hi,

I am trying to analyse the effect of the emitter resistor bypass capacitor in a common emitter amplifier configuration (using a voltage divider bias config), in particular the effect on Ib.


I understand that the RE is key for DC bias stability since; an increase in temp causes beta to increase, causing an increase in IC. An increase in IC causes more voltage to drop across RE (increased VRE). And since VB is 'fixed' due to the voltage divider network, then VBE must decrease (KVL; VR2 = VBE + VRE). This decrease in VBE causes IB to decrease which in turn decreases IC. Giving the circuit independence from beta, and hence stability.

That all makes sense to me but when I try and analyse trhe following individual parameters I'm confusing myself.

All parameters of interest:

Ib:
Vo:
Av:
IB:

If RE is bypassed.


Here's what I have come up with:

Vout: will increase (as Ie will increase because Ic would have increased because it has a more direct path to ground. Correct? Or does Vout increase because the bulk of the voltage gets dropped RC.... This now seems wrong due to the way voltage dividers work - wouldn't you want the bulk of the voltage to be dropped across RE in order for Vout to increase ?? )
Av: will increase (voltage gain, Av=Vout/Vin, and Vout would have increased.)
IB: will remain unaffected (DC signal is blocked and will flow through RE, business as usual for DC.)

My best guess regarding the effect on the ac signal into the base, Ib, is that with the bypass cap in circuit, Ib will increase. I am basing this on thinking that Ic would increase (due to less resistance.. i think but now I'm confused). And therefore Ib must also increase due to beta = delta ic/ delta ib relationship.


Thanks in advance

 

Generally VBE is 0.6 volts. It does not decrease. It stays constant. It is 0.6 volts needed to "turn on" B-E junction diode.

 

I had wondered about that. Then how is stability achieved since VB (VR2) is fixed too?

 

I had wondered about that. Then how is stability achieved since VB (VR2) is fixed too?

I will need to dig out my textbook and go over the material.

 

Emitter bypass capacitor does not affect Ib or bias. It will drastically affect the AC gain and frequency response.

 

hi,

I am trying to analyse the effect of the emitter resistor bypass capacitor in a common emitter amplifier configuration (using a voltage divider bias config), in particular the effect on Ib.


I understand that the RE is key for DC bias stability since; an increase in temp causes beta to increase, causing an increase in IC. An increase in IC causes more voltage to drop across RE (increased VRE). And since VB is 'fixed' due to the voltage divider network, then VBE must decrease (KVL; VR2 = VBE + VRE). This decrease in VBE causes IB to decrease which in turn decreases IC. Giving the circuit independence from beta, and hence stability.

That all makes sense to me but when I try and analyse trhe following individual parameters I'm confusing myself.

All parameters of interest:

Ib:
Vo:
Av:
IB:

If RE is bypassed.


Here's what I have come up with:

Vout: will increase (as Ie will increase because Ic would have increased because it has a more direct path to ground. Correct? Or does Vout increase because the bulk of the voltage gets dropped RC.... This now seems wrong due to the way voltage dividers work - wouldn't you want the bulk of the voltage to be dropped across RE in order for Vout to increase ?? )
Av: will increase (voltage gain, Av=Vout/Vin, and Vout would have increased.)
IB: will remain unaffected (DC signal is blocked and will flow through RE, business as usual for DC.)

My best guess regarding the effect on the ac signal into the base, Ib, is that with the bypass cap in circuit, Ib will increase. I am basing this on thinking that Ic would increase (due to less resistance.. i think but now I'm confused). And therefore Ib must also increase due to beta = delta ic/ delta ib relationship.


Thanks in advance

Hi,

The input impedance looking into the base depends on Re and Ce (the bypass cap here). As the resistance of Re and/or the reactance of Ce is lowered, the input impedance gets lower thus meaning that more AC base current will be seen for a given AC input voltage.

Re affects the way the impedance responds to DC, and Ce affects the way the impedance responds to AC. The DC bias point stays constant, but the AC gain goes up and the input AC base current goes up. The DC base current does not go up with Ce added.

I dont know how much you are into AC circuits yet, but the impedance of the Re and Ce leg is:
Re/(Ce^2*Re^2*w^2+1)-j*(Ce*Re^2*w)/(Ce^2*Re^2*w^2+1)

so the real part is:
Re/(Ce^2*Re^2*w^2+1)

and the imaginary part is:
-(Ce*Re^2*w)/(Ce^2*Re^2*w^2+1)

and the magnitude of the impedance is:
Re/sqrt(Ce^2*Re^2*w^2+1)

so with 'w' (the frequency) equal to 0 (which is DC) we have:
Z=Re

and with significant non zero 'w' we have approximately:
Z=1/(w*Ce)

so for DC we just have Re there, and for AC we have approximately just the cap there and it's resistance to current goes down with increasing frequency.

 

Emitter bypass capacitor does not affect Ib or bias. It will drastically affect the AC gain and frequency response.

Ok great. Thank you for clearing that up.

 

Hi,

The input impedance looking into the base depends on Re and Ce (the bypass cap here). As Re and/or Ce is lowered, the input impedance gets lower thus meaning that more AC base current will be seen for a given AC input voltage.

Re affects the way the impedance responds to DC, and Ce affects the way the impedance responds to AC. The DC bias point stays constant, but the AC gain goes up and the input AC base current goes up. The DC base current does not go up with Ce added.

I dont know how much you are into AC circuits yet, but the impedance of the Re and Ce leg is:
Re/(Ce^2*Re^2*w^2+1)-j*(Ce*Re^2*w)/(Ce^2*Re^2*w^2+1)

so the real part is:
Re/(Ce^2*Re^2*w^2+1)

and the imaginary part is:
-(Ce*Re^2*w)/(Ce^2*Re^2*w^2+1)

and the magnitude of the impedance is:
Re/sqrt(Ce^2*Re^2*w^2+1)

so with 'w' (the frequency) equal to 0 (which is DC) we have:
Z=Re

and with significant non zero 'w' we have approximately:
Z=1/(w*Ce)

so for DC we just have Re there, and for AC we have approximately just the cap there and it's resistance to current goes down with increasing frequency.

Thank you very much for your explanation. Since it is beyond my current level of understanding I will refrain from asking you a bunch of questions about it -but pls know it's to spare you, not from lack of interest! I will refer to your post later on when I do learn about imaginary numbers and such. Even without understanding it fully it still has helped, thank you

 

Thank you very much for your explanation. Since it is beyond my current level of understanding I will refrain from asking you a bunch of questions about it -but pls know it's to spare you, not from lack of interest! I will refer to your post later on when I do learn about imaginary numbers and such. Even without understanding it fully it still has helped, thank you

Hi,

Ok no problem, but remember this is a forum for asking questions so ask anything you want :and somebody here will most likely respond

To help understand just a little bit more, we can look at just two different cases where the frequency is zero (DC) and some normal test frequency like 1kHz.

Let's say we have just DC first (the bias) and Re=100 and Ce=10uf and and no AC input yet.
Let's also say that the voltage on the base is 1.7v and the base emitter drop is 0.7v, and that means we have 1v across Re. With a Beta of 100 the input resistance is approximately 100*100=10k, and with 1.7v DC on the base (and subtract that 0.7v drop) that means the base current is about 1/10000=100ua.
This is DC so the cap does not matter yet.

Next we apply an AC signal of 1v peak at 1kHz coupled into the base. Now the cap starts to react. The cap is dominate now, and the reactance is 1/(2*pi*f*C) which is:
1/(2*3.14*1000*0.000010)
which of course equals 15.9 Ohms, so now we have very roughly a 16 Ohm resistance from the emitter to ground so we have to calculate the AC base current based on that which is roughly:
ibAC=1/(16*100)=625ua peak.

Here you can see that the normal current is 100ua but went up to 625ua with the peak of the AC input.
At 2kHz, the reactance of the cap would be even lower, roughly 8 Ohms, so we would then have with the same 1v peak AC input:
ibAC=1/(8*100)=1250ua peak

and so you see as the frequency goes up, the base current goes up with a fixed input AC voltage.

Also, we went from Re=100 to a rough equivalent of 16 Ohms and then down to 8 Ohms from emitter to ground, so you can see how the approximate resistance goes down with frequency. That's what makes the bypass cap work so well, because it makes the emitter resistor look smaller as frequency goes up, and these circuits are used mainly to amplify AC signals.

Also, a rough approximation of the gain is Rc/Re for DC, but for AC at 1kHz it would be roughly Rc/(1/(w*Ce))=Rc*w*Ce=Rc*6.28*1000*Ce=Rc*6280*Ce, so you can see that the gain can go up quite a bit if we have an AC input. With Rc=100 and Re=100 the gain is about 1 for DC, but at 1kHz the gain is about 6.
Of course for very low frequencies we need a bigger bypass cap, but that depends on the application.

 

Hi,

The input impedance looking into the base depends on Re and Ce (the bypass cap here). As Re and/or Ce is lowered, the input impedance gets lower thus meaning that more AC base current will be seen for a given AC input voltage.

Should it be Re is lower and/or Ce is higher (because Zc = 1/(w*C)), the input impedance gets lower?

 

Hi,

Yes let me reword that.

As the resistance of Re gets lower or the reactance of Ce gets lower the input impedance gets lower.
As the VALUE of Re gets lower or the VALUE of Ce gets higher the input impedance gets lower.

Looks like you are catching on now, very good to hear

 

My best guess regarding the effect on the ac signal into the base, Ib, is that with the bypass cap in circuit, Ib will increase.

Depending on if you are talking ac or dc. However, the basic principle is identical.

 

Thanks for going into it further. Overall I can see what you're saying. I am not sure about how you have used some of the values in relation to one another.

With a Beta of 100 the input resistance is approximately 100*100=10k

is this based on the small signal model hybrid pi eqn ; r pi =(beta +1) * re?
Is there a difference between the notation Re and re? Or say Ib and ib? Or do both notations refer to the instantaneous (ac) value? I think I am misunderstanding something here as the resistor value hence static value of the emitter resistor is 100ohms, which I thought is notated as RE. Sorry to get caught up in notation but I'm trying to follow how you have made these calculations and relationships and need all the help I can get!?

with 1.7v DC on the base (and subtract that 0.7v drop) that means the base current is about 1/10000=100ua.

here are you saying that

IB= (VB-VBE)/r pi
have I got that right..?

and the other one I'm sorry but I can't see how you have come up with is this one..

so now we have very roughly a 16 Ohm resistance from the emitter to ground so we have to calculate the AC base current based on that which is roughly:
ibAC=1/(16*100)

I can see how you got the capacitive reactance =approx 16ohms no worries but I'm struggling with how the following = ib. i notice here too that you are using both lower case i and lower case b here.

ibAC=1/rpi*beta ?? I really don't know here, that's probably not the idea...

I know my questions are really basic and pretty much beside the point of what you are really explaining. I just can't fully understand your response and follow through your example scenarios properly unless I can make these relationships. Thanks again for your help

 

Thanks for going into it further. Overall I can see what you're saying. I am not sure about how you have used some of the values in relation to one another.

is this based on the small signal model hybrid pi eqn ; r pi =(beta +1) * re?
Is there a difference between the notation Re and re? Or say Ib and ib? Or do both notations refer to the instantaneous (ac) value? I think I am misunderstanding something here as the resistor value hence static value of the emitter resistor is 100ohms, which I thought is notated as RE. Sorry to get caught up in notation but I'm trying to follow how you have made these calculations and relationships and need all the help I can get!?



here are you saying that

IB= (VB-VBE)/r pi
have I got that right..?

and the other one I'm sorry but I can't see how you have come up with is this one..


I can see how you got the capacitive reactance =approx 16ohms no worries but I'm struggling with how the following = ib. i notice here too that you are using both lower case i and lower case b here.

ibAC=1/rpi*beta ?? I really don't know here, that's probably not the idea...

I know my questions are really basic and pretty much beside the point of what you are really explaining. I just can't fully understand your response and follow through your example scenarios properly unless I can make these relationships. Thanks again for your help

Hi,

No worries, it is understandable why you would ask these questions, which BTW i think are good questions.

First, yes we often used re for the transistor 'internal' parameter, and we might use RE for the external 100 ohm resistor, i just used Re for the 100 ohm external resistor. In fact, at no time do i refer to the 'internal' values of the transistor assuming for the moment that either they are not significant or that once you have the right way to do this you can add them yourself to get a more accurate result.

The base input resistance is based on the DC beta of the transistor and how it appears to multiply the emitter resistance Re, and this is an approximation. For example, with a base current of 10ua and emitter resistor 1k and Beta of 10, the collector current is 100ua and that means we have 110ua in the emitter, and that 'boosts' the voltage at the emitter and because of the nearly constant 0.7v base emitter drop the base also increases in voltage so it looks like a higher input resistance. To see this you have to look at both the base voltage, emitter voltage, and also the base emitter current and collector current.
Without the gain of the transistor (Beta=0) the current with a 1.7v base voltage would be:
(1.7-0.7)/1000=0.0017 amps. WITH the Beta=10 however, the emitter current is at least equal to the collector current which now could be 0.017 amps, but that produces the whole 1.7v in the emitter resistor so the base current will go down to:
ib=(1.7-0.7)/1000/10=0.00017 amps
and that will produce 0.0017 amps in the emitter resistor, which would produce approximately 1.7v at the base, and hence the base current would be about 170ua and the emitter current about 1.7ma.
That's just a DC calculation based on the base voltage and collector current and how it raises the voltage at the emitter thus reducing the base current.
Since the base current was reduced, it looked like a higher input resistance which is approximately:
Rin=Re*10 where Beta=10 here.
You can get more accurate by simply analyzing the circuit more accurately and emulating the transistor as a current controlled current source. That will be almost the same as above but more accurate.
Also, the base emitter drop was approximated as a constant 0.7v just for the example.

The calculation with the capacitive reactance was a rough idea how things change when the frequency goes up. The capacitor is emulated as a resistor and thus we see a lower impedance from emitter to ground. With a lower impedance we no longer have 1k (this example) or 100 ohms (your example) but it goes down and that reduces the input impedance as well.
To get a better solution you would just analyze this in the frequency domain with the transistor emulated as a current controlled current source or any other model you choose to use. In that case the capacitor would be in parallel with the emitter resistor Re, and that would be a frequency dependent network so you'd have to do more math. As the frequency goes up, the cap starts to dominate and makes the impedance lower and lower. At infinite frequency, the cap acts like a short circuit for AC.

 

is this based on the small signal model hybrid pi eqn ; r pi =(beta +1) * re?

you certainly can do that.

With beta sufficiently high - true most of the cases, a short cut exists: Vin = Vb = Vbe + Ie * Re, where Vbe = 0.7v (very little variance with Ib). So any change in Vin will cause a corresponding change in Ie thus Ic.

the rest follows.

 

Hello again,

Here is an approximate calculation of the input current to the base as the frequency varies, using Re=100 and Beta=100 and Ce=10uf, and constant base voltage:
ib=sqrt(4.0e-6*pi^2*f^2+1)/sqrt(4.0e-6*pi^2*f^2+100020001)

This shows that the base current increases with increasing frequency. That is because of the capacitor.
The base current starts out very low at very low frequencies like 1Hz, but then gradually increases as the frequency increases, so at 1kHz it may be 10 times more. That's the main point here, and also that means that the collector current increases and the stage gain increases.

These results were obtained by doing an AC analysis of the stage using a transistor emulated as a current source and ignoring the base emitter voltage drop.

 

Hi,

One thing i forgot to mention is that the collector current acts to produce some positive feedback to the base because it acts to raise the voltage at the emitter. This means the input impedance goes up.
If we make Re lower however, it acts to reduce the input impedance.