NpN bipolar transistor current gain

In summary: "beta" is a measure of how much voltage gain is achievable...& "gm" is a measure of how much current gain is achievable.
  • #1
Azelketh
40
0
I have been studying npn bipolar transistors and how to apply gain to the current in a circuit, yet i simply do not comprehend how altering the base - emitter current can effect the emitter - collector current,
can anyone here explain how altering the base - emitter current effects the emitter - collector current?
 
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  • #2
https://www.physicsforums.com/tags.php?tag=transistor
 
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  • #3
The base region is very thin, around 1.0 micron or less. The b-e & c-b junctions are in extreme proximity. When forward biased, the b-e jcn base emits holes towards the emitter, & the emitter emits electrons towards the base. The holes reach the emitter & recombine just as would in a diode. But because the base region is so thin, the electrons emitted from the emitter pass right through the base before having a chance to recombine, & are colected by the collector.

If an ac signal is superimposed on the b-e jcn, then the number of electrons emitted varies w/ the ac signal amplitude. Thus the number of electrons collected per unit time, collector current Ic, varies in proportion to the ac signal.

In a nutshell, a bjt operates on 3 events:

1) emission
2) diffusion/drift through very thin base region
3) collection

The bjt works, again, only because the base is extremely thin. If the base was very wide, say 1.0 mm, the electrons from the emitter would almost all recombine in the base & few reach the collector. Then it is nothing more than a pair of diodes w/ a common anode (npn). Does this help?

Claude
 
  • #4
If the electrons emitter from the emitter overshoot the base region and pass through to the collector region then what charge carries complete the current across the base emitter region?
Also how does that account for the current amplification? if the electrons from the emitter-base section are overshooting the base and ending up in the collector shouldent the current amplification be almost infintisimal as a few electrons overshooting isn't a factor of 100( or some such high number) the base-emitter current?
 
  • #5
I've referenced this before here and I think it might answer your questions.

tran10.gif


Looks like 99% of the electrons defuse across the base region.

Check out the other illustrations they provide on this subject.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html#c4"
 
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  • #6
"Transistor action" takes place as I stated above. Emitter current overshoots the base & becomes collector current. But this doesn't happen unless the b-e junction is forward biased. Emitter current, Ie, exists only when b-e jcn is forward biased. But biasing the b-e jcn forward requires some base current, Ib, as well as base emitter voltage, Vbe.

So the b-e jcn is forward biased with Ib/Vbe/Ie positive & non-zero. The base region is intentionally doped lighter than the emitter so that the forward base current is very small compared w/ the emitter current, i.e. Ib << Ie. Since Ib does not contribute to Ic, it is desirable to make Ib as small as feasible w/o degrading other parameters. So the doping density in the base is intentionally made very light, & heavy in the emitter.

When forward biased, the base is very thin, so that the volume of p-type charge carriers is very low (holes for npn device), & light population, resulting in a small number of holes injected from base to emitter, with a large number of electrons emitted towards the base.

The holes from the base recombine with electrons in the emitter. Charge neutrality is maintained with the same number of electrons entering the emitter lead from the outside world. A few electrons emitted recombine in the base. Charge neutrality is preserved as an equal number of electrons exit the base to the outside world. The overwhelming majority of emitted electrons go through the base & into the collector, where they propagate through & exit the collector lead to the outside world.

In order to keep the process going, some base current Ib is needed, as well as base-emitter voltage Vbe. "Current gain" is simply the output current Ic, divided by the input current Ib. For a "common base" configuration, Ie is the input, & current gain is slightly less than unity. The output of a bjt is a pretty good constant current source, Ic.

Current gain is Ic/Ib, aka "beta" and/or "hfe". The voltage gain, actually "transconductance" of a bjt, is Ic/Vbe, aka "gm". A bjt amplifies both current & voltage. But the gain factors are not infinite. In order to amplify a signal, a small Ib & Vbe are needed. If beta is 100, & gm is 0.10 amp/volt, then if Ic changes by 100 uA, Ib must change by 1.0 uA, & Vbe by 1.0 mV.

"Current gain" tells us how much base current is needed for a given value of collector current. So if I wish to use a bjt with a minimum beta of 50 to drive a relay requiring 500 mA, then the bjt drive network must be able to supply the bjt base with a current of 500/50 = 10 mA. If the bjt is used as a saturated switch, we overdrive the base & use more base drive current.

Anyway, in a nutshell, for a given Ic, you must input an Ib equal to Ic/beta for active region operation, more than that for saturated operation.

DId I help, or make things worse?
 
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  • #7
thanks that's cleared a lot up.
Thinking of the transistor as a switch other than an amplifier is a conceptual difference that makes a lot of difference as well, i was approaching this from the understanding that if the base current changed then the collector current would change proportionally. From your useful descriptions it seems clear that the base current acts as a switch allowing the main collector current to flow?
 
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  • #8
Well, the bjt can be used as a switch, but when used as an amplifier, don't think in terms of a switch. The Ic to Ib ratio is linear/constant over a limited range. A 5% change in Ib should result in a 5% change in Ic. The bjt can work as an on/off switch, where it takes on 1 of 2 states at a time, cutoff or saturation. In this case, the base current is above the minimum for saturation, or near zero.

In the switching mode, we don't vary Ib linearly w/ a small signal. Instead Ib is toggled between zero (or near zero) value, & fully on hard. This results in the bjt looking like a switch fully on or fully off.

Or, we can use a bjt as an amplifier. The Ib & Vbe values are established at some quiescent level (dc). Then an ac signal modulates the total base current (dc + ac), same with Vbe (dc + ac). The output current, Ic, has a dc & ac component. The ac part is an amplified facsimile of the ac input. Does this make sense?

Claude
 
  • #9
cabraham said:
A 5% change in Ib should result in a 5% change in Ic.
Claude
If Ib is already great to foreward bias the base emitter junction, allowing Ic, then how does an additional 5% Ib cause the current Ic to change by 5% as well?
 
  • #10
Because the relation between Ib, Ie, & Ic is dependent upon doping, geometry, & charge carrier distribution. The base & emitter terminals are the control electrodes of the bjt. Let's say the bjt is configured as a common emitter. If the base is driven w/ a signal voltage generator & series resistance (or the Norton equivalent, a current source & parallel resistance), an increase in the generator signal results in a corresponding increase in Ib, as well as Ie.

If Ie increases by 5%, there are 5% more electrons emitted from the emitter, which means the number collected will increase by 5%. The increase in Ib requires energy. The source inputted to the amp stage must increase its amplitude & power by 5%. This results in a stronger electric field in the base-emitter region, eventually a slight increase in Vbe, & a 5% increase in Ie.

The impedance at the base terminal is rbb' (base spreading resistance), plus r_pi (hybrid pi small signal equivalent resistance, r_pi = beta/gm), plus the emitter region bulk resistance plus external emitter resistance multiplied by (beta+1). So a 5% increase in signal amplitude, increases Ib, & Ie by 5%, to a pretty good approximation. Draw the hybrid pi circuit & compute it to verify.

Since 5% more electrons are emitted, 5% more will be collected. This will be the case until the current rises to the level where "beta droop" occurs. At high values of current, the base region gets wider, & beta gets lower. So a 5% increase in Ib results in less than 5% increase in Ic. Search using the key words "beta droop" or "Kirk effect" to see what I'm referring to.

If we are operating on the flat part of the Ic/Ib curve, the 2 are closely related. When the current is very high, beta droop occurs & the relationship gets skewed. Make sense?
 
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  • #11
http://en.wikipedia.org/wiki/Photodiode#Other_modes_of_operation
Phototransistors also consist of a photodiode with internal gain. A phototransistor is in essence nothing more than a bipolar transistor that is encased in a transparent case so that light can reach the base-collector junction. The electrons that are generated by photons in the base-collector junction are injected into the base, and this photodiode current is amplified by the transistor's current gain β (or hfe).
 
  • #12
Great, that makes sense. Cheers for your patient replies to my questions dude's.
 

Related to NpN bipolar transistor current gain

What is a NpN bipolar transistor?

A NpN bipolar transistor is a type of semiconductor device used in electronic circuits. It consists of three layers of doped material, with two layers of p-type semiconductor sandwiching a layer of n-type semiconductor.

How does a NpN bipolar transistor work?

A NpN bipolar transistor works by controlling the flow of current between the two outer layers, known as the collector and emitter, through the middle layer, known as the base. When a small current is applied to the base, it allows a larger current to flow from the collector to the emitter.

What is the current gain of a NpN bipolar transistor?

The current gain of a NpN bipolar transistor, also known as hFE or beta, is the ratio of the collector current to the base current. It is typically denoted by the symbol "hFE" and can range from a few to several hundred depending on the specific transistor and its operating conditions.

What factors affect the current gain of a NpN bipolar transistor?

The current gain of a NpN bipolar transistor can be affected by various factors such as temperature, voltage, and frequency. Additionally, the physical properties of the transistor, such as its material and dimensions, can also impact its current gain.

How is the current gain of a NpN bipolar transistor measured?

The current gain of a NpN bipolar transistor is typically measured using a multimeter in a common emitter configuration. The base is connected to the positive lead, the collector to the negative lead, and the emitter to the common lead. The resulting current ratio is then used to calculate the current gain.

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