# Frequency response of CE amplifier function of output coupling capacitor and bypass capacitor?

Discussion in 'Homework Help' started by Saviour Muscat, May 9, 2015.

1. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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Hi guys

Now I am studying the frequency response of a common emitter amplifier with output coupling capacitor and bypass capacitor. Please refer to the attached image of the Malvino. I read the text several times but I am not 100% certain that I understood in full. What I didn't understood is as follows:

"In the midband of the amplifier ,the emitter is at ac ground, and the voltage is maximum, as shown in fig 16-10b as the frequency decreases to fc, the output voltage decrease to 70.7 per cent of the maximum."

I think that the graph in figure 16-10b have missed the frequency response of the bypass capacitor shown in fig 16-9c , I think at midband frequency, the bypass capacitor is above its cutoff frequency(short) fig 16-9c and when the frequency decreased to fc (of the coupling capacitor Cout 16-10a) the gain is 0.707 of Vmax and the reactive resistance of bypass capacitor high negative feedback increases. Please can someone help me in this!

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I believe all that's being shown is that if one observes the relative voltages, in the first instance from emitter to base and in the second instance, from collector to base, the 3dB transitions occur at the same frequency, fc. The underlying assumption is that the low frequency cut-off frequencies due to the inter-stage coupling capacitors (Cin or Cout) are much lower than the fc determined by the emitter bypass capacitor.
This can readily be demonstrated by simulation of a CE amplifier and measuring the relative voltage "gains" from collector to base and emitter to base.

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3. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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I am little confused saying from relative voltage from collector to base.
What I need is the frequency response graph or by explanation, frequency vs Vout(across RL to ground) the effect of input, output coupling capacitors and bypass capacitor, from said midband frequency to fc( figtext0 attached, I don't now which cutoff frequency is, for bypass or output capacitor)

4. ### FrancescoC Member

Nov 22, 2014
30
5
Hi,
When I was a student ant Staffordshire University I did enjoy solving that kind of problems.
I have now forgotten it all.

However, I would advice you to use some simulation software and prove to yourself that there is a mistake in the book.

Regards

Francesco

5. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
82
0
Thank you for your opinion but what I need help to sort out the problem and to continue learning!

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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@Saviour Muscat
Two things to consider.
1. Terminology
You use the terms "frequency response of the bypass capacitor" & "bypass capacitor is above its cutoff frequency".
What do you mean when using those terms? The text rather refers to cutoff frequency of the particular "circuit" function - either bypass or coupling. This is an important distinction.

2. Assumptions
The text explicitly states in the opening paragraph, that the coupling circuit function lower critical frequency is assumed to be sufficiently lower than that of the emitter bypass circuit cutoff, that the coupling circuit behavior can be ignored - i.e. purely for the sake of argument / clarity in this instance, the reader is informed that the emitter bypass function is the dominant determinant of the amplifier lower -3dB (cutoff) frequency.

Last edited: May 11, 2015
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7. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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Thankyou I might solved the problem, You said to ignore the critical frequency of the coupling capacitor and solely focus on the critical frequency of the bypass capacitor.
I was confusing the Output voltage with the voltage across the bypass capacitor(Ve), and the graph is solely for Vout vs Frequency.
1.When the frequency is at midband the bypass capacitor is shorted and there is no AC negative feedback and the output voltage is at maximum.
2.when the frequency is decreased to the cutoff frequency of the bypass capacitor, a voltage will built up across the bypass capacitor(Ve) therefore AC negative feedback will appear and the output drops to 0.707Vout.
Kindly can you confirm or corrects my explanation
Many Thanks
SM

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It seems English may not be your first or native language. Apologies if I have assumed incorrectly.
If my observation is true, I will presume that some of the words you use to describe your understanding of the text are intended to be descriptive rather than technically 100% correct - at least from my perspective.
In any case, I think you now have a better understanding of what is written in the text book.
So I'll leave it at that, unless you have further need of clarification on the topic.

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9. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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Can you be so kind to show how Vout is 70.7% when the bypass capacitor is at cutoff frequency with no too much complex mathematical prove?
I appreciate so much your help
SM

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well simply voltage divider.

$Vout = Vin * \frac{Xc}{Rth + Xc}$

Also do you know that $Z =Rth + Xc = \sqrt{Rth^2 + Xc^2 }$

So we have

$Vout = Vin * \frac{Xc}{\sqrt{Rth^2 + Xc^2 }}$

And for frequency where Rth = Xc we have :

$Vout = Vin * \frac{Xc}{\sqrt{Rth^2 + Xc^2 }} = Vin * \frac{1}{\sqrt{1^2 + 1^2 }} = Vin * \frac{1}{\sqrt{2}} = Vin *0.707$

And Xc = Rth are equal for frequency:

$F = \frac{1}{2 *\pi * Rth * C }$

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11. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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I understood the last derivation, the problem is the Vout for fig16-10b
I am waiting to for reply from you!

Last edited: May 13, 2015
12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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@Saviour Muscat
Since the output lower cut-off frequency indicated in fig 16-10b depends on several component values in the amplifier topology as well as the particular transistor parameters, your request for a "not too complex mathematical proof" may be difficult to accommodate. It all depends on what you interpret as "not too complex" ...

Last edited: May 14, 2015
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13. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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I understood JonyC derivation! but can you attempt for the derivation or description for show how Vout is 70.7% when the bypass capacitor is at cutoff frequency of fig 16-10a and fig16-10b

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But 0.707 is 70.7 in percentage.

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15. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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Yes 70.7% per cent of Vout

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I attempted this because it is of interest to me rather than your zealous demand for a response.

As I stated earlier in post #12 this is not a trivial exercise.

I will state only the final equation I derived and the meaning of the individual terms in relation to the circuit.

The low side -3dB cut-off frequency where the output has dropped to 70.7% of the mid-band value is given by ...

$\text{\omega_c=\sqrt{\frac{1}{C_{e}^2R_e^2R_x }\{ {\frac{R_e^2}{R_x}+2R_e-R_x} \}} \smal{radians per sec}}$

where
$\text{R_x=\frac{R_{th}}{\beta}+\frac{1}{g_m}}$

and

$\text{R_{th}=R_1 || R_2 || R_G}$

Further simplification is possible to give the frequency in Hz ...

$\text{f_c=\frac{1}{2\pi R_eC_e}\sqrt{$$\frac{R_e}{R_x}+1$$^2-2}}$

And finally, provided Re>>Rx, one can further simplify to ...

$\text{f_c=\frac{1}{2\pi R_xC_e}}$

... which is probably not a surprising outcome.

So the lower -3dB point is conveniently determined by this circuit ...

... with the proviso Re>>Rx.

This latter outcome was presumably the general point of comments in the text book in relation to fig 16-9.

Hopefully another member can confirm my result but I wouldn't hold my breath. I'm happy to post a complete solution for those who may be interested, but this may not be particularly informative given the simplicity of the final result based on some reasonable assumptions.

Last edited: May 15, 2015
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17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I'll second that!

Confirmation to follow.

I solved the circuit using Shekel's method:

I knew your expression was much simplified compared to the raw output from Mathematica, and I didn't want to spend as much time as you probably did simplifying, so I decided to use some numerical values for the circuit components to see if I would get the same value for the 6 dBV down frequency. I noticed while playing around with some values that for certain combinations of RG, RE, re and β, the very low frequency gain is not even 6 dBV down, so there is no ω for which the gain is 6 dBV less than the high frequency gain. When this combination of component values occurs, your expression gives an imaginary result, as does my expression. The values I have used in my example calculations are very near the boundary for the anomalous hehavior:

I derived a symbolic solution for the 6 dBV down frequency from the expression for the gain:

My value is quite different than yours for these component values. For component values not near the odd behavior boundary, my value is much nearer yours. I spent quite a while figuring out why the difference. I discovered that if I make a couple of changes to your variable Rx, I could get an exact match between my value and yours:

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18. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks Electrician. I truly appreciate your insights and comments. Very helpful to note that your approach gives some good agreement with mine - particularly as you analysed the effect of simplification on my part. Interesting observation concerning the anomalous case - a matter that evaded me.
I'm not sure the TS will appreciate our efforts and mental gymnastics but it was an interesting exercise nonetheless.
Cheers!

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19. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Electrician why 6dBV ??

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20. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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Sorry for the trouble I caused I appreciate all of your effort to teach me in the right way thanks very much !
Saviour Muscat