# BJT amplifies current or voltage?

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#### alphacat

##### New Member
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.

#### bailey45

##### New Member
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.

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#### alphacat

##### New Member
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.

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#### BrownOut

##### Banned
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.

#### BrownOut

##### Banned
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.

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.

#### crutschow

##### Well-Known Member
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.

#### alphacat

##### New Member
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.

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#### BrownOut

##### Banned
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.

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#### BrownOut

##### Banned
Is*e^(VBE/Vth) (Vth=KT/q) but that's not a common way to do it.

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

#### crutschow

##### Well-Known Member
Actually, it is very common, and will be found in most any engineering text.
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.

#### BrownOut

##### Banned
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.

#### MrAl

##### Well-Known Member
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.

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).

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#### alphacat

##### New Member
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).

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#### crutschow

##### Well-Known Member
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.

#### Claude Abraham

##### Member
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?

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#### BrownOut

##### Banned
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).

Be very careful not to get confused by thinking that the OP's equations for collector current is the same as equation 2 stated without alpha. His equation calcualtes collector current directly as an approximation of base diffusion current. The constant in his equation is abbreviated as "Is":

IC = Is*e^vbe/vt
IE = IC/α

That isn't the same equation as (2) given above and ignoring alpha.

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#### Claude Abraham

##### Member
Be very careful not to get confused by thinking that the OP's equations for collector current is the same as equation 2 stated without alpha. His equation calcualtes collector current directly as an approximation of diffusion current. The constant in his equation is abbreviated as "Is":

IC = Is*e^vbe/vt
IE = IC/α

That isn't the same equation as (2) given above.

Is, or Ies, same thing. The "Is" notation is used when describing diodes. There is only 1 junction in a diode, 2 terminals. A bjt has 2 junctions, b-c, & b-e. Since the emitter & collector regions have different doping levels, they have differing scaling currents. The Ebers-Moll equations specify "Ics" for the c-b scale current, & "Ies" for that of the b-e junction. I am not the least bit "confused". The alpha factor cannot be omitted. The collector current depends on alpha all the time for any bjt.

If the base region was so wide that no electrons (npn polarity) make it through the base to the collector, the equation Ic = alpha*Ies*exp((Vbe/Vt)-1), still holds. The result is zero Ic, since alpha would be zero. Without the alpha factor, the Ic value predicted by the equation is wrong. You can never omit alpha. The exponential without alpha is the emitter current.

Collector current is simply the emitter current times alpha. That is the most basic relation. Of course, Ie cannot be established w/o Ib & Vbe. Ib & Vbe are indispensable since Ie couldn't exist w/o them. Hence eqns 1) & 3) account for the functional relations involving Ib & Vbe.

All 3 eqns are important. All 3 variables, Ib, Vbe, & Ie, are indispensable and all necessary to produce Ic. The OP question was if the bjt is amplifying current or voltage. I answered that it amplifies BOTH. No other answer remotely makes any sense at all. Eqn 1) describes the bjt current gain, & eqn 2) its voltage gain. Eqn 3) describes the physics of bjt action at the individual charge carrier level. This is well known for over half a century. No one has ever successfully challenged this well established bjt model. BR.

#### BrownOut

##### Banned
First of all, Is is not the same as Ics. Neither is it the same as the diode coefficient.

Secondly, the equation that calculates collector directly does not come from the ebbers-moll model. Instead, it uses the difussion profile in the base as an approximation to collector current. It's a very common way to calcualte collector current as a function of base-emitter voltage. But it's not ebbers-moll

Thirdly, I'm not concerned about your confusion. I'm concerned about the student that's learning transistors, and hoping that he doesn't get confused into thinking his equation is the same as eqn2 and ignoring alpha.

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#### Claude Abraham

##### Member
First of all, Is is not the same as Ics.

Secondly, the equation that calculates collector directly does not come from the ebbers-moll model. Instead, it uses the difussion profile in the base as an approximation to collector current. It's a very common way to calcualte collector current as a function of base current. But it's not ebbers-moll

Thirdly, I'm not concerned about your confusion. I'm concerned about the student that's learning transistors, and hoping that he doesn't get confused into thinking his equation is the same as eqn2 and ignoring alpha.

As far as "Is & Ics & Ies" go, I don't follow. In a diode, Id = Is*exp((Vd/Vt)-1). "Is" is simply the scaling current of the p-n junction, which depends on temperature strongly, dimensions of the junction, doping profile, etc.

In a bjt, "Is" is usually understood to be "Ies". In the forward active region of operation, the b-c junction is reverse biased so that "Ics" is small compared to Ic. Hence Ic = alpha*Ies*exp((Vbe/Vt)-1). Often, rather than write "Ies", some just use "Is". It is the scale current of the b-e junction. Eqn 2) is indeed the Ebers-Moll eqn. Of course, eqn 1), Ic = beta*Ib, is NOT Ebers-Moll, nor is eqn 3).

"Is" & my "Ies" are simply the scaling current, or "reverse saturation current" of the b-e junction. They are the same thing. Look it up.

Also, you defined Ic as Is*exp( ), then Ie as Ic/a. That is not correct. Ie = Is*exp( ), & Ic = alpha*Ie. First the b-e junction is forward biased. Ib, Vbe, & Ie are established. The holes from the base move towards the emitter, the electrons move from emitter to base, & Vbe is present. No Ic exists at the start. Then a moment later, electrons emitted from the emitter transit through the base & continue into the collector. But a few holes from the base enter the emitter, & a few electrons in the base recombine & never reach the collector. The holes frpm base to emitter & the electrons from the emitter recombined in the base do not contribute to Ic. Thus alpha is defined as Ic/Ie. Ie comes first, then a moment later Ic is established. Ic is almost equal to Ie for a good device, not quite equal.

This is well established. Ie produces Ic, not vice-versa. Ie comes first along with Ib & Vbe. Then Ic follows. Ic is defined by alpha*Ie at the device level.

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#### BrownOut

##### Banned
I did look it up. They are not the same thing. Ies is given by ebbers-moll, which is a very accurate model but is not being used by the OP's equation. There are more than one way to calculate IC. They are two different models using two different coefficients. Both coefficients are a functions of geometry, carrier mobility and doping levels, among other things.

Writing IE=IC/α is completely valid, and identical to your verision of IC=αIE. It's not a matter of what happens first, the equaions follow simple algebraic manipulation without losing validity.

Using a different model for analysis does not in any way attempt to invalidate ebbers-moll. It remains a relative accurate model to use, if one wishes. However, the equations that the rest of us are using gives very good results too.

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