# Can a bipolar transistor work with Ib < 0?

1. Jan 9, 2012

### fbs7

I came up with this doubt last week.

Consider a NPN transistor. Usually we bias it so that current enters through the collector and base, and leaves through the emitter, right? That allows the base to form some holes, which recombine with the electrons emitted by the emissor and that way the bugger works.

Now, say I set my my circuit like this: I set a current source of 2 mA that goes to a 10kΩ resistor which then goes to the collector. Then I set a current drain that pulls 1 mA from a 10kΩ resistor which is connected to the emissor. And I set the base on the ground.

So Vcb > 0, and Vbe > 0, so the thingie is in forward active mode. But Ic = 2 mA and Ie = 1 mA, so Ib must be -1 mA... the current is flowing out of of the base.

If the current flows out of the base, then there are no holes generated there, and the transistor should cut off... but it can't cut off because it's in forward active mode, and also because there is nothing in the general Ebers-Moll model that actually prevents Ib from being negative...

How to resolve this dilemma?

2. Jan 9, 2012

### fbs7

Hmm... a diagram may help.. :)

https://www.physicsforums.com/attachment.php?attachmentid=42587&stc=1&d=1326146397

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Last edited: Jan 9, 2012
3. Jan 9, 2012

### jim hardy

a common base amplifier has current gain approximately 1.
the transistor will do its best to make collector current very nearly same as emitter current.
they'll differ by ~1/hfe
which for a 3055 is maybe 5%?

you're trying to force too many variables.
what is nature of those current sources?

but your picture shows 100 ohm resistors.

it is not clear what you have in mind.

4. Jan 9, 2012

### fbs7

That's correct. For any configuration that puts the transistor in active forward mode, Ib is small and Ic ≈ Ie.

What I'm trying to do is figuring out what happens if I use external circuits to force Ic = 2 mA and Ie = 1 mA... whether the transistor will still be in active mode or not.

I will fix the resistors to 10 k.

ps: about the nature of the current source, I'm not sure it matters; this is a conceptual problem, and conceptually I should be able to build a circuit that will force 2 mA into the transistor; same for the current sink (it's just a current source of -1 mA ).

Last edited: Jan 9, 2012
5. Jan 9, 2012

### jim hardy

now i understand , it's sorta a thought experiment.

the question boils down to what happens if you force excess current through a transistor's collector junction.

well it seems to me the transistor will attempt to block half of your 2ma and collector voltage will rise to nearly 30 volt supply, starving your current source.

BUT - because an ideal current source can produce infinite volts,
in thought experiment land you're not limited to 30 volts.....

so i am going to estimate the following:

from 3055 datasheet,

section "off characteristics"
leakage at 100volts and base cut off is 1ma at normal test condition, which is about all you need ,
and it's 5 milliamps at 150 degC.

i assert, for sake of discussion, the voltage at collector will rise quite high as transistor tries to hold current at 1ma
and your 1ma excess current will heat the junction , P= V X I and V is large now,

until the collector junction's reverse leakage current becomes enough to allow the extra milliamp to flow, and it shouldn't take a lot of heat,

at which time the transistor "softens" and your scenario changes.
What began as irrestible force (current source) meets immovable object(collector junction) changes to dueling ideal current sources.

That's my speculative answer to the hypothetical question.
What do you think will happen?
are you equipped to try it? one experiment is worth a thousand expert opinions...

6. Jan 9, 2012

### yungman

Remember the basic of BJT is the emitter current setup the collector current, not the other way around. In your example, you have 1mA pulling from the emitter, so all the transistor can pull from the collector is 1mA minus the base current as Ie=Ic+Ib. So the collector cannot sink more than that. You put a 2mA current source, the voltage will rise and keep rising until the transistor burn ( assume the 2mA current source supplied by HV source) OR the current source will saturate and stop pumping out 2mA.( for example the current source is a PNP with emitter tie to +15V. the collector on the transistors will rise to +14.9V and can go no further, the PNP become saturated and current reduced).

There is no way you can have the circuit like you want.

7. Jan 9, 2012

### jim hardy

thanks Yungman ...

i feel better now.

old jim

8. Jan 9, 2012

### fbs7

Yeah, I think you guys are right. Once Ie is set at 1mA, Ic will be αf.1 mA, so the only way to force a higher current through Ic is by punching through the C-B junction.

So in the real world the 2mA current source will increase its voltage until it stops being a current source, and is then just another voltage.

I was finding this an interesting situation because I modeled it in SPICE, and the thing actually ended with Ic = 2 mA, Ie = 1 mA and Ib = -1 mA... but that's most probably because current sources are not exactly realistic...

Thank you both! :^)

Last edited: Jan 9, 2012
9. Jan 9, 2012

### yungman

I am not familiar with Spice, I don't know what model they use for the current source. I honestly don't know when the NPN collector voltage break down, where the current goes. I assume it goes to the base as the structure is that the base is really in between the collector and emitter so the Vcb is the key. Maybe Pspice said the extra 1mA is the break down current of the collector base junction and that 1mA flow from collector to the base.

You know, it actually make more sense about the Pspice result as I type this!!!! Now, I am just talking out loud.

10. Jan 9, 2012

### fbs7

Oh, you should try it, Yungman... LTspice IV - SPICE with a quite usable GUI. Google will find it in a minute - and it's free.

11. Jan 12, 2012

### nikhil khatri

hey i dont now how to post so i am asking here.
I was confused actually how a transistor can provide gain. as we know that energy can not be created.so from where this extra energy is coming.
I know that dc sources used in baiseing are providing this energy.
but how it is happening.

12. Jan 12, 2012

### fbs7

That's my take on this: the Vbe voltage between base and emitter controls how much current goes through the collector and emitter, and the transistor is built such way that the Vbe voltage causes a very small current between base and emitter. Therefore, the final effect is that a small current in the base causes a big current in the collector and emitter, the ratio of which is what is what they call forward gain (βf = Ic / Ib ).

The tricky part is the first paragraph above... why should Vbe control how much current goes through Ic? When people explain that, they usually give a pretty picture like

I think the actual effect is quite complicated, and it takes someone smarter than me to say they truly understand it, but this is how I made peace with this: in the diagram above (of a NPN transistor), EB junction is directly polarized, right? So plenty of electrons are emitted from the emitter, pulled in by the higher voltage in Vb (electrons like to go to higher voltages).

But, because the base is thin, only a few recombine with the base current, and the majority end up collected by the collector instead (as it has an even higher voltage than the base). So, as Vbe increases, Ibe increases a little bit, and Ic and Ie increase a lot. So it seems like Ic is controlled by Ib. We then call that apparent effect that Ic is controlled by Ib, as being the forward gain of the thingie.

That's how a see it, at least. I'm sure some quantum physicist is snickering at my explanation, but that's how I made peace with it :)

13. Jan 14, 2012