# Calculating the power dissipated by a resistor on the emitter of a BJT

#### SantaCruz1

Joined Jul 27, 2017
6
I don't understand how I would draw the hybrid pi model for this circuit. Any help would be greatly appreciated. #### WBahn

Joined Mar 31, 2012
26,398
The Hybrid-pi model is a three-terminal subcircuit that goes in place of the transistor. Your DC supplies get turned off, and low-frequency capacitors are replaced with shorts.

You need to show your best attempt from here.

#### SantaCruz1

Joined Jul 27, 2017
6
The Hybrid-pi model is a three-terminal subcircuit that goes in place of the transistor. Your DC supplies get turned off, and low-frequency capacitors are replaced with shorts.

You need to show your best attempt from here.
I should have stated my question better. I think I'm getting thrown off by the capacitors in the circuit. What exactly constitutes "low-frequency". When creating my equivalent circuit, I've tried converting Ce to a resistance value.
Ce = 1/jwC = -31.831i

Then I calculate Ze = Re||Ce = 6.27233-30.5429i
ER = Ze||RL = 14.35-22.7915i

Now I try to calculate the current of the emitter.

b= gm*rpi = 30

Ib = (Vcc-Vbe)/(Rb + (1+b)*ER) = (24-0.7)/(1950+ (30+1)*(14.35-22.7915i) = 8.94+2.64i mA

Ic = B*Ib

Ie = Ic+Ib

I then use a current divider twice. Once to calculate the current going into Re||Ce, Iz = Ie*[(RL)/(Re||Ce+RL)]

And then again to find the current in Re, Ire = Iz*[(Ce)/(Ce+Re)]

Pre = (Ire)^2*Re

Obviously I'm getting the wrong answer. I just don't see where I'm going wrong.

Last edited:

#### WBahn

Joined Mar 31, 2012
26,398
How are you getting that the impedance of Ce is -j31.831 Ω ?

#### SantaCruz1

Joined Jul 27, 2017
6
How are you getting that the impedance of Ce is -j31.831 Ω ?
I apologize, I was looking at the next problem that asks for the value at 50kHz.
It should be -j(1/2pi*f*C) or -159.155j

I'm going to edit the original post. Does the math seem to be on the right track?

#### WBahn

Joined Mar 31, 2012
26,398
I haven't look at your math -- after seeing the very first computation being so far off it didn't make much point to continue.

To answer your other question, a capacitor is a "low frequency" capacitor if its impedance in the frequency range of interest is low enough that it can be considered to be a short circuit to a reasonable approximation.

Here you have the capacitor having the same magnitude of impedance as the resistor it is in parallel with, which indicates that this is on the edge of a change in frequency response. You have to be careful assuming it is a low frequency capacitor at this frequency.

Note that your problem asks for the average power dissipating in a resistor and you have two signals superimposed, one at DC and the other at 10 kHz. What do you know about how powers add in this situation?

#### SantaCruz1

Joined Jul 27, 2017
6
I would calculator the complex power (S=VI*) then use the real part of it

#### WBahn

Joined Mar 31, 2012
26,398
Shouldn't the DC component become the magnitude of the of the AC sinusoidal signal?
Uh... no.

What are you basing that claim on?

If what you are claiming is true, then the AC signal at the output is ONLY dependent on the magnitude of the DC component. Then what is supposed to happen as the magnitude and frequency of the input signal changes?

#### SantaCruz1

Joined Jul 27, 2017
6
Well what about using the DC component as the magnitude of the AC sinusoidal. The voltage would oscillate about the DC value.

#### WBahn

Joined Mar 31, 2012
26,398
Well what about using the DC component as the magnitude of the AC sinusoidal. The voltage would oscillate about the DC value.
The voltage is always going to oscillate about the DC value -- what other option is there?

In general, even making such an attempt it very poorly defined. Consider the output in this circuit -- what IS the magnitude of the DC component at the output? Hint -- you have a "very large" DC blocking capacitor between the amplifier and the load!

There is no basis for claiming or using the DC component as the magnitude of the AC signal. In this case, you have a specified magnitude of the AC input signal -- it is 1 Vpp at 10 kHz.

#### SantaCruz1

Joined Jul 27, 2017
6
So the only voltage going through Re is the AC input signal? Does that mean Vre = Vrl because they're both in parallel?

#### WBahn

Joined Mar 31, 2012
26,398
So the only voltage going through Re is the AC input signal? Does that mean Vre = Vrl because they're both in parallel?
No, Re has both a DC and an AC current going through it. The load only has an AC signal in it because the DC signal is blocked.

For DC, Re is NOT in parallel with Rl. But for AC (signals at sufficiently high frequency) they ARE in parallel to a good approximation.

Remember, splitting the analysis into DC and AC components is nothing more than superposition. The presence of reactive elements changes the distribution of voltages and currents at different frequencies, but the total signal in any one component is the sum of the signals at all the various frequencies.