View Full Version : Common base BJT amplifier

anik321

10-25-2008, 06:57 PM

Hey guys. I am trying to understand this common base circuit.

This is what I understand:

1. It is a common base circuit

2. There is a resonant circuit with a small signal source at the emitter.

3. qualitatively I know that at resonance, the impedance will be the lowest, therefore Vout (collector voltage) will be the lowest at this resonant frequency. (lower impedecence->more current through collector/emitter->more voltage drop across Rc.

I just do not understand how to go about starting this analysis. Can you guys help? How can I numerically calculate the resonant frequency?

Here is the circuit: (http://img296.imageshack.us/my.php?image=cbaseld7.jpg)

http://img296.imageshack.us/img296/8404/cbaseld7.jpg (http://img296.imageshack.us/my.php?image=cbaseld7.jpg)

http://img296.imageshack.us/my.php?image=cbaseld7.jpg

http://img296.imageshack.us/my.php?image=cbaseld7.jpg

hgmjr

10-25-2008, 07:49 PM

Take a look at the material on series RLC circuits contained here (http://www.allaboutcircuits.com/vol_2/chpt_5/2.html) in the AAC ebook.

hgmjr

hgmjr

10-25-2008, 08:10 PM

Take a look at this material on series RLC resonance (http://www.allaboutcircuits.com/vol_2/chpt_6/3.html) as well.

hgmjr

anik321

10-26-2008, 02:53 AM

Thank you,

So I have calculated the resonant frequency at which the impedance of the cap and the inductor cancels each other. This works out to be 10khz. (10068khz to be exact)

Here is my DC analysis. (I left out V1, and shorted V2 and V3 to ground)

2V - 0.7 - (Ie*100) - (Ie*1.2k) = 0

Ie ~ 1mA

Vcollector = 10- 1mA(1k) = 9V.

How do I obtain the small signal transfer function to draw bode plots? I prefer using the T model, but i am not sure how to include the inductor.

anybody?

hgmjr

10-26-2008, 03:01 AM

Thank you,

So I have calculated the resonant frequency at which the impedance of the cap and the inductor cancels each other. This works out to be 10khz. (10068khz to be exact)

That is the frequency I calculated also.

Here is my DC analysis. (I left out V1, and shorted V2 and V3 to ground)

2V - 0.7 - (Ie*100) - (Ie*1.2k) = 0

Ie ~ 1mA

Vcollector = 10- 1mA(1k) = 9V.

Your calculations appear to be correct to me.

hgmjr

Audioguru

10-26-2008, 03:08 AM

The series LC is a short at resonance.

The AC voltage gain of the transistor is Rc/(RE + Re). Rc is 1k. RE is 100 ohms. Re is the internal Re of the transistor which is about 26 ohms at 1mA. Therefore the voltage gain is 1000/(100 + 26)= 7.94 times.

anik321

10-26-2008, 03:16 AM

The series LC is a short at resonance.

The AC voltage gain of the transistor is Rc/(RE + Re). Rc is 1k. RE is 100 ohms. Re is the internal Re of the transistor which is about 26 ohms at 1mA. Therefore the voltage gain is 1000/(100 + 26)= 7.94 times.

Thank you guys.

Audioguru,

Your calculations even at resonance do not take into account the 1.2k resistor all the way at the bottom.

Are you saying at resonance (when the L and C does not offer any impedance, all the emitter current flows to ground through my ideal signal source and nothing through the 1.2k? Is it safe to assume this for signal sources?

what happens when you are NOT at resonance? how can I sketch a bode plot of this?

anik321

10-26-2008, 03:20 AM

also, my collector voltage at resonance do come extremely close to your gain of 7.94. It is in fact 7.84. (spice)

I am simply trying to understand what happens to the 1.2k and why is it not taken as part of the emitter resistance when calculating Rc/(RE+re)

I the cap and L i am guessing would also give just a 20db*2=40db/dec roll off.

steveb

10-26-2008, 03:31 AM

I just do not understand how to go about starting this analysis. Can you guys help? How can I numerically calculate the resonant frequency?

How do I obtain the small signal transfer function to draw bode plots? I prefer using the T model, but i am not sure how to include the inductor.

anybody?

It sounds like you understand how to use an equivalent circuit model and you just want to know how to include the inductor. Since you are assuming a small signal AC analysis, you can use the Laplace Transform approach. Replace all inductors with impedance sL and all capacitors with impedance 1/sC. Then, use your normal circuit analysis to find the transfer function output/input. Your transfer function then depends on the complex frequency "s". To get the numerical answer, Matlab is the best choice. You can directly enter the transfer function and get the bode plot. Without Matlab, the math is tedious, but can still be worked out.

This is the basic approach, but if you need more detail, just let us know what part needs clarification or more detail.

anik321

10-26-2008, 03:37 AM

I am OK at best with small signal models.

but the math when you replace inductors with sL and caps with 1/sC gets very tedious. To the point where I cannot envision when the gain is real or not etc.

Is it reasonable to not be able to 'sketch by hand' a bode plot of a transfer function this complicated?

When I look at the circuit, spice simulations give me exactly when I picture in my head. and infact I can easily do it without deriving a transfer function. (i will post a Vmag vs. frequency graph soon)

so yes: my problem is simply deriving the full small signal transfer function.

steveb

10-26-2008, 03:55 AM

I am OK at best with small signal models.

but the math when you replace inductors with sL and caps with 1/sC gets very tedious. To the point where I cannot envision when the gain is real or not etc.

Is it reasonable to not be able to 'sketch by hand' a bode plot of a transfer function this complicated?

Yes, you are right, the math is tedious. That's why I recommend using Matlab. Matlab lets you just enter the transfer function and then do the bode plot. It is best to get the transfer function is a form of a ratio of two polynomials, and then use the tf command.

If you don't have Matlab, then it is possible to get a bode plot by hand, if you can work out the poles and zeros and use the bode plot graph procedures. Since I haven't actually worked out your problem, I don't know how difficult this would be in your case. Sometimes it's easy, sometimes not. The order of the system does not look too large, and usually a second or third order system can be worked out. But, yeah, it's a pain. It's good to work through a few of these by hand as a student. Then, once you know you can do it, avoid it like the plague and use Matlab. :D

anik321

10-26-2008, 09:04 PM

Please take a look.

SOLUTION (http://www.mediafire.com/?sharekey=eefc42c8ea1d28d8d2db6fb9a8902bda)

The gain at resonant frequency comes close to what spice gives

Hand calculations: 8.85

Spice:7.9

Is the analysis right? Can someone please look at it?

How can I obtain the phase from the transfer function (Vo/Vs) i have provided above?

steveb

10-26-2008, 11:47 PM

I think there is a mistake in your work. I derived the gain relation and calculated the gain at the resonant frequency, but I got a gain of 8. One of your relations looks incorrect. You say that Vin/Rin=ie, but this is the input current.

As far as the phase, you must calculate the complex value of A=Vo/Vi and then take the magnitude as sqrt(Im(A)^2+Re(A)^2 and the phase as atan(Im(A)/Re(A)). Note that Im(A) is the imaginary part and Re(A) is the real part.

When you are at the resonance frequency, the gain is purely real, so A is the magnitude of the gain and the phase is zero.

anik321

10-27-2008, 01:00 AM

steveb,

I also get 8 only when I do not take the 1.2k resistor to be in parallel with the (100+re). Then my Rin is simply:

Rin= jwL+ 1/jwC + 126 (just 126 at resonance)

H(jw)=Rc/Rin = 1k/126 ~8

if you take the 1.2k to be in parallel with the 100ohms then 1.2k//100=113

Rin= jwL+ 1/jwC + 113 (just 113 at resonance)

H(jw)=Rc/Rin = 1k/113 ~8.84

steveb

10-27-2008, 01:05 AM

You seem to be making an honest effort to learn this, so I thought I would post my work, since it could help you. Please double check my work. I make just as many mistakes as the next guy, and I'm a little rusty on this stuff.

I was taught to use signal flow graphs (SFG) and Mason's gain formula, and I don't expect they teach that anymore, but you should be able to understand the mathematical relations in the SFG. The nodes are variables and the arrows are gain terms. If two or more arrows go into a node, then that is an addition operation. For example the right-most arrow show the equation Vo=-alpha*Rc*Ie. The left-most relation shows (with two arrows) the equation Ii=(Vi-Ve)/(sL+1/sC).

(Note: as shown in a later post, this work has an sign error in the gain formula: CB amplifier is non-inverting)

Ron H

10-27-2008, 04:27 AM

With ideal L and C, the impedance at node 4 is zero at resonance, so signal current will flow through it, but will not affect the emitter (and collector) current. If you are trying to do a Bode plot (Ic vs frequency), then the 1.2k ohm resistor will enter into your calculations.

steveb

10-27-2008, 01:17 PM

When you are at the resonance frequency, the gain is purely real, so A is the magnitude of the gain and the phase is zero.

I have to correct myself here. The phase is not zero, but 180 degrees due to the negative sign.

Actually, it's good I made this mistake because it lets me point out an issue with using the inverse tangent function to find the phase. You need to keep track of the correct angle quadrant when you use the inverse tangent function to find the angle. Some programing languages have an "atan2" function which does this automatically.

Ron H

10-27-2008, 02:56 PM

I have to correct myself here. The phase is not zero, but 180 degrees due to the negative sign.

Actually, it's good I made this mistake because it lets me point out an issue with using the inverse tangent function to find the phase. You need to keep track of the correct angle quadrant when you use the inverse tangent function to find the angle. Some programing languages have an "atan2" function which does this automatically.Steve, did you got the sign wrong? Common-base amplifiers are noninverting.

steveb

10-27-2008, 06:04 PM

Steve, did you got the sign wrong? Common-base amplifiers are noninverting.

Yes, you are right. Oops! :eek: I've so rarely used CB that I forgot. Thanks for correcting this.

I see that I defined the direction of Ie in the equivalent circuit to be in the wrong direction. That flipped the sign in the math.

As I said, I'm a little rusty. :)

anik321

10-27-2008, 09:54 PM

to be honest, i cannot quite figure out where my calculations have gone wrong. but a gain of 8 sounds more correct.

The Electrician

10-27-2008, 10:15 PM

to be honest, i cannot quite figure out where my calculations have gone wrong. but a gain of 8 sounds more correct.

Ron H in post #16 explained it.

When the frequency is equal to the resonance frequency of the L/C circuit, the L/C combination may be replaced by a short. You can then see that the 1.2k resistor is simply an additional load from the input to ground. Since the source is assumed to be a zero output impedance device, the 1.2k load does nothing when it's connected directly to the source, as it is when the L/C circuit is at resonance.

Off resonance, the impedance of the L/C circuit combined with the 1.2k forms a voltage divider.

So, at resonance, the 1.2k does nothing. Off resonance, it has an effect.

steveb

10-27-2008, 11:50 PM

to be honest, i cannot quite figure out where my calculations have gone wrong. but a gain of 8 sounds more correct.

I believe your mistake is where you say Ie=Vs/Rin. This would only be true if the 1.2K resistor was not there. You should instead say that Ie=Vs/Rin-I3 where I3 is the current in the 1.2K resistor. Then you need further relations for I3, such as I3=V3/1.2K where V3 is the voltage across the 1.2K resistor. Then you continue until all unknowns are related to know parameters and end up with an equation for Vout/Vin.

The SFG I posted basically does this, although I made a sign mistake.

The previous posts are basically saying the same thing in a different way.

Another way to say it is your gain is Rc/Rin, which includes the 1.2 K resistor, even at resonance. If you look at my gain formula at resonance, it works out to Rc/Re3 which does not depend on the 1.2K resistor.

Ratch

10-28-2008, 01:55 AM

To the Ineffable All,

I maintain that the 1K2 resistor will not affect the resonant frequency and will reduce the output of all frequencies by the current division rule 1200/(1200+100+26) = 90% . It is simply a shunt that diverts current from the transistor. So we have for a transfer function Vo/Vi = 0.9*1000/(100+26+Ls+1/Cs) = 900Cs/(LCs^2 + 126Cs +1) . Notice at resonance we can substitute w^2 = 1/LC into the transfer function and it reduces to 900/126 = 7.14 . The Bode plots are shown below.

Ratch

steveb

10-28-2008, 02:36 AM

To the Ineffable All,

I maintain that the 1K2 resistor will not affect the resonant frequency and will reduce the output of all frequencies by the current division rule 1200/(1200+100+26) = 90% . It is simply a shunt that diverts current from the transistor. So we have for a transfer function Vo/Vi = 0.9*1000/(100+26+Ls+1/Cs) = 900Cs/(LCs^2 + 126Cs +1) . Notice at resonance we can substitute w^2 = 1/LC into the transfer function and it reduces to 900/126 = 7.14 . The Bode plots are shown below.

Ratch

No, the gain of 7.14 is certainly incorrect, provided that Rc=1K and Re3 = 125 ohms. If you look at the transfer function I derived, and correct for my mistake with the negative sign.

The gain is Av=Rc/Re3 at resonance. At the resonant frequency Wo=1/sqrt(LC):

s^2 = (jWo)^2 = -Wo^2 = -1/(LC)

Hence, the denominator of the transfer function is very simple at resonance.

I do agree with your comment that the 1.2K resistor does not affect the resonant frequency.

Ron H

10-28-2008, 02:44 AM

To the Ineffable All,

I maintain that the 1K2 resistor will not affect the resonant frequency and will reduce the output of all frequencies by the current division rule 1200/(1200+100+26) = 90% . It is simply a shunt that diverts current from the transistor. So we have for a transfer function Vo/Vi = 0.9*1000/(100+26+Ls+1/Cs) = 900Cs/(LCs^2 + 126Cs +1) . Notice at resonance we can substitute w^2 = 1/LC into the transfer function and it reduces to 900/126 = 7.14 . The Bode plots are shown below.

RatchI simulated the actual circuit in LTspice, and the gain at resonance was 7.946, which validates the opinion that I and others have stated, namely that the 1.2k resistor has no effect on the voltage gain at resonance.

Ratch, If you wanted to plot current gain from the voltage source, the indeed approximately 10% of the signal current flows through the 1.2k resistor at all frequencies, but the current gain would be 0.9.

Ratch

10-28-2008, 03:15 AM

steveb,

No, the gain of 7.14 is certainly incorrect, provided that Rc=1K and Re3 = 125 ohms. If you look at the transfer function I derived, and correct for my mistake with the negative sign.

The gain is Av=Rc/Re3 at resonance. At the resonant frequency Wo=1/sqrt(LC):

Did you correct for the shunting effect of the 1K2 resistor? My last posting shows why it drains 10% of the current going into the transistor. So, at resonance, 0.9*1000/126 = 0.9*7.94 = 7.14 . Are we arguing about whether the 1K2 resistor has any effect on the gain? It surely does because it subtracts from the gain current.

By the way, I have a hard time following the analysis you posted. I think your result is too complicated, because you got off course somewhere. Perhaps it would be simpler for you to critique my simpler result to see where I went wrong.

Ratch

Ratch

10-28-2008, 03:25 AM

Ron H,

I simulated the actual circuit in LTspice, and the gain at resonance was 7.946, which validates the opinion that I and others have stated, namely that the 1.2k resistor has no effect on the voltage gain at resonance.

Ratch, If you wanted to plot current gain from the voltage source, the indeed approximately 10% of the signal current flows through the 1.2k resistor at all frequencies, but the current gain would be 0.9.

I would check your simulation again. How can you explain why it has no effect on gain at resonance? What if the shunt resistor was 126 ohms instead. That would mean that 50% of the current does not flow through the transistor. I still cannot why the gain does not drop to 90% of what it would be if no resistor were present.

Ratch

Ron H

10-28-2008, 03:33 AM

Ron H,

I would check your simulation again. How can you explain why it has no effect on gain at resonance? What if the shunt resistor was 126 ohms instead. That would mean that 50% of the current does not flow through the transistor. I still cannot why the gain does not drop to 90% of what it would be if no resistor were present.

RatchWasted current through R1 has no effect on the voltage gain at resonance. The LC impedance at resonance is ZERO - ZIP - NADA. The voltage source resistance is zero at all frequencies. Think of the LC circuit as a piece of wire. It doesn't matter how low the shunt resistor is. At resonance, the voltage at the junction of R1, R3, and C1 is identical to the voltage of the source.

steveb

10-28-2008, 03:38 AM

steveb,

Did you correct for the shunting effect of the 1K2 resistor? My last posting shows why it drains 10% of the current going into the transistor. So, at resonance, 0.9*1000/126 = 0.9*7.94 = 7.14 . Are we arguing about whether the 1K2 resistor has any effect on the gain? It surely does because it subtracts from the gain current.

By the way, I have a hard time following the analysis you posted. I think your result is too complicated, because you got off course somewhere. Perhaps it would be simpler for you to critique my simpler result to see where I went wrong.

Ratch

Yes, the full effects of the 1.2K resistor are included in my analysis. My analysis is probably hard to follow because I use a SFG approach and Mason's gain formula. Most people are not familiar with this, so I understand. However, the formulae are very simple and all effects are included. If you like, I can write out the individual formula used to make the SFG, but it includes all voltage loading effects.

I think the error in your thinking is that the current gain is not the same as the voltage gain. At resonance, the input voltage directly drives the 1.2K resistor with no effective source impedance. Hence, the diverted current has no effect on the voltage gain.

I've checked my work a couple of times now, and it correlates with a spice simulation very accurately.

Ron H

10-28-2008, 03:40 AM

Here is a plot of the gain at the junction of R1, R3, and C1.

Ratch

10-28-2008, 04:08 AM

steveb and Ron H,

Yes, I see what you are talking about now. I should have included R1 in my transfer analysis. I know about signal flow graphs, but I am very rusty, and never have used them much. Thanks to both of you for the correction.

Ratch

steveb

10-28-2008, 04:19 AM

No Problem Ratch. It's very easy to make mistakes in these problems. I'm not sure, but we may have lost the person who posted the question. Hopefully he sees the that using discussions, thinking, theory and simulation tools, together, eventually leads to the answer.

Ratch

10-28-2008, 04:50 PM

steveb,

... It's very easy to make mistakes in these problems. ...

Yes, I see my mistake now. I did not take into consideration how much the 1K2 resistor increases the current from the input voltage. 126||1200 = 114 ohms instead of 126 ohms. 1000/114 = a gain 8.77 . Multiplying by the shunt factor of 1200/(1200+126) = 0.9049774 for a gain of 8.77*0.9047774 = 7.94 which is the same as 1000/126 .

Ratch

steveb

10-29-2008, 02:40 PM

How do I obtain the small signal transfer function to draw bode plots?

anybody?

I thought I would add some information about second order bandpass systems such as this. The second order bandpass system is so common that you should learn the standard form of the transfer function. I've attached a PDF which shows the basic form of the transfer function and makes it easy to identify gain, center frequency and bandwidth.

Whenever you see that you have a second order system with "s" in the numerator and a quadratic of "s" in the denominator, try to put it in the standard form. If you can, then you know it's a bandpass and you can pick out formula for gain, center frequency and bandwidth which allows you to design the circuit by tuning parameters in an intelligent way, rather than just randomly changing numbers in a simulator such as spice.

Using the formula shown, you would end up with the following for your particular circuit.

Gain: Ao=7.94

Center Frequency: fo=10.07 kHz

Half Power Bandwidth: B=1.31 kHz

hgmjr

10-30-2008, 02:44 AM

Steveb inspired me to take a stab at my own solution to this problem.

I believe that the factor \alpha needs to be included in the final AC gain expression as I have shown in my solution.

hgmjr

PS. I have made the changes to the original write-up per the suggestions by electrician in subsequent replies to this thread. The updated version can be found in this reply (http://forum.allaboutcircuits.com/showpost.php?p=95767&postcount=43) further along in this thread.

Ratch

10-30-2008, 03:13 AM

hgmjr,

Steveb inspired me to take a stab at my own solution to this problem.

You certainly did a very complete and exhaustive analysis.

I believe that the factor needs to be included in the final AC gain expression as I have shown in my solution.

To be complete, yes. But assuming that the current value of the emitter is the same as the collected does not change the response all that much. Your final figure for the gain at resonance is 7.92 vs 7.94 shows that the gain is slightly less due to the loss of current to the base circuit.

Ratch

steveb

10-30-2008, 03:16 AM

I believe that the factor \alpha needs to be included in the final AC gain expression

hgmjr

Yes, you are absolutely correct. I started my analysis with the alpha included, and for some reason lost it in the following steps. It was probably just an old subconscious habit to approximate alpha=1, but it's better to keep alpha in the equations.

Good work, by the way!

The Electrician

10-30-2008, 04:20 AM

Steveb inspired me to take a stab at my own solution to this problem.

I believe that the factor \alpha needs to be included in the final AC gain expression as I have shown in my solution.

hgmjr

You certainly get A++ for quality of presentation!

Now, in order to be even more complete, as a SPICE analysis would probably be, you need to take into account at least 4 low-frequency two-port parameters. The values I get from the data sheet are (for an emitter current of 1mA):

hie = 3800 ohms

hre = .00013

hfe = 110

hoe = .0000085 mho (as they used to call them)

Converting to common base parameters, we have:

hib = 34.2287 ohms

hrb = .000160965

hfb = -.9909936

hob = .000000076564 mho

The Electrician

10-30-2008, 04:57 AM

Steveb inspired me to take a stab at my own solution to this problem.

I believe that the factor \alpha needs to be included in the final AC gain expression as I have shown in my solution.

hgmjr

On page 5 of 6, at the top of the page, you say, "The term aR1 is the product of a (I'm using a for alpha) and the collector resistance." Looking at your schematic, it appears that R2 is the collector resistor, so you should replace R1 by R2 in the sentence quoted and propagate that change through the rest of your equations.

I call R2 the collector resistor, and not the collector resistance, because normally the term "collector resistance" refers to an intrinsic property of the transistor, namely, the reciprocal of the hob parameter, the output admittance.

At Eq. 2.2, you say:

gm =IE/Vi

"...it is possible to interpret equation 2.1 as the transconductance of the CB amplifier in Figure 1."

The expression you have given (Eq. 2.1) is the transconductance from the input where Vi is applied, to the emitter of the transistor. The signal has not passed through the transistor at this point, and this expression would not normally be considered the transconductance of the complete amplifier.

The transconductance of the amplifier should be expressed like this:

gm=Ic/Vi

so if you multiply your Eq. 2.1 by alpha, that would be the transconductance of the CB amplifier.

Then your Eq. 2.6 becomes:

Vo

--- = gm R2

Vi

a familiar expression one remembers from vacuum tube days.

These were the two problems I noticed with a quick read through, otherwise it think it's right on.

steveb

10-30-2008, 11:35 AM

Now, in order to be even more complete, as a SPICE analysis would probably be, you need to take into account at least 4 low-frequency two-port parameters. The values I get from the data sheet are (for an emitter current of 1mA):

Converting to common base parameters, we have:

hib = 34.2287 ohms

hrb = .000160965

hfb = -.9909936

hob = .000000076564 mho

You are correct of course, and a good designer needs all of this in the general case. Still, the question becomes how precise does the model have to be to understand this circuit and predict it's performance.

His analysis includes alpha=hfb and hib. It is clear that hib is critical and is highly dependent on temperature. Including alpha represents a 0.1 % correction to the answer. Now including hob would be less than a 0.01% correction.

We could also say that the frequency limitations of the transistor should be considered. However, CB has very good frequency response and the speed limitations of the transistor would have small effect at 10kHz.

Compare these effect to the uncertainty in transistor parameters, temperature dependence of hib and resistor tolerance (1%), and you can make estimations of what is important. There is always a fine line in modeling as far as what to include and exclude.

The Electrician

10-31-2008, 01:59 AM

You are correct of course, and a good designer needs all of this in the general case. Still, the question becomes how precise does the model have to be to understand this circuit and predict it's performance.

His analysis includes alpha=hfb and hib. It is clear that hib is critical and is highly dependent on temperature. Including alpha represents a 0.1%correction to the answer. Now including hob would be less than a 0.01% correction.

I wouldn't really expect the results from including all the h parameters to give a result any more useful in the real world than the simple approximations.

It is an exercise in circuit analysis.

If you take alpha to be 1, and use .025 mV to calculate re, then the gain at resonance is 8.000. If you include an alpha of .99 as hgmjr did, you get a gain of 7.92. This is a correction of 1%, not .1%, isn't it?

The value of 1/hob is about 13 megohms which is probably the source of your comment that including it would be less than an .01% correction. Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base. The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance. In the case of the amplifier in this thread, at resonance the source impedance is zero, and the output resistance of the transistor is not 13 megohms, but rather about 300,000 ohms, which means including it will give about a .3% correction.

EDIT: The previous paragraph says "base" where I should have said "emitter". Thanks to Ron H for pointing this out. Here's the corrected paragraph.

-----------------------------

The value of 1/hob is about 13 megohms which is probably the source of your comment that including it would be less than a .01% correction. Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the emitter. The parameter hob is the admittance when the emitter is open circuited, or driven from a very high impedance. In the case of the amplifier in this thread, at resonance the source impedance is zero, and the output resistance of the transistor is not 13 megohms, but rather about 300,000 ohms, which means including it will give about a .3% correction.

------------------------------

Including the reverse voltage transfer ratio, hrb, gives about a .1% correction.

steveb

10-31-2008, 03:11 AM

I wouldn't really expect the results from including all the h parameters to give a result any more useful in the real world than the simple approximations.

It is an exercise in circuit analysis.

Yes, i agree with all of your points. It's important to be able to estimate all of these parameter effects and include those that are relevant. Even when they are not significant, it is good to know what order of magnitude they are at (which I failed at that time ;)).

I didn't mean my comments to imply that i disagreed with you, but just wanted to add to your points by talking about that fine line we walk when deriving relations.

hgmjr

10-31-2008, 11:24 AM

It took me a while but I think I have address all of the issues raised by electrician. Thanks electrician for taking the time to read through my tome and catching the mis-steps. All of your comments were right on target.

Enjoy,

hgmjr

Ron H

10-31-2008, 02:37 PM

I wouldn't really expect the results from including all the h parameters to give a result any more useful in the real world than the simple approximations.

It is an exercise in circuit analysis.

If you take alpha to be 1, and use .025 mV to calculate re, then the gain at resonance is 8.000. If you include an alpha of .99 as hgmjr did, you get a gain of 7.92. This is a correction of 1%, not .1%, isn't it?

The value of 1/hob is about 13 megohms which is probably the source of your comment that including it would be less than an .01% correction. Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base. The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance. In the case of the amplifier in this thread, at resonance the source impedance is zero, and the output resistance of the transistor is not 13 megohms, but rather about 300,000 ohms, which means including it will give about a .3% correction.

Including the reverse voltage transfer ratio, hrb, gives about a .1% correction.My understanding of hob is that it is the output admittance when the base is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.

The Electrician

10-31-2008, 08:51 PM

My understanding of hob is that it is the output admittance when the base is grounded, not open...[/B].

Aren't you making the same mistake I did? Since the base is already grounded, it's the condition of the emitter that's relevant. hob is the output admittance when the emitter is open

The h parameters are mixed parameters, unlike the Z and Y parameters. Two of them are measured with the output port shorted, and the other two are measured with the input port open.

A good reference for this sort of question is the venerable General Electric Transistor Manual. I've attached the relevant page from the 7th edition.

... and that hob is strongly dependent on the impedance at the emitter.

I said: "Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base."

Of course, what I meant to say is "...impedance at the emitter.", since the base is already grounded. :-(

And, of course, I also misspoke when saying "The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance." It should be "...when the emitter is open circuited..."

I'm just not used to grounded base amplifiers, so I had grounded emitter on the brain.

The actual numbers I gave are for the emitter open (~13 Meg) and emitter grounded (~300k).

Ron H

10-31-2008, 09:06 PM

Aren't you making the same mistake I did? Since the base is already grounded, it's the condition of the emitter that's relevant. hob is the output admittance when the emitter is openYou're the one who said the base was open. I was just correcting that.

The h parameters are mixed parameters, unlike the Z and Y parameters. Two of them are measured with the output port shorted, and the other two are measured with the input port open.

A good reference for this sort of question is the venerable General Electric Transistor Manual. I've attached the relevant page from the 7th edition.

I said: "Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base."

Of course, what I meant to say is "...impedance at the emitter.", since the base is already grounded. :-(

And, of course, I also misspoke when saying "The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance." It should be "...when the emitter is open circuited..."

I'm just not used to grounded base amplifiers, so I had grounded emitter on the brain.

The actual numbers I gave are for the emitter open (~13 Meg) and emitter grounded (~300k).I had sorta figured that you had common emitter on the brain. I just didn't want to leave your errors uncorrected for future readers. Thanks for straightening it all out.

The Electrician

10-31-2008, 10:51 PM

You're the one who said the base was open. I was just correcting that.

What were you intending to correct; the "base" part or the "open" part?

The first half of your sentence seemed a little odd to me because you were "correcting" the "open" vs "shorted" question (even though you were wrong on that point), but you were not correcting the "base" vs "emitter" issue. I couldn't understand why you were "correcting" one error but not the other. Then in the second half of your sentence it is clear that you knew that "emitter" was the right word. So, I thought you had inadvertently said "base" in the first half when you really meant "emitter".

When you are exhibiting an error somebody else made, the way to make sure your readers don't think you are also making the same error is to put "(sic)" right after the error.

Had you said "My understanding of hob is that it is the output admittance when the base (sic) is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.", it would have been clear that you knew that "base" was the wrong word to use there, that it was my error you were displaying.

Ron H

10-31-2008, 11:18 PM

What were you intending to correct; the "base" part or the "open" part?

The first half of your sentence seemed a little odd to me because you were "correcting" the "open" vs "shorted" question (even though you were wrong on that point), but you were not correcting the "base" vs "emitter" issue. I couldn't understand why you were "correcting" one error but not the other. Then in the second half of your sentence it is clear that you knew that "emitter" was the right word. So, I thought you had inadvertently said "base" in the first half when you really meant "emitter".

When you are exhibiting an error somebody else made, the way to make sure your readers don't think you are also making the same error is to put "(sic)" right after the error.

Had you said "My understanding of hob is that it is the output admittance when the base (sic) is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.", it would have been clear that you knew that "base" was the wrong word to use there, that it was my error you were displaying.

You said,

Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base. The parameter hob is the admittance when the base is open circuited...I said,

My understanding of hob is that it is the output admittance when the base is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.If you compare what you said with what I said, where is the ambiguity in my corrections? And you said I was wrong on the open vs shorted question. What is wrong with the first half of my sentence? And I did correct the base vs emitter issue, in the second half of my sentence. Are you still mixing CE and CB in your mind?:confused:

EDIT: perhaps you were thinking that since the circuit is common base, the fact that the base is grounded goes without saying, but that is the issue i was addressing in the first part of my sentence.

The Electrician

11-01-2008, 02:04 AM

EDIT: perhaps you were thinking that since the circuit is common base, the fact that the base is grounded goes without saying, but that is the issue i was addressing in the first part of my sentence.

I think that's probably it. When you said, "...output admittance when the base is grounded, not open...", the "not open" part would seem to suggest that you considered the possibility that the base might not be grounded. But this wouldn't be possible if the circuit is a grounded base circuit, so I figured you must be referring to the input port, and inadvertently said "base" instead of "emitter".

Ron H

11-01-2008, 02:12 AM

I think that's probably it. When you said, "...output admittance when the base is grounded, not open...", the "not open" part would seem to suggest that you considered the possibility that the base might not be grounded. But this wouldn't be possible if the circuit is a grounded base circuit, so I figured you must be referring to the input port, and inadvertently said "base" instead of "emitter".Whew! I'm glad we got that settled! I thought I was going nuts. :)

anik321

11-03-2008, 07:27 PM

The response from you guys is certainly impressive. It is crazy how in depth you can analyze a circuit if wished to.

Thank you hgmjr, your solution has been most helpful in helping me understand. I do have a few more questions regarding this particular problem but I will save it for the time being.

Thank you.

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