BJT amplifies current or voltage?

[BrownOut]

So, you may reasonably ask yourself this; if the author of a serious engineering text writes the following:

IC=Is*e(---)

why would he write Is if he really means Ies? He already has the coefficient that references the emitter, why would he change the notation? I doesn't make any logical sense. I doesn't make any mathemetical sense. It's very, very common in mathematics that if there are two terms, and one term does not significantly affect the result, that terms is just left out of the equation for simplification. The remaining terms are retained exactly as they were before the equation is simplified.

No! If the author means Ies, then he writes Ies. He doesn't arbitraily change the notation. These are two different coefficients.

 

[MrAl]

In the case of the relay, wouldn't it be more wise to have the transistor to act as a switch rather than as an amplifier?
For example, we used to work with a 5V @ 80mA relay, and since we wanted the relay driver to have low power dissipation, we used a 5V power source (VCC) and a BJT switch, that way the BJT only dissipated 80mA*VCE_SAT.
If we would have used it as an amplifier, we would have needed to forward bias it, first need to use around a 7.5V power source, and the BJT's power dissipation would be around 2.5V*80mA (2.5V is the VCE when the BJT is forward biased, and is found as closer as possible to the center of the high gain region).

Hi,


The switch could be looked at as simply a saturated amplifier.
The equation Ic=B*Ib is also used to simplify that, even though it
might be used as a switch. Of course then we would want to take
a close look at the saturation characteristics too as i think you noted.

For example, how to drive a relay that requires 40ma coil current
from a 10v supply?
Say we have a transistor with min gain of 40. How much base
current do we need to drive the NPN transistor properly and what
resistor could be used as a pullup to drive the transistor base from
the 10v supply?

The simple equation to use is Ic=B*Ib again, and we know Ic=0.040
and B=40, so rearranging the equation:
Ic=B*Ib
Ib=Ic/B
so
Ib=0.040/40=0.001 amps
so we double that to make sure the transistor stays in sat, to 0.002 amps.

Now to answer the question about what value resistor to pull up the base
with to the 10v supply in order to keep the relay on.
Since the voltage is 10v, the resistance would be 10/0.002 or 5k.
Note here we didnt even consider the base emitter voltage and still
came up with a reasonable design.
If we wanted to consider the base emitter voltage though, it wouldnt
be hard as we take that as a constant equal to about 0.7v, or to be
sure and make things even simpler, 1v...
Now we have 10v-1v=9v, and 9v/0.002a equals 4.5k base pull up.
Here, we estimated the base emitter voltage and didnt have to consider
it's entire equation.
If we wanted to make sure this was going to work, we could look at
what happens when the base voltage changes by say 0.1v and see
what effect this has, but we can already see that the small changes in
base current wont have too much effect so this is mostly ignored.

Now repeat the above process using the exp(Vbe/VT...) equation and
see the difference.

Happy Fourth to you all too!

 

[Claude Abraham]

Regarding the bjt as a switch, i.e. in saturated region of operation, you must use the FULL E-M eqn. It is:

Ic = alpha_n*Ies*exp((Vbe/Vt)-1) - Ics*exp((Vbc/Vt)-1).

Ie = Ies*exp((Vbe/Vt)-1) - alpha_i*Ics*exp((Vbc/Vt)-1).


The extra terms involving "exp((Vbc/Vt)-1)" & "Ics" are to account for the extra component of current due to the reverse mode of operation of the fictitious upside down bjt. In saturation both b-c & b-e junctions are forward biased. Using the 1-term approximations for Ic & Ie gives the wrong answer by a country mile.

The bjt is modeled as 2 devices, 1 in the normal or forward mode, with "alpha_n", & 1 in the inverse mode with "alpha_i". The collector is the emitter for the upside down bjt. The 2nd term subtracts from the 1st giving the correct value of current.

Believe me, I'm not making this stuff up. It's been known since 1954. Ics, Ies, alpha_n, & alpha_i, are as old as the hills. Trust me. Happy 4th again.

 

[BrownOut]

Believe me, I'm not making this stuff up. It's been known since 1954. Ics, Ies, alpha_n, & alpha_i, are as old as the hills. Trust me. Happy 4th again

I agree. The one term solution may be used to determine if the transistor is in saturation or not. ( if the B-C junction if forward biased ) Once it's determined that the transistor is staurated, that solution is no longer useful.

 

[MrAl]

Hi,



Of course then we would want to take
a close look at the saturation characteristics too as i think you noted.

Sometimes we have to read between the lines.

Happy Fourth to you all too!

 

[alphacat]

Hi,


The switch could be looked at as simply a saturated amplifier.
The equation Ic=B*Ib is also used to simplify that, even though it
might be used as a switch. Of course then we would want to take
a close look at the saturation characteristics too as i think you noted.

For example, how to drive a relay that requires 40ma coil current
from a 10v supply?
Say we have a transistor with min gain of 40. How much base
current do we need to drive the NPN transistor properly and what
resistor could be used as a pullup to drive the transistor base from
the 10v supply?

The simple equation to use is Ic=B*Ib again, and we know Ic=0.040
and B=40, so rearranging the equation:
Ic=B*Ib
Ib=Ic/B
so
Ib=0.040/40=0.001 amps
so we double that to make sure the transistor stays in sat, to 0.002 amps.

Now to answer the question about what value resistor to pull up the base
with to the 10v supply in order to keep the relay on.
Since the voltage is 10v, the resistance would be 10/0.002 or 5k.
Note here we didnt even consider the base emitter voltage and still
came up with a reasonable design.
If we wanted to consider the base emitter voltage though, it wouldnt
be hard as we take that as a constant equal to about 0.7v, or to be
sure and make things even simpler, 1v...
Now we have 10v-1v=9v, and 9v/0.002a equals 4.5k base pull up.
Here, we estimated the base emitter voltage and didnt have to consider
it's entire equation.
If we wanted to make sure this was going to work, we could look at
what happens when the base voltage changes by say 0.1v and see
what effect this has, but we can already see that the small changes in
base current wont have too much effect so this is mostly ignored.

Now repeat the above process using the exp(Vbe/VT...) equation and
see the difference.

Happy Fourth to you all too!

Thank you very much
I just love reading your posts here

 

[MrAl]

Thank you very much
I just love reading your posts here

Hi again,


You're very welcome, and i enjoy reading your questions and the discussion
between other members too.