# Some basic doubts regarding transistors characteristics

1. Nov 15, 2013

### ShreyasR

1) While plotting input characteristics of a transistor, we choose to keep output voltage constant and plot the variation of input voltage and current. Why do we choose to keep output voltage constant and not current?

2) While plotting output characteristics of a transistor, we choose to keep input current constant and plot the variation of output voltage and current. Why do we choose to keep input current constant and not voltage?

I have tried answering these but never able to convince my lecturer:

Me: The transistor is a current amplification device. So it is more meaningful to keep the input current constant to plot the output characteristics. (I wasn't able to say why output voltage is chosen to be constant)

Lecturer: Even if you keep the input voltage constant to plot the output characteristics, you will get a similar curve for output characteristics... Still why do you choose the input current to be constant? And if u say the transistor is a current amplification device, why is it so?

Me: Its because when we first studied the transistor action, we considered the emitter current IB which branches into IC and IB... IB is the current due to the electrons which recombine with the holes in the base region which is small in size and lightly doped so just a few electrons recombine with the holes(considering NPN transistor)... And the ratio of collector current to emitter current is a constant (current amplification factor α), and then we derived that IC/IB = β which is the common emitter current gain. So it is basically the current which is amplified. So a transistor is a current amplification device.

But he was not satisfied with my answer. He asked me to find out. So someone please reply so that i can convince my lecturer and more importantly, me. :tongue:

Last edited: Nov 15, 2013
2. Nov 15, 2013

### AlephZero

There is no "deep" reason for this, except convenience. The usual way to model the small signal behavior of a transistor as a 4-terminal (2-port) network is using hybrid or "h" parameters, where the independent quantities are the input current and output voltage.

So, to measure the h-parameter input characteristic, you make the small-signal output zero, which means you make the output voltage constant. To measure the output characteristic, you make the small-signal input zero, which means you make the input current constant.

You could use a impedance (Z parameter) network or an admittance (Y parameter) network instead. But the conversion between the parameters of the different networks is standard theory, and the h parameter network is most convenient, because the parameters have "simple" interpretations for a transistor.

http://en.wikipedia.org/wiki/Two-port_network
http://en.wikipedia.org/wiki/Bipolar_junction_transistor#Small-signal_models

3. Nov 15, 2013

### cabraham

Actually base current IB is much more than just the recombined electrons in the base region. There are 3 major components of base current, namely injection, transport, and displacement.

When an external power source forward biases the b-e junction, holes in the base and electrons in the emitter are subjected to an electric field which exerts forces on them. The electrons in the emitter move towards the base, whereas holes in the base move towards the emitter. The base holes moving towards the emitter constitute a current, namely the injection part of IB. Just as emitter electrons are injected into the base, so are base holes injected into the emitter. Once these base holes enter the emitter, they recombine with electrons in the emitter. This is injection current. The base is doped much lighter than the emitter so that current density of electrons from emitter greatly exceeds hole density from base.

Once emitted from emitter, electrons enter base. Most survive the trip through the base but a few recombine with holes in the base region and ionize the atoms. Eventually for every electron recombined in base, one electron exits the base lead. This part of the current IB is the transport component. Transport current is much less than injection current for silicon devices.

At high frequencies, the capacitance formed by the base-emitter depletion region requires large current to rapidly charge and discharge. This is displacement current. The injection and transport components of IB are present at low frequency as well as high freq, but displacement current increases with frequency.

The frequency where the current gain, β, of the device, falls to unity (0 dB), is the transition freq, fT. Above this frequency the bjt has current gain < 1, and is of less use. A lower β means that more base drive is needed for a given collector current.

IB consists of 3 components, injection, transport, and displacement. At dc and lf, injection is the largest, transport is small, as is displacement current. As frequency increases, hf, the displacement current increases until it overcomes the other 2. Then β decreases w/ rising freq. Eventually at a freq of fT the β value is 1 and the device is less effective as an amplifier since it has unity or less current gain.

Did this help? BR.

Claude

4. Nov 16, 2013

### anhnha

Can you explain why it creates IB not IE? I don't see how these electrons and holes go out of base to constitute IB.

5. Nov 17, 2013

### ShreyasR

Thank you AlephZero! That made sense... But the above paragraph quote is not really clear to me... Why is it that the hybrid parameter model has simple interpretations for a transistor?

I actually don't find it very sensible to analyse a transistor using h parameter model. The main reason being: In an amplifier, when i connect the signal generator to the input, it is I who decide what is the input voltage is. So why should i consider it to be a function of input current and output voltage? (As in h parameters) And out of the four variables: Vin, Iin, Vout and Iout, All that i know is Vin, and the h parameters. So how do i theoretically calculate 3 unknowns (Vout, Iin and Iout) when i have only two equations from the h parameters?

6. Nov 17, 2013

### ShreyasR

And Thank you for your reply cabraham. Now I know something that i did not know. But I have the same question as anhnha. I dint get a clear picture of injection current. and also how β decreases with frequency. Can you please send me a link regarding this phenomenon?

7. Dec 16, 2013

### cabraham

The external source and its associated E field exert force on free charges in both emitter and base regions, creating both currents, Ie as well as Ib. Picture the base region. It is p type Si (for an npn bjt device). It has a light distribution of p-type acceptor atoms diffused into the silicon base region during fabrication.

The emitter region has a heavy distribution of donor atoms diffused into it during fabrication. In the base, holes are the majority carrier devices, they are very mobile since an abundance of holes are present in the p type base. In the emitter, there is an abundance of electrons, as they are majority carriers. But the number of electrons per unit volume in the emitter greatly exceeds the number of holes per unit volume in the base. This is intentional so that Ib is much smaller than Ie. During fabrication, the producer deliberately injects many n type ions into the emitter, and much fewer p type ions into the base.

When an external signal source E field is brought into the b-e junction, all charges in base and emitter feel this E field, and a force is exerted on these carriers. But the emitter region has many more free charges than does the base region. So a lot of electrons drift from emitter to base, but only a small number of holes drift from base to emitter.

The holes from the base cross the junction and diffuse into the emitter region. But the emitter region is n type while holes are p type charges. They are minority carriers, and have low mobility. They do not get far because they recombine with n type atoms in the emitter and ionize them positively. Eventually an electron enters the emitter lead to restore charge neutrality. Because the emitter region is deep, nearly all holes entering the emitter from the base are captured by n type atoms and ionization occurs. This migration of holes from base to emitter is injection base current, a large component of total base current.

For every hole emitted from the base, an electron exits from the base lead maintaining charge neutrality. When a hole in the base is removed from its parent atom, that atom is negatively ionized, so that an electron is eventually expelled from the base lead, preserving charge neutrality. Otherwise the current would cease immediately due to charge barrier build up.

But the electrons emitted from the emitter undergo a different scenario. If the base region were thick, nearly all electrons would eventually recombine with p type acceptor atoms in base region. Atoms ionize, and electrons are expelled from base lead. This component of base current is the transport components. If the base were very thick, transport is roughly equal to injection currrent. But the collector current is near zero, and we have 2 diodes in series. Not very useful.

But the base region is intentionally made very thin. When the emitter injects electrons into the base, they diffuse, but because the base is so very thin, they get yanked into the collector before they can recombine with base atoms. The collector base junction is reverse biased, and the E field in that region attracts electrons. So the number of electrons emitted that get captured by base acceptor atoms is very small. Those electrons that do recombine ionize base atoms, and an electron is expelled from the base lead to keep charge neutrality.

This is the transport part of Ib, which is much smaller than the injection part of Ib. In a nutshell, the E field from the external source exerts forces on base holes and emitter electrons. Base holes are fewer in density than emitter electrons. A small number of holes move from base to emitter, and a large number of electrons move emitter to base. This hole motion is the injection base current.

When a tiny fraction of emitted electrons recombine in the base, the electrons exiting base lead as a result of ionization, is the transport base current component. Transport current is much smaller than injection current. At dc and low freq, these are the 2 major parts of Ib.

But as freq increases, displacement current increases. At hf, charges move between base and emitter to charge and discharge the junction. The faster the freq, the greater the charging current. This component is displacement current, and it imcreases with freq.

When injection current is larger than displacement, then beta is primarily determined by injection current. So at dc and low freq, beta is near constant. But eventually there is a high enough freq where beta starts dropping. This is the freq where the displacement current is roughly equal to innjection current. As freq increases further, beta drops with freq.

We describe this action by labeling the freq where beta has fallen to unity as the "transition freq", ft. At ft, beta = 1, at 0.1*ft, beta = 10.

I hope I've helped.

Claude

8. Mar 23, 2015

### Nithin singh

Hello,
Why have you considered the motion of both hole and electron while explaining the components of base current .Aren't they meant to be same (beacause motion of hole is just theoretical way to explain the motion of electron ). Thanks in advance !