Hi...

I'm now trying to calculate the curent gain for the same amplifier but now without Ce capacitor, meaning unbypassed Re.

I'm having some issues trying to find the input current.
I'm reading this link at the "CE stage with Re unbypassed" but I can't understand the Inpu impedance

They say that impedance is base impeance in parallel with biasing resistors. And hen they say that base impedance is Zb = hie + (hfe + 1)*Re. I don't understande this equation in particular! Can you help understanding it?

 

Hi...

I'm now trying to calculate the curent gain for the same amplifier but now without Ce capacitor, meaning unbypassed Re.

I'm having some issues trying to find the input current.
I'm reading this link at the "CE stage with Re unbypassed" but I can't understand the Inpu impedance

They say that impedance is base impeance in parallel with biasing resistors. And hen they say that base impedance is Zb = hie + (hfe + 1)*Re. I don't understande this equation in particular! Can you help understanding it?

What is the current in RE? It's Ib+Ic isn't it? But since Ic = β Ib, the current in RE is Ib+β Ib = (1+β)Ib, and the voltage across RE is RE*(1+β)Ib. In a similar manner calculate the voltage across hie. Then the voltage at node B is equal to the sum of the voltages across hie and RE. The current into the base is Ib, so divide the voltage at node B by the current into the base and you will have Zb.

 

But the input curent shouldn't be Iin = I_Rbb + I_b. Looks like hat what you're telling me is that, the so called input current, is from node B to the right, right? Or am I misunderstanding you?

Editing;

Fist of all, let me ask the following:

when we are considering the hybrid model, isn't (Rb1//Rb2) in parallel setup with (hie + Re)?
So I would say that the input impedance, that I call Zin, would be Zin = (Rb1//Rb2)//(hie + Re). But probably I'm wrong!

 

But the input curent shouldn't be Iin = I_Rbb + I_b. Looks like hat what you're telling me is that, the so called input current, is from node B to the right, right? Or am I misunderstanding you?

Iin is called Ii in this image. Ii splits between Rbb and hie, but Ib only goes into the base of the transistor.

Editing;

Fist of all, let me ask the following:

when we are considering the hybrid model, isn't (Rb1//Rb2) in parallel setup with (hie + Re)?
So I would say that the input impedance, that I call Zin, would be Zin = (Rb1//Rb2)//(hie + Re). But probably I'm wrong!

What they show on the page you linked is: Zi = RBB || (hie + hfe RE), and since RBB is Rb1||Rb2, you are almost correct.

 

Ok, so I'm doing the following for the current gain. Please help me confirming my calcs:

 

Ok, so I'm doing the following for the current gain. Please help me confirming my calcs:

View attachment 86414


View attachment 86415

I think your problem may consider the Iout in the expression \(Ai=\frac {Iout}{Iin} = \frac{\frac{Vout}{Rout}}{\frac{Vin}{Rin}}\) to be the current in Rload, rather than the total current out of the collector, so you should really be calculating \(Ai=\frac {Iload}{Iin} = \frac{\frac{Vout}{Rload}}{\frac{Vin}{Rin}}\)

The output current Io splits into the current I(Rc) and I(Rload), so you need to account for that by using the current divider rule at the junction of Rc and Rload.


Also, at the link you already gave, http://www.zen22142.zen.co.uk/Theory/tr_model.htm, the procedure under the heading "CE stage with RE unbypassed", doesn't have an Rload like you do.

You will have to deal with the fact that you have both Rc and Rload when calculating Av. The expression you have for Av is \(-\frac {hfe*Rload}{hie+Re(1+hfe)}\)

This is derived from an assumption that Vo = -Io*Rload, but that is only true if there is no Rc. In your circuit the output load is Rc||Rload.

You need to correct your calculation for Av.

 

Yes, I think the main problem is interpreting what stands for iOUT and vOUT and rOUT...

I think my teacher considers iOUT as current across RLoad, rOUT is when we consider RLoad = infinite (open circuit) and a voltage source vX is palced at the open circuit. The input voltage source, Vsig,is replaced by it's internal impedance. That way, rOUT = Rc because Ib = 0, therefor hfe*Ib = 0 and the current Ix (from Vx) flows sollely across Rc. That's why rOUT = Rc.

About vOUT, as we have a parallel setup between Rc and RLoad, vOUT = V_Rc = V_RLoad. I think I'm correct here!


Edited:

I have done this for the current gain. In the second line of the image, where I have "aI - ....." it should be "aI = ...."

 

Please try not to mix, the the amplifier output resistance with load resistance.
Output resistance is a resistance seen from the load perspective when we looking back into the amplifier output terminal from the perspective of the load resistor.



And your Av equation is wrong. Once again you have forgotten about RL.

 

ok.

aV = (hfe*Ib*(Rc||Rload))/(hie + (1+hfe)Re)

And the rest of the equation is correct? The one that has rIN/RLoad????

 

ok.

aV = (hfe*Ib*(Rc||Rload))/(hie + (1+hfe)Re)

Remove Ib from the equation.

And the rest of the equation is correct? The one that has rIN/RLoad????

Looks good.

 

Oh, thanks god and thank you...

My mistake to let Ib there... It simplifies with the Ib in the denominator!

 

Ok, 2 more questions and I think I'll call it an end...

1 - When we want to calculate the hie parameter, which is Vcc/Ib, this Ib value is the AC, DC or AC+DC value of Ib?

2 - When we recalculate the value of Rb1 for Maximum Possible Compliance, the values that we use for Ic or Ib are AC, DC or AC+DC values?

 

Ad1. Are you sure that Hie = Vcc/Ib ??
Ad2. DC

 

Ad1. Are you sure that Hie = Vcc/Ib ??
Ad2. DC

At least was what our teacher told us, that hie = Vt/Ib where Vt = 25mV...

 

In this case you need to use a DC base current in your calculation.

 

In this case you need to use a DC base current in your calculation.

So for my case would be:

hie = 12/(4.187/458.7) = 2.7kΩ

Isn't this value to high?

I've seen here that hie = Vbe/Ib.

Also seen this here:

perhaps it is helpful to use words instead of symbols only.
Therefore: hie (some call it h11) is the so called "short circuit input resistance". Here, short circuit means "signal short" (Vce=const).
This parameter hie is in direct relation to the dc bias base current Ib and, thus, to the dc collector current Ic.
Because of the known relation between Ic and Ib and because of Shockleys equation (resulting in the transconductance g=Ic/Vt, Vt=thermal voltage)
you can measure Ic, calculate the transconductance g and than use the equation hie=hfe/g.

I don't know what to consider!

 

Why you think that hie is to high ??

All this equations will give you exactly the same answer, so do not bother about them.

hie = (hfe + 1) * re = ( 458.7 + 1) * 25mV/4.196mA = 2.7kΩ where

re = Vt/Ie ≈ 25mV/Ie

hie = hfe/gm = 458.7/(4.187mA/25mV) = 2.7kΩ

 

Ok, thank you...

I saw in some forum someone saying that hie should be just a couple of dozens of ohms. Anyway I already sent this assignment to my teacher...

I just didn't was capable of confirming my calcs with LTSpice in terms og voltage and current gain! Values were quite different.

 

I just didn't was capable of confirming my calcs with LTSpice in terms og voltage and current gain! Values were quite different.

Which model of a transistor you have chosen ?
Because 2n2222 has a Hfe in rang 200A/A not 460A/A

 

Which model of a transistor you have chosen ?
Because 2n2222 has a Hfe in rang 200A/A not 460A/A

I've used BC547C that in LTSpice has 458.7 for hFE. Also used 0.6V for Vbe...