# what happens if emitter-base & base-collector junctions are forward biased with 0.7V?

Discussion in 'General Electronics Chat' started by devanandiamin7, Feb 9, 2012.

1. ### devanandiamin7 Thread Starter New Member

Apr 14, 2007
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I have a doubt regarding BJT. what happens if both emitter-base & base-collector junctions are forward biased with 0.7V ? will BJT be in saturation region? What will be the voltage across the collector and emitter and Ic ?

-Devanand T

2. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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First draw a circuit and prove that u can do that.

3. ### DickCappels Moderator

Aug 21, 2008
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You could make that with, say, and NPN transistor, by grounding the base and running the collector and emitter to a positive power supply through separate resistors.

Hmmm... maybe it will be saturated because the emitter and collector will be a roughly the same voltage. The exact voltage depends on the design of the particular transistor.

4. ### wmodavis Well-Known Member

Oct 23, 2010
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A transistor can be biased as you propose. If it is it will not operate in the normal fashion where a small base current can be used to controll a larger collector/emitter current.

Question for you. Can you define 'saturation region' in a fully and technically accurate way? When you do/can do that then see if the BJT meets the technical criteria of that definition when both junctions are forwaed biased. If it does not meet the necessary conditions of properly defined saturation that will give you the answer.

If biased as you propose, the voltage across the collector and emitter and Ic can be determined by standard loop/node analysis and/or measured with suitable instruments.

5. ### Ron H AAC Fanatic!

Apr 14, 2005
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The power supply needs to be negative for an NPN.

6. ### crutschow Expert

Mar 14, 2008
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Connecting the collector and base together of a transistor is sometimes used to emulate a near ideal diode characteristics. For that connection both the base-emitter and collector-emitter junctions are forward-biased if the collector-base voltage is more positive than the emitter (NPN). Then the transistor is operating on the edge of the saturation region and the voltage will stay near the base-emitter voltage of 0.7V, depending upon the current.

7. ### DickCappels Moderator

Aug 21, 2008
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An aside: I was thinking PNP since positive power supplies are prevalent.

8. ### hobbyist Distinguished Member

Aug 10, 2008
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Alright lets logically think this thru.

A transistor is properly biased (NPN) when the base is more positive than the emitter,
and the collector is more positive than the base.

Under these conditions you have the base emitter junction acting as a diode in the forward biased way, the base collector junction is acting as a reversed biased diode, due to the purposely designed thickness of the emitter collector junction a barrier is formed and bla bla bla..

It works as a current amplifier.

Now if the base is made more positive than the collector, then there is a collector base forward bias, which causes the transitor to leave its linear amplifier curve.

One way this happens is if the collector resistor is to large and the base voltage is to large, causing a significant voltage drop across the collector resistor, thereby lowering its collector voltage with respect to the base voltage.

Under these conditions, the base emitter junction acts as a forward biased diode, with the collector almost being open, which means is, very little current will flow into the collector, and most of the current from ground will flow out the base, this is not saturation, but a wrongly biased transistor.

Saturation is a transistor that is at the top of its linear curve, with maximum current flowing, where any increase in base current has little to none effect of the collector current.

So a forward biased base emitter, and a forward biased collector base junction can not enter saturation, but only act as a base emitter diode, with the collector almost opened.

This is just my hobby experiance (NON professional) opinion.

9. ### Ron H AAC Fanatic!

Apr 14, 2005
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Your experience has led you astray.
Consider an NPN transistor used as a switch, with the emitter grounded and the collector tied to a positive supply (Vcc) through a resistor (Rc). If sufficient base current is applied to drive the collector voltage to less than ≈0.2V, the transistor is saturated. If the base current is increased further, the current through the resistor cannot increase, so the excess base current flows through the base-collector junction. Thus, both base-emitter and base-collector junctions are forward biased. Note that collector current is maximum, being limited to (Vcc/Vsat)/Rc.

10. ### Mark_T Member

Feb 7, 2012
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As a by-the-by if you short the source and drain of a JFET you get a very low leakage diode. See a 'PAD 5' for example. If you do the same with an NPN tansistor the emitter will have a lower Vf and most current will flow through the emitter, but it will still lokk like a diode.

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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Actually, most of the current (typically more than 90%, according to simulations) will flow through the collector. The b-c junction is typically larger than the b-e junction, so b-c will hog most of the current.

12. ### hobbyist Distinguished Member

Aug 10, 2008
773
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Yeh, the voltage drop across the transistor from collector to emitter, is a case of saturation. when the base voltage and collector voltages are properly met.

I wasn't refering to biasing for saturation, but trying to explain how a improperly biased transistor could cause the effects I described.

Case in point, there were times I would design CE stages then as a quick check on the biasing I would measure key voltages, and there are times where I find the base voltage divider pretty well out of range.

So I know I'm getting to much current out the base causing the drop.

I remember one time I had my meter connected to the base terminal, and I calculated all the values I needed but still was getting a wrong collector voltage, and base voltage was too low,

so from there I began to change out resistors, so going over the calculations and looking over the resistors, I noticed I had the wrong collector resistor in there insted of red being the multiplier, I had the yellow, wich is of course 100 times larger,

so finding the fault I pulled out the collector resistor, and as expected the base voltage dropped further, due to all the current flowing thru the base emitter diode, I then replaced the collector resistor with the proper one of lower value, and I got the calculated values, checking the base, emitter and collector showed proper voltages,

as a take on it of curiosity I put the wrong value back in again, and checked the voltage from collector to base, this value was very small, then I kept the meter there at the collector, and put back the proper value, the collector base voltage raised significantly, where the base was less positive than the collector as it shoukld be.

From there I realized the meaning of a transistor being bilateral,

and not unilateral, which explains a lot to me about the hybrid model,

where the output has some affect on the input as well as the input on the output.

That experiance has taught me a lot more understanding about how a transistor operates, and how the proper biasing is needed, for the transistor to work linearly,

as well as input impedance is all about, I found that without the proper value of collector resistor, can also affect the imput impedance of a stage, because if the collector is not positive enough the base emitter becomes more isolated in current flow, and it's like a signal source attached to a diode to ground.

In that condition, beta times RE is no longer viable, for calculations, why, because beta deals with collector current taken from the emitter current, so the base resistance looks higher due to the collector current flowing away from the base and into the collector terminal, but if the collector side is not taking the required current, then that leaves the full emitter current to flow out the base terminal, hence the base resitance is now lowered due to the excess base current.

Therefore under the extreme condition of the high collector resistor mistakenly put in the circuit, did cause the transistor to be more like a BE diode FW biased, with collector almost opened ended,

however because the collector was not opened physically a small current did flow through to cause the voltage drop to occur,

but the transistor was not saturated, because the CE drop was no where near ground, but just very low for the expected calculated values.

13. ### Ron H AAC Fanatic!

Apr 14, 2005
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Given everything you described, I think the transistor was saturated. In a saturated transistor with an emitter resistor, Vc will not be near ground, but Vce will still go to near zero volts. The key here is where you connect your black probe. It needs to be on the emitter, not on ground.

14. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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why don't u both bread board and blow one ....!

15. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
Yeh, thats right,

Saturation is across CE junction, I think I did have the probe to ground, when I detected a fault with the wrong resistor value at the collector, so I would be measuring voltage drop across RE mostly.

Thanks for pointing out the simple mistake in my measurement process.