Transistor Base/Collector Current mystery

Adder_Noir

Hi,

I've been learning about transistors lately through the Horowitz & Hill book. I'm surprised to see that when used as a current source the collector voltage only changes with load voltage with an ideal transistor.

When I say I'm surprised I mean that in non-ideal models the base current seems to vary with changes in collector voltage/current. I believe this is related to something called the Early Effect and changes in hfe caused by collector voltage change.

So my question is how can collector voltage affect base current? I thought base current dominated everything and set the emitter current, therefore I thought it would be totally independant of Vc and Ic.

Would I be right in thinking of a transistor used as a current source as a 'current-stifler' from the collector's perspective? In the respect that regardless (assuming non-saturation) of Vc's magnitude the current stays set, and in a real model this doesn't quite hold true? Is it possible to bugger a transistor by setting the current too low and collector voltage too high?

This is probably the most interesting thing I've discovered since I was a young child. I had no idea electronics was so absorbing!

berkeman

I don't think you have the cause and effect right. The collector current is set by the base current, and yes, the Early Effect causes Ic to vary slightly with Vc.

Think of how a curve tracer works when scanning a BJT. The base current is set at some value, and Vc is swept (for example from 0V to some higher voltage less than BVceo), and then Ib is stepped up some, and Vc is swept again. The result of this repeated stepping and sweeping is the characteristic "fan" shape plot that we're used to picturing when thinking of BJTs and FETs.

Adder_Noir

So one could think of the collector current as being set by the base voltage. Is that about right?

olebuckeye

Yes Adder Noir, I think you've got it. Base voltage decides emitter to collector current. You do not want very much base current, only a voltage. The base potential acts like a variable barrier to the emitter to collector current. I like to think of it in plumbing terms in that the base is like a valve that controls the flow of the emitter to collector current. Too much base current equals a burned out transistor. The reason I used the plumbing anology is that I was taught tube theory and they were sometimes described as valves and even had the schematic reprsentation of a V for valve. Transistors are not that different at the macro level.

Adder_Noir

Quality, thank you

cabraham

The key to understanding Early effect is as follows. The regions in a bjt known as "base, collector, & emitter" are not exact in terms of where one stops and another starts. Take the base and collector regions. This junction is reverse biased. A "space charge zone" exists in between the base and collector. This region is also called the "depletion region", or "transition zone" . The width of this region is not an exact value. It is strongly influenced by Vbc, the base-collector voltage, and by temperature.

At room temp, with the base-emitter junction forward biased, the b-c space charge region has a specific width. If you examine the characteristic curves, Ic vs. Vce, with Ib as a parameter, you will see a family of curves. When Vce is more than 1 volt or so, the collector current flattens out to an almost horizontal line, approximating constant current source behavior. It is not, however, perfectly horizontal, but sloped upward. That is, for a given base current, if Vce is increased while Ib is held constant, Ic increases. Thus Vce, or more exactly, Vcb, the reverse bias potential on the b-c junction, influences Ic slightly. This is Early effect.

Early effect is due to "space charge region modulation". As Vce (Vcb) increases, the b-c depletion region, or space charge region gets wider. This is the case with any reverse-biased p-n junction. When the b-c depletion region increases in width, it impinges upon both the collector region and the base region. The base region is very very thin to begin with. When the depletion zone extends into the already thin base region, the base region becomes even thinner. Semiconductor physics text will explain this in detail along with detailed math.

When the base gets thinner, the forward current gain, hFE, increases. A thinner region contains the same density of dopant atoms, but now with reduced volume. For a given forward bias, and electric field strength, the number of carriers injected from the base into the emitter decreases due to the reduced volume of the base region. But the number of carriers emitted from the emitter hardly changes. Thus the same value of emitter current can be attained with a smaller value of base current, so that hFE has increased. But, the data sheet characteristic curves are generated by holding base current, Ib constant. The emitter current, Ie, will increase with the same value of Ib due to increased hFE inevitable with increased Vce.

In a nutshell, Ic is strongly determined by the forward bias on the b-e junction, and weakly influenced by Vce.

Adder_Noir

That's an amazing reply thanks a bundle. That's being printed off and is going in my notes!

I see, just what one would need for a good current source.

Urmi Roy

Why does the same slanting of output characteristics curve not occur for CB configuration...(essentially,the same things are happening in bothe CE and CB)

sophiecentaur

Just reading that question 'from cold' it strikes me that you are sort of assuming the transistor 'knows' where it is. The characteristics if devices will have to apply, always, under the same driving conditions. 'Flaws' in the transistor characteristics (departures from the ideal) will have different levels of effect in different configurations but the transistor still has to be doing the same thing in response to what you do to it.

cabraham

Base current Ib, base-emitter voltage Vbe, & emitter current Ie, are all needed to obtain collector current Ic. Ib/Vbe/Ie are mutual & inter-related. Which one "sets Ic" is a chicken-egg vicious circle. All 3 are involved.

The key word is "same". The slant on the curves is much less for CB configuration. This is becuse CB has the base at a fixed potential wrt ground, and the input signal is at the emitter. Thus the transfer ratio is given by:

Ic = alpha*Ie.

The CB current gain, alpha, changes very little w/ Early effect. If Early effect increases beta from 100 to 120, the corresponding increase in alpha is much smaller. Since alpha = beta/(beta+1), then a beta increase from 100 to 120 results in an alpha increase from 0.9901 to 0.9917. Beta increased by 20%, but alpha increased by a mere 0.165%.

Thus the slope of the CB curve is almost horizontal.

Did I help?

Claude

Urmi Roy

Looking at this intuitively....in CB, the input is the Ie,the output is the Ic...now,increasing the Vbe,the emitter-base (E-B) current increases....due to the early effect,the base region has become very small....so the Ib decreases,resulting in a corresponding increse in Ic....but Ic is basically a part of Ie (Ie=Ic+Ib),and since Ib has reduced in magnitude,Ic is now a bigger fraction of Ie than usual...so now,alpha is bigger.
(But I guess my explanation still doesn't explain why the change in alpha is much smaller than the change in beta).

On the other hand,in CE,the input is the Ib,and the output is the Ic...on increasing the value of Vce,the reverse bias on the C-B junction increses,causing the early effect....due to this,there is a large reduction in Ib,and a corresponding increase in Ic...thus,as I said before,the Ic is now a larger fraction of Ie than what it would be wothout the early effect....thus there is a certain change in alspha (it incresases slightly)...but since Ic is now much bigger than Ib,the value of beta is much bigger....is this right?

(looking at all this in the intuitive way really helps me to understand,please refer to this issue in 'intuitive' terms).

cabraham

But I did explain it, in post #6 above. Please reread post #6, and then if you need clarification, let me know. As I stated in post #10, the reason alpha changes to a much lesser degree than beta, is simply because the math relation is

alpha = beta/(beta+1). Set beta to some starting point, 100, 80, 125, whatever. Compute alpha. Now increase beta by 10%. Recompute alpha, and it will have changed very little. It's just the nature of the beast.

"Intuitively" is hard to do with semiconductor physics. The thermal energies, bandgaps, fields, covalent bonding, doping, carrier concentrations, saturation velocities, etc. are too much to intuitively grasp. One needs to study the principles, do the math, and rely on the semiconductor producers' info since they know it best. That is my advice.

Claude

sahil_time

I have a small question. How can you Hold the base current constant. I mean because of base-width modulation the base current is bound to change right?

Averagesupernova

I guess the obvious answer would be to feed the base with a current source. Do you have a specific configuration in mind? Most transistor circuits are about configuration to get the desired result.

sophiecentaur

For example, a 100V source, connected to the base via a 10MΩ series resistor would put 10μA into the base (as near as you could measure) whatever Vbe turns out to be (with the emitter grounded).
There are more convenient ways of doing it but brute force and ignorance can give you a 'current source'.

sahil_time

Ok...! Actually i just thought that considering a common emmiter configuration...if at all Vce is increased...keeping Vbe a constant. Then the base width decreases...so there should be some decrease in the Base current. Thats what i thought.

Also i had a question....that if u consider Early Effect in a common Base configuration then with increase in Vcb and keepin Vbe constant, the emmiter current increases..How?
How can Vcb affect the emmiter current at all?

cabraham

How can you keep V_BE constant? A constant voltage source directly across the b-e junction will likely destroy the bjt. A bjt cannot withstand "voltage drive", that is why a bjt is always "current driven".

To view Early effect, you can hold I_B constant & vary V_CE. Or you can hold I_E constant & vary V_CE, for the common base configuration. The current gain here is alpha, not beta. Alpha will change very little due to Early effect as I stated above. BR.

Claude.