Author Topic: Gain calculation of common emitter with emitter feedback  (Read 1160 times)

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Offline Qmavam

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Gain calculation of common emitter with emitter feedback
« on: April 21, 2017, 10:06:03 PM »
I have ran the calculations for a common emitter circuit from this page.
http://www.electronics-tutorials.ws/amplifier/emitter-resistance.html

 However the circuit I have has emitter feedback to the base. I don't know how to
figure that into the calculation
 In the attachment you will see the amp I calculated from the page on the left, and
the amp I am using on the right. I want a gain of 4.5 on the amp on the right.
I also want to keep the 500 ohm collector resistor.
 What is the purpose of the feedback capacitor?
      Thanks, Mikek

 

Offline danadak

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Re: Gain calculation of common emitter with emitter feedback
« Reply #1 on: April 21, 2017, 10:49:20 PM »
That cap is 15 ohms at 1 Khz, so essentially forcing signal at emitter
to be same at base, hence gain would be forced to 1.



Regards, Dana.
 

Online Ian.M

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Re: Gain calculation of common emitter with emitter feedback
« Reply #2 on: April 21, 2017, 10:55:47 PM »
No.  Its a bootstrapping capacitor.  It raises the AC input impedance by virtually eliminating the loading of R8.  If the source impedance is low it has little to no effect on the gain (assuming the output is taken from the collector).
« Last Edit: April 21, 2017, 11:38:22 PM by Ian.M »
 

Offline Hensingler

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Re: Gain calculation of common emitter with emitter feedback
« Reply #3 on: April 21, 2017, 11:04:12 PM »
No.  Its a bootstrapping capacitor.  It raises the AC input impedance by virtually eliminating the loading of R8.  If the source impedance is low it has little to no effect on the gain (assuming the output is taken from the collector).

In the given circuit at high frequency it also effectively halves the emitter resistor and so doubles the amplifier gain.
 

Offline Qmavam

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Re: Gain calculation of common emitter with emitter feedback
« Reply #4 on: April 21, 2017, 11:09:20 PM »
That cap is 15 ohms at 1 Khz, so essentially forcing signal at emitter
to be same at base, hence gain would be forced to 1.



Regards, Dana.

  I know thats not correct, because I have built several with gains of 10 to 15.
Note; you do have a 1.2k resistor between the cap and the base, so I the base and emitter
won't necessarily be the same.
 

Offline Qmavam

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Re: Gain calculation of common emitter with emitter feedback
« Reply #5 on: April 21, 2017, 11:14:35 PM »
I should have added I'm using this from 500kHz to 10Mhz, so the 10uf is basically zero ohms.
does that change the gain calculation to, Gain = RL/(R9+R10) ?
 
                                             Thanks, Mikek
 

Offline Hensingler

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Re: Gain calculation of common emitter with emitter feedback
« Reply #6 on: April 21, 2017, 11:26:57 PM »
I should have added I'm using this from 500kHz to 10Mhz, so the 10uf is basically zero ohms.
does that change the gain calculation to, Gain = RL/(R9+R10) ?

A 'zero ohm' capacitor effectively connects R9 and R10 in parallel not series)
 

Online Ian.M

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Re: Gain calculation of common emitter with emitter feedback
« Reply #7 on: April 21, 2017, 11:37:53 PM »
No.  Its a bootstrapping capacitor.  It raises the AC input impedance by virtually eliminating the loading of R8. If the source impedance is low it has little to no effect on the gain (assuming the output is taken from the collector).

In the given circuit at high frequency it also effectively halves the emitter resistor and so doubles the amplifier gain.
Yes.  I forgot the effect of it shunting the emitter resistor with R9.  Thanks for pointing out my mistake.
 

Offline vk6zgo

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Re: Gain calculation of common emitter with emitter feedback
« Reply #8 on: April 22, 2017, 09:00:14 AM »
It's years since I did this crap, but the gain formula for an  common emitter amp with unbypassed
emitter resistor is:- G = gmRc/1+gmRe

1 is very much smaller than gmRe , & can be ignored, so the formula looks " near as dammit" to be G = gmRc/gmRe.

Dividing top & bottom by gm gives us G = Rc/Re,so for your first circuit  G=500/62, or  8.0645, just as you say.

In the standard bypassed configuration the bypass C makes Re  look very, very, close to zero ohms for ac, so we get G = gmRc/1+0 , or G = gmRc.

In your second circuit, I would guess* that any additional feedback appearing in the base circuit due to the connection is minimal, because of the presence of R8, so the formula reverts to
Rc/Re, where Re is R9 & R10 in parallel, so G = 500/55, or 9.09, not 4.5.

*Only a guess-------as I said, I've haven't done this stuff for years.
 

Offline Hero999

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Re: Gain calculation of common emitter with emitter feedback
« Reply #9 on: April 22, 2017, 10:00:37 AM »
Ian and Hensingler are correct. C4 is a bootstrapping capacitor. It increases the input impedance by coupling the signal at the base to the emitter, via R8 and connects R9 in parallel with R10
(hence the gain is doubled in this case as R9 = R10) at higher frequencies. If I remember rightly, it's mentioned in the Art of Electronics.

The input impedance is roughly R8/(the gain of the emitter follower - 1). Ignore RC, refer to Wikipedia for the formula to work out the gain of the emitter follower.
 

Online Ian.M

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Re: Gain calculation of common emitter with emitter feedback
« Reply #10 on: April 22, 2017, 10:41:23 AM »
Don't forget R7, the upper resistor in the base bias potential divider.  For AC, its in parallel with R8 multiplied by the bootstrapping factor 1(1-emitter_follower_gain), and also you must consider the input impedance of the transistor itself.
 

Offline Qmavam

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Re: Gain calculation of common emitter with emitter feedback
« Reply #11 on: April 22, 2017, 12:08:03 PM »
It's years since I did this crap, but the gain formula for an  common emitter amp with unbypassed
emitter resistor is:- G = gmRc/1+gmRe

1 is very much smaller than gmRe , & can be ignored, so the formula looks " near as dammit" to be G = gmRc/gmRe.

Dividing top & bottom by gm gives us G = Rc/Re,so for your first circuit  G=500/62, or  8.0645, just as you say.

In the standard bypassed configuration the bypass C makes Re  look very, very, close to zero ohms for ac, so we get G = gmRc/1+0 , or G = gmRc.

In your second circuit, I would guess* that any additional feedback appearing in the base circuit due to the connection is minimal, because of the presence of R8, so the formula reverts to
Rc/Re, where Re is R9 & R10 in parallel, so G = 500/55, or 9.09, not 4.5.

*Only a guess-------as I said, I've haven't done this stuff for years.
I used 220 ohms for R8 and R9, the gain was 4.3, although after I put the diode detector load on it, the gain dropped to 3.8.
 I need to recheck my circuit detector, it shouldn't load it that much.
 

Online Ian.M

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Re: Gain calculation of common emitter with emitter feedback
« Reply #12 on: April 22, 2017, 01:19:50 PM »
I used 220 ohms for R8 and R9, the gain was 4.3, although after I put the diode detector load on it, the gain dropped to 3.8.
 I need to recheck my circuit detector, it shouldn't load it that much.
The electrolytic is probably pretty crappy at RF - put a 0.1uF ceramic directly across it.
 

Offline vk6zgo

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Re: Gain calculation of common emitter with emitter feedback
« Reply #13 on: April 22, 2017, 02:13:26 PM »
I had a look at http://www.electronics-tutorials.ws/amplifier/emitter-resistance.html, & to say that I was "underwhelmed" is understating it.

Initially,they showed a BJT with no Emitter resistor, but with a voltage divider bias supply.
The "bottom" resistor only had to drop 0.6 volts, so would have been vanishingly small compared to the "top" one .
The upshot would be an extremely low input impedance.

They then went on to a "reasonable" discussion of  emitter resistors, (without, however, addressing the above point), then on to emitter bypassing, but paid a lot of attention to the effect of the capacitor at higher frequencies, I quote:-

"one of its main disadvantages is that at high frequencies the capacitors reactance becomes so low that it effectively shorts out the emitter resistance, RE as the frequency increases."

In my innocence, I thought that it was supposed to effectively do that across the frequency range of interest!

"The result is that at high frequencies the reactance of the capacitor allows very little AC feedback control because RE is shorted out which also means that the AC voltage gain of the transistor is greatly increased driving the amplifier into saturation."

If they really mean "high frequencies", this is bollocks, as the high frequency gain is much more dependent upon the CR time constant of the Collector load & circuit stray capacitances, so this "saturation" is extremely unlikely to happen.
Another point is the apparent assumption that the emitter bypass cap was a perfect capacitor, & wouldn't start to look a bit "inductive" at higher frequencies.




 

Online Ian.M

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Re: Gain calculation of common emitter with emitter feedback
« Reply #14 on: April 22, 2017, 03:23:03 PM »
Due to the complexity of the calculations for the bootstrapped common emitter amplifier with emitter degeneration, unless you *HAVE* to plough through them for academic credit, I strongly recommend using a SPICE package to do the analysis.
I prefer to use the free LTspice (http://www.linear.com/designtools/software/) for this sort of stuff.

There are three key analyses you need to run:
  • .op to get the DC operating point so you can see if you've FUBARed the biassing,
  • .ac for the small signal AC node voltages and currents over the frequency range you specify
  • .net to get the two port network parameters between the specified input source connected to the input port, and the output port.  This is actually a sub-analysis of the .ac one, and we only need it for Zin and Zout
 

For a TLDR didn't read the EE101 textbook introduction to Two-port Network Parameters, see: https://global.oup.com/us/companion.websites/9780199339136/student/app/app_c

For a TLDR intro to setting up LTspice .net analyses with realistic source and load impedances see: http://courses.ee.sun.ac.za/HFTegniek_414/JB/notas/Two%20port%20parameters%20in%20LTSPICE.pdf

I've set up your circuit and run the analyses with an infinite source impedance (using a current source) and no load on the output.


The NPN model is a near ideal NPN BJT with Hfe=100 (see http://ltwiki.org/index.php5?title=Q_Bipolar_transistor) and to get results that resemble what you've breadboarded, you'll need to start off by finding a SPICE model for whatever transistor you are actually using, then add realistic parasitics (ESR and inductance) for C4, and realistic source and load impedances and coupling capacitors.

I'm seeing an input impedance of 1.14K and an unloaded voltage gain of 4.43.  N.B. your biassing is FUBARed - the collector voltage is sitting too close to the positive supply so it will clip on anything other than very small signals.  Increase R8 and R9 to get the collector to about 1/2 Vcc or till the collector current reaches the current the transistor's Hfe is specified at (whichever comes first) then adjust R10 to get the correct gain.

Zipped sim and plot settings attached.
« Last Edit: April 22, 2017, 04:25:01 PM by Ian.M »
 

Offline Qmavam

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Re: Gain calculation of common emitter with emitter feedback
« Reply #15 on: April 22, 2017, 10:33:44 PM »
  Thanks for all the time you put into that.
 I have had a few iterations on R values, don't think I ever had the exact values in your model.
Obviously with the collector voltage at 23.04, that can't work.
btw, I'm using a BC547B transistor.

The working circuit I have has R7=12.4k, R8=2.7k, R 9=220 R=220 and R= 500 ohms.
For R8, I installed a pot and adjusted until I got distortion at both peaks at the same amplitude.
 The collector is 14.41V, the base is 4.61V and the emitter is 3.95V.
 The circuit does give me just over 15Vpp. This is with an 11K load,
which I should have removed, it is a remnant from the diode detector circuit.
When I add the rest of the diode detector it is an additional 7k load.
 You are correct about the 10uf, the original circuit did have a 0.1uf bypassing the 10uf.
                                                           Mikek
 

Offline Qmavam

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Re: Gain calculation of common emitter with emitter feedback
« Reply #16 on: April 23, 2017, 02:14:58 PM »
I used 220 ohms for R8 and R9, the gain was 4.3, although after I put the diode detector load on it, the gain dropped to 3.8.
 I need to recheck my circuit detector, it shouldn't load it that much.
The electrolytic is probably pretty crappy at RF - put a 0.1uF ceramic directly across it.
    Hmm, the electrolytic was more than crappy. I was having intermitent distortion on
one of the peaks. At first I thought it was my transistor getting warm, after a couple of shots of freeze mist
I found  it was the 10µf capacitor. I put it on my cap tester, it looked ok at 100Hz, 1000Hz, 10Hz, but when I went to 100kHz
it looked like a resistor. The part was from Mouser and is a tiny Nichicon 10uf aluminum electrolytic. 10µf 16V 85*C.
 I put in a larger 10µf cap and the problem went away.
                                    Mikek
 

Offline MrAl

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Re: Gain calculation of common emitter with emitter feedback
« Reply #17 on: April 25, 2017, 06:55:07 PM »
Hi,

The cap is a short for AC but open for DC, so for biasing the cap is open circuit but for amplification it is short circuit at these frequencies.

You could replace the transistor with a current controlled current source with gain (Beta) maybe 50 and go from there.  That would allow you to use a regular analysis technique like Nodal analysis.

One thing missing however is the impedance of the input source voltage.  We need that to actually calculate the gain because that plays a part in the overall gain of the circuit.

Using an impedance of 1k ohm for the input source resistance, at a frequency of 1MHz the gain calculates to 6.34 with some usual assumptions like the base emitter voltage is a constant 0.7 volts and the internal base resistance is 1 ohm, and the Beta of the transistor is a constant 50.


« Last Edit: April 25, 2017, 07:35:33 PM by MrAl »
 

Offline danadak

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Re: Gain calculation of common emitter with emitter feedback
« Reply #18 on: April 25, 2017, 10:24:56 PM »
All caps have parasitics, like L which make them ineffective at high frequency.

Bulk caps, plain vanilla electrolytics, run out of gas at moderate frequencies, << 100 Khz.
Tantalums better, polymer tanatalum better better, MLC best.


Pay attention to the actual datasheets, in any given technology not all caps perform the
same, largely dictated by construction, dielectric.


Regards, Dana.
 


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