# BJT colpitts transconductance value

Discussion in 'General Electronics Chat' started by KCHARROIS, Apr 8, 2013.

1. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
8
Hello,

The colpitts oscillator circuit has come up many times is this forum but I can't find one regarding transconductance of the circuit. (Circuit is attached)

So lets say initialy the circuit is not oscillating, and I have 1.7V going into the base of the transistor. ≈0.7V will be dropped accross the diode leaving 1V accross R2. If R2 were to be equal to 1KΩ, Ie would equal 1mA there for I can assume that gm small signal equals 1mA/Vbe(26mV) = 0.038, correct me if I'm wrong. Once the circuit begins to oscillate, and that theres 156mV signal going into the base 1.156V will be accross R2 meaning now that I will have a Gm large signal of 1.156mA/Vbe(26mV) = 0.044??? Is this correct?

I'm reading a book, and its showing calculation for the axact same circuit below. Initialy circuit isn't oscillation R2=1kΩ at 1V giving an mho of 0.038. When it start oscillating he says that theres 0.156mV signal going into the base makings 1.156V accross R2 giving mho of 0.044 but the book says Av=gm/Gm, gain was equal to 3.2 gm=0.038mho which means that Gm would have to ve equal to 0.011875mho.

How does this make sense I calculate 0.044mho and the book says 0.011875 mho?

Thanks

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Where did this information come from ...?

"Once the circuit begins to oscillate, and that theres 156mV signal going into the base 1.156V will be accross R2 ....."

3. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
8
The actual circuit is attached below and the information/calculation aswell:

VRe = 1.7 - 0.7V = 1V
IRE = 1V/1000Ω = 1ma
gm = 1ma/0.026mV = 0.038 mho

Gain = (gm*RL)(1/n) 1/n is step down between caps C1/(C1+C2) = 1/31
(0.038*2700)/31 = 3.2

Since gain equals 3.2 the author presumes a reasonable prediction that the input amplitude will be 1/gain = 0.312 and that its approximately Gm/gm there fore meaning Av = gm/Gm.

From a graph gm/Gm = 0.304 which gives an approximate amplitude of 156mV.

And then output voltage will be 156mV*0.312*0.038*2700 = 5.0Vp

He then notes that Gm = 0.312 * 0.038mho = 0.011856mho which I don't think makes sense.

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4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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"He then notes that Gm = 0.312 * 0.038mho = 0.011856mho which I don't think makes sense."

I've no concerns with the author's conclusion. What is your concern?

One expects the large signal Gm to be lower than the small signal (pre-oscillatory) gm. Otherwise the oscillation amplitude would not stabilize to give a "pure" sine wave output.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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As a further clarification, your original post stated that the signal amplitude into the base would be 156mV. As this is a common base configuration the base is at earth potential for the purposes of AC behavior. The 156mV AC feedback component is effectively applied to the emitter in this topology - rather than to the base.

6. ### Brownout Well-Known Member

Jan 10, 2012
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GM will vary linerarly IAW the equation: GM = iC/VT. iC is IC/aV, at the minimum. So, substuting: GM= IC/(aV*VT) = gm/aV.

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7. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
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My concern is I dont understand how he came about to saying that Gm = 0.011856 mho. Where did the formula Av = gm/Gm come from?

Thanks

8. ### Brownout Well-Known Member

Jan 10, 2012
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Did you not read my post?

9. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
8
Sorry should have read/looked better but I don't see how you came up with this you show at the end of your post that GM= IC/(aV*VT) = gm/aV
but where did VT go? Where did you get this information from?

Thanks

10. ### Brownout Well-Known Member

Jan 10, 2012
2,373
1,003
little gm is IC/vT, so look at it this way: GM = (IC/vT) * 1/aV. The first factor is just little gm, so that results in gm/aV.

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11. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
8
Great this makes sense, is there a book or a website that you got this information from that I culd reference too?

Thanks

12. ### Brownout Well-Known Member

Jan 10, 2012
2,373
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Not really. I just fooled around with the math. I learned transistors from Microelectronics Curcuits by Sedra and Smith about 20 years ago.

13. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
8
Well I've decided to analyze the circuit differently and I came up with these calculated values which matched the actual circuit values within a few millivolts. So I'm wondering if someone can look at my calculated values and see if they agree with the calculations I did.

Thanks

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14. ### Brownout Well-Known Member

Jan 10, 2012
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Your attachment is impossible to read. However here's how I do it. hfe will be lowest near cutoff. As an estimate, I calculate that the peak amplitude to be the difference between the Q-point and cutoff. No bessel functions or large signal analysis needed. And it comes out pretty close.

15. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
302
8
If you open the picture and click on it, it zooms in making it readable...

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Not sure I understand this - would you be kind enough to elaborate with some numbers perhaps. Thanks.