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# Small Signal Input to a BJT

http://img528.imageshack.us/img528/3466/schematiccharkz4.png (included schematic and characteristic)

-The amplifier uses a BJT whose characteristic I provided has a different load line.-
The amplifiers load line connects between 4 mA on the Ic axis and 8 V on Vce. The amplifiers operating point occurs where the load line crosses Ib = 20 microA.
-Rc = 8V / 4mA = 2000 ohms
Vcc = 2mA * 2000 ohms + 4V = 8V
Rb = (8-.7)/20 microA = 365000 ohms

Given a small signal input:
What would be the net resistance in the signal input ciruit, the amplitude of the base signal current, the max. and min values of the total base current and collector voltage, and the gain of the amplifier?

Small signal:
Signal amplitude = 3mV
source resistance = 4KOhms
Transistors AC beta = 150
Transistors re = 40 ohms
Coupling capacitor = 1 microF
Signal frequency = 1 MHz

I am working on this one... So far I was wondering if using the equation for Rin = 1/(1/R1+ 1Rb) would be correct??

1/(1/365KOhms + 1/4Kohms) = 3956.64 Ohms?? I think I am way off though..

is the amplitude of the base signal current: Ib = (Vcc-.7)/Rb?? that seems incorrect to me..

would the voltage gain be Av = -Rc/Re   =>   -2KOhms/40 Ohms = -50 Ohms??
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wolfinator
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1 Solution

Commented:
What about the characteristics of the transistor?
Shouldn't that figure in somehow?

Do you know how to find the beta for the transistor from the load line?
Have you seen or heard of the hybrid pi model yet?
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Commented:
Maybe you don't have to worry about the hybrid pi model yet if they give you a value for re.

I think perhaps you are supposed to include the effects of re in the base resistance.

That would be something like 1/(1/4k + 1/365k + 1/(beta*Re)).

The current amplitude would be Vin/Rb.     Vin = 3 mV.

I think you have the correct gain equation, but the units are wrong.
Voltage gain is dimensionless.
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Author Commented:
aren't Re and re different than eachother?? b/c this problem gives the transistors re..

I believe I found beta to be 100 in the previous problem(I lost the piece of paper I had all my work on).. So, I guess that would be  1/(1/4KOhms + 1/465KOhms + 1/(100*40Ohms??))  ??? I get 1989.1 Ohms for this.

the current amplitude would then be 3mV/365KOhms??  something doesn't seem right there

and so the voltage gain is correct, it just is -50 with no units??

The other two questions from above though, how do I go about finding the max and min values of the total base current and collector voltage???

Thank you
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Commented:
re and Re are the same.

>> the current amplitude would then be 3mV/365KOhms??
something doesn't seem right there

The base current amplitude should be Vin/Rb.    You found Rb = 1989.1 ohms.

The max and min values of the base current are the operating point value plus and minus the amplitude from above.

You have to find the max and min values of the collector voltage from the load line after you figure out what the base current is doing.

=====================================================
>> would the voltage gain be Av = -Rc/Re   =>   -2KOhms/40 Ohms = -50 Ohms??

This is true sometimes.  If Rc/Re is much less than beta for example.
But it is probably not true here.
And the "ohms" always cancel to give a dimensionless number.

Once you find max and min values of the collector current from the load line, you will be able to find the gain.

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Commented:
Correction:

Once you find max and min values of the collector VOLTAGE from the load line, you will be able to find the gain.

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Author Commented:
so Rin is the same as Rb??
Rin = 1/(1/4k + 1/365k + 1/(100*40)) = 1981.1 Ohms

amplitude of base signal current = Vin/Rb = 3m/1981.1 = 1.514 microA ?

i found the operating point to be at (2mA, 4V)   ..so I am kind of confused as to what I am adding 1.514microA to..

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Commented:
You were given the quiescent/steady-state value of base current:

>>  The amplifiers operating point occurs where the load line crosses Ib = 20 uA

So max Ib = 21.514 uA    and    min Ib = 18.485 uA
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Author Commented:
oh ok..haha i was wondering if I was looking too far into it actually..

-so now that I see the base current fluctuating 1.514 microA, and given the load line from 4mA to 8V how do I go about finding the max and min values of the colector voltage(isn't that Vce?)
wouldn't Vce be V -IRc??

once I find the max and min of the collector voltage do I use the equation -IcRc/Vin?
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Commented:
The collector voltage is the output voltage, i.e. the voltage at the collector.

It is (Vcc - IcRc)  ==>  (Vcc - B*Ib*Rc)   where B = beta.

Once you find the amplitude of Vc, you can divide by Vin to find the overall gain.
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Author Commented:
should I be using the Beta provided(the transistors AC beta = 150??)  and will I be using the coupling capacitor or the signal frequency at all for anything??

I still can't figure out the minimum and maximum values of the collector voltage..

do I just use (Vcc-B*Ib*Rc)   and use  (8v-100*18.485microA*2kOhms)  & (8v-100*21.514microA*2kOhms)
-Or should I have been using 150 as the beta for this entire problem??
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Commented:
For the Load Line Figure you posted at the beginning of the problem:
Ib =20 uA  ==>  Ic =2 mA     and     Ib =30 uA  ==> Ic =3 mA

This clearly indicates a beta of 100.

Of course the load line for this problem is different, but it looks to me like the load line will be parallel to the one in the figure and beta would still  be 100.

If the transistor data is not consistent with the curves, you have a problem.

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Author Commented:
ok..then it is definitely a beta of 100..

for the minimum and maximum collector voltage do I just use (Vcc-B*Ib*Rc) from what i posted before??
-using  (8v-100*18.485microA*2kOhms)  & (8v-100*21.514microA*2kOhms)??

the min and max would be 4.303v, 3.6972  ..that's not making sense to me though
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Author Commented:
sorry for all the problems this is causing, i think it is almost finished...just this collector voltage stuff and the gain(which we have a solution to almost anyway)..
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Commented:
That looks correct.

The thing to realize is that the voltage at the collector varies around the operating point.

The bypass capacitor on the output blocks the DC compontent, so the output voltage oscillates around zero.
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Author Commented:
Is the voltage gain going to be a negative number??

Should I be using -IcRc/Vin as the equation??

(-2mA*2kOhms)/3mV??
= -1333.33
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Author Commented:
or if it's Vc/Vin... Do I use
Vc = 3.8972-4.303 = -.4058

-.4058/3mV = -135.26 ?? hopefully..
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Author Commented:
are you sure I don't use 150 as the Beta in these problems... I mean, it is one of the givens in the small signal

Transistor's AC Beta = 150

Shouldn't I use 150 when calculating Rin, and that would change my values for everything else?
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Commented:
>> Is the voltage gain going to be a negative number??
Yes.

>> Should I be using -IcRc/Vin as the equation??
Yes, but only the part that is varying, and that depends on Ib, beta, the
input voltage and the input resistance.

>> Vc = 3.8972-4.303 = -.4058

Is 3.8972 the operating point??  I thought it was supposed to be 4.00 V.
This is the maximum output so the sign is wrong.
And you get the maximum output when the input is negative, so the gain is
negative as well.

I don't know how to reconcile the discrepancy between the given value of beta (150) and the given load line.

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Author Commented:
ooh..so I should use the operating point of 4V as the Vc?? Thanks for clearing that up, if that's the case..

I would then use

-4V/3mV = -1333.33V??
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Commented:
>>  the min and max would be 4.303v, 3.6972

If that means (4.000 + 0.303) for the max   and (4.000 - 0.303) for the min

The gain would be (+0.303V)/(-3mV)   or    (-0.303V)/(+3mV)  ==>  101

And the volts still cancel.
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Author Commented:
you lost me.. I got 4.303V and 3.6972V from
(8v-100*18.485microA*2kOhms)  & (8v-100*21.514microA*2kOhms)

I thought maybe I was supposed to subtract the minimum from the maximum collector voltages when doing Vc/Vin, but I guess that's wrong..

Should I just use the (operating point/Vin) for the gain??
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Author Commented:
I got really confused on this operating point situation..should I be using the difference b/w the minimum and maximumvalues of the collector voltage or should I ust 4V from the operating point??
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Commented:
The output operating point is 4.000 V.

This is the voltage on the collector sits when there is no input signal.
It is also the average value and midpoint for a sinusoidal input.
But it has nothing to do with the small signal gain.

You need to develop this sort of equation:

Vc = (8V  - Ic*2k)
= (8V - B*Ib*2k)                                     B is beta
= (8V - (20 + B*Vin/Rin*sinwt)*2k)        Rin is the equiv input resistance

Then you can break the equation into steady state and time varying terms:

Vc = [8V - 20uA*2k] - [B*Vin/Rin*sinwt*2k]

The first term is the operating point:   [8V - 20uA*2k] = 4 V

The second term is the gain:             - [Vin/Rin*sinwt*2k]      (note the minus sign)

Rin is the hardest thing to calculate.
It is a combination of the source resistance, the base bias resistance, and the equivalent base input resistance of the transistor.
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Author Commented:
Didn't we calculate Rin earlier in the problem??

Rin = 1/(1/4kOhms + 1/365kOhms + 1/(B*Re))   => 1/(1/4kOhms + 1/365kOhms + 1/(100*40)) = 1989.1 Ohms

-[Vin/Rin*sinwt*2k]
-[3mV/1989.1 Ohms*sinwt*2k]    how can I find w and t
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Commented:
The Rin calculation is correct.  That is how you figure the relationship between
the input voltage Vin and the time varying part of the base current.

Sorry for the confusion about sinwt.  I didn't write it properly.  It should be
sin(w*t)    where w is omega (the frequency of the input sinusoid) and t is time.
You don't have to figure them out.

The way to avoid confusion is to seperate the steady-state and time-dependent
parts of the problem.  Things become clearer once you include the sin term
explicitly.
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Author Commented:
isn't the signal frequency given in the problem as 1 MHertz??

So would I just exclude the time part of the problem and use the -[Vin/Rin* sin(w)*2K]??
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Commented:
Yes the amplitude of the output is the coeeficient of the sinusiod, but you left out beta:

-[B*(Vin/Rin)*2K]*sin(w*t)     Note that it has units of volts -- ohms cancel

The gain would be
-[B*(Vin/Rin)*2K]/Vin  =  -[(B*2K)/Rin]      which is dimensionless.

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Author Commented:
so the amplifier voltage gain is
-[(B*2k)/Rin]
[(-100*2kOhms)/1989.1Ohms] =  -100.55??

If so, thank you for all of your incredible help..

I realized I have a follow up problem to this question, but it shouldn't be too bad..

I am going to post it and then link it to here if you could help me with that. It deals with finidng the mid band and low frequency of this amp. I can find that with an RLC circuit and such, but have never done it with a BJT.. Thank you.
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Author Commented:
Here is the link to the follow up..thanks for all the help and hopefully you can help me with this one also..

http://www.experts-exchange.com/Other/Miscellaneous/Q_22516313.html
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