about 6 to 7V if you connect it backwards. :lol:

 

[ljcox]

You shouldn't worry about the "normal" base-emitter voltage anyway, the base of a transistor is driven by current, not voltage.

Not true. The collector current is an expotential function of the base - emitter voltage. The charge in the base - emitter region is proportional to the Vbe. If the charge is excessive, the transistor is in saturation. If not, it is in the active region.

Certainly Ic = hFE * Ib is a very useful relationship, Ib is required to maintain the b - e charge.

Len

 

[audioguru]

Hi Len, down under. D'day mate!
When is a transistor fed from a voltage source? An emitter follower? Maybe a low circuit gain transistor with an emitter resistor that swings with the input voltage.
A transistor with a high circuit gain is fed from a current source which is the impedance of the source or maybe the collector resistor of a preceeding transistor stage.
If you used a low impedance voltage source to feed a high circuit gain transistor then its Q-point biasing would be all over the place due to variations of Vbe between transistors or temperature. It would also have an exponential response causing very high distortion. A current input would be much more controllable and more linear. :lol:

 

[theimperia]

As far as I know, the Vbe junction is a diode, so the formula would be:

Ie = Is * exp(Vbe / n*Vt)

Ie - emitter current (approximately equal to collector current)
Is - in the order of 10^-14
Vbe - the Vbe voltage
n - can be between 1 and 2
Vt - approximately equal to 0.024 Volts

That the info from an electronics class, I didn't actually look for Is or n in the datasheet.

Usually, you would look at the Vbe vs Ic graph in the datasheet to know the operating point (aka Q point) of the transistor.

You should rely much on those values though, as it was said earlier, coz you'll get huge changes due to temperature, and the actual transistor (same part number, from the same manufacturer and the same batch will give totally different results).

If you're using it as an amplifier, make the feedback define the gain and the bias...

 

[ljcox]

Hi Len, down under. D'day mate!
When is a transistor fed from a voltage source? An emitter follower? Maybe a low circuit gain transistor with an emitter resistor that swings with the input voltage.
A transistor with a high circuit gain is fed from a current source which is the impedance of the source or maybe the collector resistor of a preceeding transistor stage.
If you used a low impedance voltage source to feed a high circuit gain transistor then its Q-point biasing would be all over the place due to variations of Vbe between transistors or temperature. It would also have an exponential response causing very high distortion. A current input would be much more controllable and more linear. :lol:

G'day audioguru
You're missing the point mate.

Say you connect a voltage source between the base and emitter of a NPN transistor and a ammeter between the collector and say a +10 Volt supply.

Start the experiment with the Vbe = 0 Volt then gradually increase it while watching the Ic.

At about 0.5 V (for Si) Ic will start to flow. CAREFULLY increase the Vbe and record it and Ic. Stop when Ic reaches say 10 mA. (assuming the transistor can dissipate at least 100 mW safely)

Now plot a graph of Ic vs Vbe. You will find it is expontial.

Now say you connect this transistor as an emitter follower with a 1k emitter resistor and set the base voltage (about 5.6 V) such that Ic = 5 mA. If you measure the Vbe, you will find it is exactly the same as the Vbe on your graph for Ic = 5 mA.

Len

 

[audioguru]

Yeah, Len.
An emitter-follower has 100% negative feedback so the tiny amount of exponential change in the base-emitter voltage is nearly negegible.
In a common emitter amp without an emitter resistor providing negative feedback, a low impedance voltage input would result in severe distortion at the output.With the output swinging less than half the total supply voltage, the distortion is 15% as shown here. The distortion increases with a higher input/output.

 

[ljcox]

Yeah, Len.
An emitter-follower has 100% negative feedback so the tiny amount of exponential change in the base-emitter voltage is nearly negegible.
In a common emitter amp without an emitter resistor providing negative feedback, a low impedance voltage input would result in severe distortion at the output.With the output swinging less than half the total supply voltage, the distortion is 15% as shown here. The distortion increases with a higher input/output.

We're talking at cross purposes. You're referring to the AC situation, whereas I'm talking about the fundamental theory of transistor operation - the DC bias voltages and currents.

In a common emitter amp with or without an emitter resistor providing negative feedback, the collector current is an expontential function of the Vbe. Vbe is the DC voltage across the base - emitter. Certainly the AC signal has to be a small fraction of this voltage in order to prevent distortion - as you said.

In any circuit where the transistor is in the active region (ie. not in cutoff or saturation) the collector current is an expontial function of Vbe.

Len

 

[audioguru]

Hey Len,
I haven't tried it but maybe some linear IC's work like this:
1) Connect the collector of the common emitter 1st transistor to the emitter of a common base 2nd transistor.
2) Maybe the collector current of the 2nd transistor will be linear with a linear voltage into the base of the 1st transistor. I think the distortion will cancel. :lol:

 

[Roff]

Is the voltage across a resistor caused by the current through it, or is the current caused by the voltage across it?.
.
.
.
The only difference is that this is a linear relationship, while in a transistor or diode it is exponential. I think this is the point ljcox is making.

 

[eblc1388]

Is the voltage across a resistor caused by the current through it, or is the current caused by the voltage across it?.

Mm.. Interesting. I'm thinking of a resistor with a current flowing through it being cool down gradually until it becomes a superconductor.

So in this case we have current but no voltage. We also have the case for voltage but no current for perfect insulator.

Therefore, following the argument, superconductor "must" therefore exist and it does.

1. voltage, no current (insulator)
2. voltage, some current (resistor)
3. no voltage, current (superconductor)

 

[ljcox]

Hey Len,
I haven't tried it but maybe some linear IC's work like this:
1) Connect the collector of the common emitter 1st transistor to the emitter of a common base 2nd transistor.
2) Maybe the collector current of the 2nd transistor will be linear with a linear voltage into the base of the 1st transistor. I think the distortion will cancel. :lol:

Good morning audioguru,
I don't know the answer to this question. To answer it, you would need to do the maths.

Len