The BJT has a large input resistance, known as rΠ, with magnitude of order of kΩ.
Moreover, it also has a large output resistance, known as ro, which is usually even larger than rΠ.
Since both of its Rin and Rout are large, then when a BJT is used as an amplifier, for example a Common Emitter amplifier, the amplifier is considered to be a transconductance amplifier, meaning the input signal is voltage (vBE).

So why is it said (at least in the Semiconductor's Basics course) that the BJT amplifies current, if its input signal is voltage?

Moreover, in what I managed to read so far about BJT amplifiers, they always calculated the collector current using VBE, not IB.

 

The BJT is considered a current amplifier because Collector or Emmiter currents are a function of input or base current. (Ic = Ib*gain, Ie=(Ib*gain)+1) The voltages in the circuit are determined components connected to the collector or emitter.

 

The thing is that In all of the design examples I've read, in order to calculate the collector current, they didnt use beta (what you called gain), but used: Is*e^(VBE/Vth) (Vth=KT/q).
I dont see any beta here.

Moreover:
1. A current amplifier doesnt have a large Rin (as the CE amplifier).
2. The input signal of a current amplifier is current not voltage.

 

The resistances you're talking about ( ie rΠ ) are derived from the transistor's forward current characteristics. Essentially, the transistor is a current amplifier, with an active forward gain usually specified as hFE or β. the small-signal transconductances and such are abstracted from the current gain specifications, and the transistor's specific geometry, doping levels, carrier mobility, etc.

Being able to derive quantities for transconducatance and input resistance makes it possible to construce a mathematical model convenient for analysis for specific input sources and conditions. It's just as valid to contruct a small signal model wherein the output is a current controlled by an input current.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Remember, the transistor's collector current is controlled by current through the forward-biased base, emitter junction ( i.e. Diode ) Current thru a diode is alwasy an exponential, such as the one you're talking about. Thus, although the current is given as a function of voltage, what we're really talking about is the current that arises as a result of the applied voltage.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

As BrownOut notes, the BJT base-emitter input looks like a forward biased diode, thus it has a nonlinear function of voltage to current. Rin is a small-signal characteristic for a particular input bias condition. It will vary significantly with a different bias condition.

It may be technically correct to calculate the collector current using the equation Is*e^(VBE/Vth) (Vth=KT/q) but that's not a common way to do it. Don't know where you've read these design examples but in most usual engineering design, beta is used to calculate collector current of a BJT transistor, at least for non-RF circuits.

__________________
Carl
Curmudgeon Elektroniker

 

I understand, thanks

I've also just read that IB = Is/β * e^(VBE/Vth) (when forwared biased), so it works out great.

Still, a CE amplifier for example, is not a natural current amplifier since you cant apply a current source on its input, since the diode needs an applied voltage to start conducting.

 

Actually, you can apply a current to a diode to forward bias it. Current is ultimately dependent on a voltage, so in effect, there must always be a voltage for there to be a current, even it it's a small voltage. A CE amp is, at it's most elementary mode, still a current amp, though it's normally biased using a voltage with associated resistive dividers.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Actually, it is very common, and will be found in most any engineering text.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

In may be common in engineering texts, but it's not common for it to be used when doing circuit design, which was the point I was trying to make. There are many common equations in engineering books that are seldom if ever used by an engineer once he's in practice.

__________________
Carl
Curmudgeon Elektroniker

 

In one sense, I agree, but in another I disagree. There are many different tools available to engineers to analyze and design circuits. For me personally, I don't do alot of analysis in the classical sense, but I keep the governing equations in the back of my mind, and use them to guide me in my designs, along with empirical design techniques. It is very important, IMO, for the engineer to master these analitical methods, however, to get the most out of his designs, although he'll 'streamline' the way in which he uses them.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Hi there,


The transistor is said to be 'current controlled' because that is the
main way the transistor operates when looked at from a certain point
of view where you can see the base current vary widely when the
base emitter voltage only varies slightly. It's almost like saying that
an LED is current controlled, because the main feature to look at
in most cases is the current, not the voltage.
This doesnt mean that voltage is totally out of the picture, because
of course there has to be enough voltage to start with, but if you
picture the current source on the input as theoretical then you dont
really think as much about the input voltage because it goes to
whatever it needs too. In fact, in many designs you would regard
that small voltage change as a disturbance rather than something
you design for.
There are views that look at the voltage vs collector current as you
already noticed. That's an attempt to make some sense out of
how the transistor operates with respect to its input voltage rather
than the current, but for many devices like this they are often
characterized by their dominate features rather than the smaller
features that change only a little. This also aids in the design
sometimes where you can ignore certain features and concentrate
more on others.

Part of engineering is noting both the small and the large, but knowing
when the small can be mostly ignored. The nature of the application at
hand is the most important dictator of how you should look at a circuit
element...sometimes the voltage is very important (log converter)
and sometimes it is not important at all (AC amplifier).

 

Thank you very much friends
You helped me out a lot on this.

By the way, if using a current source as an input to the BJT, then this current source must have a very large source resistance, since the input resistance of the BJT is also quite large (as was said it, it - Rin - depends on the DC collector current).

 

Kohms are not usually considered a particularly high input impedance.

It's not hard to make an active current source with many kohms of apparent output impedance. For example the collector impedance of a typical small silicon transistor is in the tens of kohms.

__________________
Carl
Curmudgeon Elektroniker

 

A bjt amplifies BOTH current & voltage. It wouldn't be too useful otherwise. That's what seperates active circuits from passive ones. A transformer can increase voltage while decreasing current, or vice-versa. Transformers are passive so the net power gain cannot exceed unity. A bjt is active & offers current gain & voltage gain both greater than unity in unison.

Regarding Ic, there are 3 equations.

1) Ic = beta*Ib.

2) Ic = alpha*Ies*exp((Vbe/Vt)-1).

3) Ic = alpha*Ie.

Equation 3) describes transistor action. Ie is the emitter current. When an npn bjt emitter injects electrons towards the base, they go straight through the base region since it is very thin, and get collected by the electric field in the collector base region. Alpha is very close to 1 for a good transistor. If the base region was not thin, most of the electrons would recombine in the base and never reach the collector. In that case, alpha is low, much less than unity. Thus Ic is determined by Ie and alpha. A low alpha value makes the device just 2 diodes back to back.

Equation 2) displays the voltage gain properties of a bjt, and eqn 1) shows its current gain. By the way, either alpha or beta always appears in all equations. Eqn 2) is usually stated w/o the alpha, but remember that Ies*exp((Vbe/Vt)-1) is not the collector, but the emitter current. An *alpha* factor must be multplied into the Ie to get Ic. Thus Ic = alpha*Ies*exp((Vbe/Vt)-1).

Eqn 1) describes "current gain". Eqn 2) describes "voltage gain" or "transconductance". Eqn 3) describes the "physical transistor action". All 3 of these functional relations are all important. Every electrical device in the universe requires both current & voltage for its operation. Some devices are better suited to be driven from a high impedance source, or "current driven". Others are better driven from a low impedance source, hence "voltage driven".

All 3 eqns are important. Did I help?

__________________
Claude