Class A amplifier problem

[cartman007]

Hi everyone

I have a couple of questions regarding this question.


Ok so in order to sketch the wave for all the points "A to J" I first need to know if it clips or saturates.
I was not given the Gain of either DC or AC so I figured that I am going to have to calculate that first.

This is how I got the DC gain.
Vb = Vcc*(R2/R1+R2) = 20(10000/10000 + 39000) = 4.08V
VE = VB - VBE = 4.08 - 0.7 = 3.38V
Rth = (39000*10000 / 49000) = 7960ohm
IB = 4.08/7960 = 512uA
IE = 3.38 / (180 + 820) = 3.38*10^-3 A
IE/IB = BetaDC = 3.38*10^-3/512*10^-6 = 6.6 Gain.

Then I would have to find the QPoint to se if it swings right... I ma not sure.
VC = Vcc -Ic.Rc = 20 - (2700)(3.38*10^-3) = 10.87v
Vce = Vc - Ve = 10.87 - 3.38 = 7.494V <--- I have no idea if this is true. It does not make sense to me. This would make more sense to me Vcc = Vc + Ve + Vec ; 20 - 10.87 - 3.38 = Vec = 5.75V

So QPoint is Qv = 5.75V and Qi = 3.38*10^-3 A

r`e = 25mV / IE = (25mV / 3.38*10^-3) A = 7.396 Ohm
Av = Rc/r`e+RE1 = 2700 / (7.396+180) = 14.4

I am not quite sure if any of this is correct and where to go from here?

So assuming I have done everything correct. It will mean that my maximum Peak will be 7.2V and the supply can give that. Although the lower peak will but cut off.

I am not sure if this is done correct... feel free to bash the work.

Thanks

 

[Pjdd]

You got things almost right. Here are a few points to reconsider:

Rth = (39000*10000 / 49000) = 7960ohm
IB = 4.08/7960 = 512uA
IE = 3.38 / (180 + 820) = 3.38*10^-3 A
IE/IB = BetaDC = 3.38*10^-3/512*10^-6 = 6.6 Gain.

Beta is an intrinsic property of the transistor and cannot be derived this way. It's taken either from the manufacturer's data or from actual measurement.

Ib flows through R1 in addition to the current through R2. This modifies Vb from your basic calculations but with proper bias resistor values, the change is usually small. Since no beta value or the transistor type number is given, you are probably expected to ignore the difference.

Then I would have to find the QPoint to se if it swings right... I ma not sure.
VC = Vcc -Ic.Rc = 20 - (2700)(3.38*10^-3) = 10.87v
Vce = Vc - Ve = 10.87 - 3.38 = 7.494V <--- I have no idea if this is true. It does not make sense to me.

This is correct and makes perfect sense.

This would make more sense to me Vcc = Vc + Ve + Vec ; 20 - 10.87 - 3.38 = Vec = 5.75V

So QPoint is Qv = 5.75V and Qi = 3.38*10^-3 A

This makes no sense.

r`e = 25mV / IE = (25mV / 3.38*10^-3) A = 7.396 Ohm
Av = Rc/r`e+RE1 = 2700 / (7.396+180) = 14.4

I am not quite sure if any of this is correct and where to go from here?

This would be correct except that R6 (8k2) is in parallel with Rc for AC signals. The effective load for AC is 2k7||8k2. You should use this to calculate the voltage gain and the maximum possible output swing. I'll leave it to you to do the modified calculations and see if it will clip.

N.B.: Other factors come into play for real-world amplifiers. For example, r`e can be regarded as constant only for very small signals. With large signal swings as in your problem, it changes as IE changes. The effect is swamped out to some extent by its being in series with the much larger 180-ohm resistor. But it's something to keep in mind for the future.