# Finding the lower critical frequency for an amplifier

Discussion in 'General Electronics Chat' started by epsilonjon, Jun 5, 2012.

1. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Hello

I have a quick question about deriving the lower critical frequency for the output circuit of an amplifier. Thevenizing the output circuit we get this:

The output voltage can be written as

$V_{out} = \frac{R_L}{R_L + R_C +\frac{1}{j \omega C}}V_{coll}$

so just concentrating on the magnitudes we get

$V_{out} = \frac{R_L}{\sqrt{(R_L + R_C)^2 +X_C^2}}V_{coll}$

I think I am right in saying that the lower critical frequency is defined as the frequency at which the voltage has decreased by a factor of 1/√2 (i.e. 3dB)? So should I not solve the above equation for $X_C$ in terms of $R_L$ and $R_C$ when $V_{out} = \frac{1}{\sqrt2}V_{coll}$?

My book states the answer as

$f_{cl(output)} = \frac{1}{2 \pi (R_C + R_L)C_3}$

i.e. $X_C = R_L + R_C$, but this does not give an attenuation of -3dB does it?! The answer I get is

$X_C = \sqrt{R_L^2 - R_C^2 - 2R_LR_C}$

Last edited: Jun 5, 2012
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
The mid-band output factor K would be given by

$K=\frac{R_L}{R_C+R_L}$

With the capacitance C3 coming into the picture the factor K' would be

$K'=\frac{R_L}{R_C+R_L-j{X_C}_3}$

or

$|K'|=\frac{R_L}{\sqrt{$${R_C+R_L}$$^2+{{X_C}_3}^2}}$

The output will drop by 3dB when

$\frac{|K'|}{K}=\frac{1}{\sqrt{2}}$

or

$\frac{R_C+R_L}{\sqrt{$${R_C+R_L}$$^2+{{X_C}_3}^2}}=\frac{1}{\sqrt{2}}$

or

$2$$R_C+R_L$$^2=$$R_C+R_L$$^2+{{X_C}_3}^2$

or

$$$R_C+R_L$$^2={{X_C}_3}^2$

& finally

${X_C}_3=\sqrt{$$R_C+R_L$$^2}=$$R_C+R_L$$$

3. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Ahhh yes how stupid of me! I was forgetting that it is the gain relative to the midrange gain which is the important quantity.

Last edited: Jun 5, 2012
4. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Sorry but I have another question on the same topic, but this time regarding the bypass circuit. Say we have a common-emitter amplifier as follows:

We can write the overall gain of the amplifier as

$\frac{V_{out}}{V_{in}} = \frac{V_{out}}{V_c} \frac{V_c}{V_b} \frac{V_b}{V_{in}}$

so when any one of these three individual gains reduces from its midrange value by a factor of 1/√2, the overall gain will have reduced by roughly 3dB, and the frequency at which this occurs is called the lower critical frequency?

I think the bypass circuit only affects the middle one of these, $V_c / V_b$? If we make the approximation $I_c = I_e$ then we have

$\frac{V_c}{V_b} = \frac{I_e(R_L ||R_C)}{I_e(r'_e + (-jX_c||R_E))} = \frac{R_L||R_C}{r'_e + (-jX_c||R_E)}$

At a midrange frequency, $X_c=0$, and the gain is

$\frac{V_c}{V_b} = \frac{R_L||R_C}{r'_e}$

My method to find the lower critical frequency for the bypass circuit would be to set the magnitude of the first expression equal to 1/√2 times the midrange gain, and then solve for $X_c$ to find the frequency. Would this work?

My book does it another way, by Thevenizing the circuit connected across the bypass capacitor, but it seems incorrect to me.

Thanks once again

Last edited: Jun 6, 2012
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
1,088
This approach will not work.t Try to find resistance seeing from C2 capacitor terminal.

6. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Can you explain why the approach will not work, then maybe I will understand better what is going on?

Thanks.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Using say the hybrid-pi model (as I attempted) one could come up with something akin to

$\frac{v_{out}}{v_{in}}=$$\frac{v_b}{v_{in}}$$ $$\frac{v_{be}}{v_b}$$ $$\frac{v_{out}}{v_{be}}$$$

One can then formulate these individual components in terms of the mid-band and low frequency versions.

The problem then is twofold .....

Firstly, the presence of the frequency dependent negative feedback in the emitter leg which contributes to the complexity of the individual terms. The first term Vb/Vin including the input and bypass capacitances is particularly difficult to deal with.

Secondly, how does one apportion the contribution at the low frequency -3dB from each of the three terms when they are expressed as a ratio of their low-frequency to mid-band transition?

8. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Thanks for your reply. In hindsight I can see that the bypass capacitor does feature in Vb/Vin since you have to include the input resistance at the base.

My thinking was that if the midrange gain is

$(A_v)_{mid} = (\frac{V_{out}}{V_c})_{mid} (\frac{V_c}{V_b})_{mid}(\frac{V_b}{V_{in}})_{mid}$

then when any one of the individual gains decreased by 1/√2, the new gain divided by the midrange gain would be

$\frac{A_v}{(A_v)_{mid}} = \frac{\frac{1}{\sqrt2}(\frac{V_{out}}{V_c})_{mid} (\frac{V_c}{V_b})_{mid}(\frac{V_b}{V_{in}})_{mid}}{ (\frac{V_{out}}{V_c})_{mid} (\frac{V_c}{V_b})_{mid}(\frac{V_b}{V_{in}})_{mid}} = \frac{1}{\sqrt2}$

so

$20log(\frac{A_v}{(A_v)_{mid}}) = 20log(1/ \sqrt2) = -3dB$

Am I right in thinking that this does not work because when one gain decreases, so too will the other ones? I can understand that, but then I don't see how the method of analyzing each circuit separately (input/bypass/output) works. In the book I'm reading they analyze each circuit whilst assuming that the capacitors in the other two circuits are shorts. Do the effects somehow cancel each other out perfectly when you do that?

Anyway, if I look at the bypass circuit (whilst shorting C1) after a bit of work you get

where Rth = R1||R2||Rs.

This is where I get stuck cos I don't see exactly how this circuit affects the gain of the amplifier. It was easier with the input and output circuits since they have an obvious attenuating effect.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
1,088
The math involved to find frequency response for Ce capacitor is not easy one.
This is why we use approximation method. In this approximation method we need to find resistance seeing from Ce capacitor terminal.
If we looking into emitter terminal from Ce capacitor perspective, we see two ways for the current to flow.
First is through Re resistor. And we have the second path through re = 1/gm plus (R1||R2||Rs)/(β+1).

Rse = (R1||R2||Rs)/(β+1) + re

So as we can see Ce capacitor see two parallel connected resistors

Re||Rse
.

But in reality we ignore Rs and R1||R2 and re << RE we end up with this equation

$Fd = \frac{1}{2*\pi*re*Ce}$

In reality the frequency response will be close to this one

And we can find this equations by using hybrid-pi model.

File size:
13.8 KB
Views:
188
10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
When one does the complete small signal analysis it turns out that the relevant poles in the overall amplifier gain transfer function correspond extremely well to those found using the piecemeal short and open circuit time constants approach followed in your book. It's one of those fortuitous occurrences / quirks that make the analysis a lot easier.

11. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Thanks for both your replies. I guess I will just have to accept it until I move on to a higher level book.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
You don't have to accept anything on our say so. The proof is within the grasp of anyone with the (essentially) algebraic skills and circuit analysis skills - such as nodal analysis. It's then a matter of pure perseverance and recognition of where a few simplifying adjustments can be made.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I thought it possibly worth documenting a comparison of the notionally exact factors [poles] for the low frequency response vs the pole values obtained using the approximate methods discussed earlier in this thread. I've used the schematic of post #4 by way of illustration. The results are correct to the best of my knowledge but the algebra became quite complex so I wouldn't discount the likelihood of errors.

File size:
48 KB
Views:
17