I was looking to hear some other examples besides the voice and speaker case.
It makes it much more interesting to learn about all these types of amplifiers when you treat the load not just as a resistor nor the input as a sine wave.

 

Sorry, the equation I wrote is, once again, a different method of calculating collector current. The coefficient is different. I spent the entire morning yesterday going back over this just to make sure of what I am saying. It's a very common method of analysis.

Different equation/different coefficient. yeah, no brainer.

EDIT: I"m glad we've had this discussion. It forced me to go back over stuff I haven't looked at in decades.

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

 

Sources: Photo detector, antenna, thermocouple, etc.

Loads: motor windings, indicator light (ie - led), another amplifier stage, etc.

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

 

Thank you very much

 

If you're not confused enough yet, look at a hartley or colpitts oscillator. The feedback network is both the source and the load!

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

 

I just meant that this started to turn into a discussion of whether or
not Ies=Ics or not, when in some forums they are equal and in others
they are not.
I didnt meant that statement as an end all of all end's all either, so
if you wish to keep 'discussing' this thing that actually has two answers,
then please go ahead. I think it's somewhat interesting anyway.

I have to respond to another post too but i need to reply again.

 

Hi again,


I think you are getting something from this thread and that's good.
I just want to offer another point of view...

Think about this:
You can make an amplifier from either a voltage dependent current
source (VDCS) *OR* a current dependent current source (CDCS), both of
which will work. To amplify a voltage with the VDCS you only need apply
the input voltage, then convert the output to a voltage with a resistor or
in the load itself. To amplify a voltage with a CDCS, you first need to
convert the input voltage to a current (with say a resistor) and then
convert the output to a voltage also with either a resistor or in the
load itself.

When the transistor is looked at as an VDCS we think about it in
those terms and design accordingly. When it is looked at as a CDCS we
think about it differently and use different equations.

For example, it's much faster to use Ic=B*Ib when we want to drive
a relay that requires 100ma and we know the min gain of the transistor
is 100...we know we can get by with only 1 or 2ma base current.
In this case, it would have been much more abstract to think about it
in terms of what the base voltage is doing while the collector current
changes...almost a waste of time. We'd have to calculate the base
current anyway.

 

Transistors can be used to provide voltage gain, current gain or both ( or neither ) It can be used as an "impeadance transformer" to accept a signal with high source impeadance and deliver into a low impeadance load. It can be used as a switch. It can be used to transform a signal to one of a different shape. It can be used as a muliplier or as a voltage reference. Sometimes, the "input" might not even be electrical, as in the case of thermal input.

The transistor is one of the most amazingly versital devices ever invented!

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

 

Ok, 1 point at a time is fair. Regarding Ics & Ies, they are NOT equal in general. The collector region is doped lighter than the emitter. So the b-c jcn has a different "Is" reverse sat current than the b-e jcn. In Shockley's diode eqn, Id = Is*exp( ), the "Is" is that scaling current, the reverse sat value. In a diode there is but 1 junction, so "Is" describes the scaling current for that 1 junction.

In a bjt, "Is" differs because there are 2 junctions not just 1 like the diode case. Since the doping densities differ, we must clearly differentiate betwen Ics & Ies, the reverse sat currents, or scaling currents if you prefer, for the 2 junctions, c-b, & b-e.

Regarding alpha, every text which *derives* the E-M relation has no alpha in the Ie eqn, and includes alpha in the Ic eqn. So, Ies*exp( ) is the emitter current. The collector current is alpha*Ies*exp( ).

Any text which presents Ies*exp( ) as collector current is simply rounding alpha to 1, a good approximation. If "Is" is used instead of "Ies", then it is *understood* that "Is" is referenced to the *b-e* junction, not the b-c junction. Since bjt's are usually operated in the active region, this is understood, since the b-c junction is reverse biased. When used as a saturated switch, however, both Ics & Ies come into play since BOTH the b-c & b-e junctions are forward biased. You cannot just use "Is" in that case. You must acknowledge the scale currents for both junctions, seldom equal due to differing doping levels. When you say "in some forums they're equal & in some they aren't", well, whatever.

Happy 4th of July to all.

__________________
Claude

 

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).

 

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.

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

 

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!

 

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.

__________________
Claude

 

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.

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

 

Sometimes we have to read between the lines.

Happy Fourth to you all too!