Talk:Bipolar transistor biasing

Latest comment: 2 years ago by Wtshymanski in topic Practical amplifier circuit may not work
Article milestones
DateProcessResult
November 4, 2007Peer reviewReviewed

Peer Review

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I'd like to request any editor whos had a look at this page to please go to the peer review link above and add your suggestions to improve the article. Thanks, xC | 19:28, 2 November 2007 (UTC)Reply

Learn to talk

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"... Q-point, is the point on the output characteristics that shows the DC collector–emitter voltage (Vce) and the collector current (Ic) with no input signal applied." What on EARTH does 'shows' mean??? Totally meaningless sentence.

Auto Peer Review

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The following suggestions were generated by a semi-automatic javascript program, and might not be applicable for the article in question.

You may wish to browse through User:AndyZ/Suggestions for further ideas. Thanks, xC | 19:30, 2 November 2007 (UTC)Reply

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Practical amplifier circuit may not work

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The practical amplifier circuit may not work for two reasons.

The first, alluded to in the article, is that the output transistors may go into thermal runaway. It would be good if the articled showed the rigorous way to calculate the minimum resistance needed. I'm doing this on the top of my head but the constraint should look like VCE*gm/(1+gm*Rdegen)*(2.2mV/degC)*ThetaJA < 1. That is, an increment in power causes a temperature change proportional to ThetaJA. This in turn causes a decrease in the VBE needed for fixed bias - by approximately 2.2mV/degC. Multiplication by the effective transconductance, gm/(1+gm*Rdegen), then provides the change in collector current. This is turn causes a change in power proportional to VCE. Rdegen should be chosen so that in the worst case, the thermal/power feedback is less than 1.

As regards the 2.2mV/degC...this confuses a lot of people. If one simply takes the derivative of the equation Ic=Is*exp(q*Vbe/(k*T)), one would expect the VBE to need to increase with temperature for a fixed current bias. However, Is also has a temperature dependence and this more than compensates for the T in the exponential.

The second issue why the circuit may not work is that it is not compensated. There should be a capacitive feedback from the output to the input of the common-emitter (PNP) stage. This is known as Miller compensation. On top of that, the emitter followers may oscillate - especially driving capacitive loads. This is covered in Gray & Meyer, a standard textbook on circuit design; it's relatively easy to derive too. ArtKalb (talk) 21:21, 26 July 2022 (UTC)Reply

Wikipedia is not a textbook. The amplifier might also have terrible impedance mismatch and low gain at 22 GHz. Notice the article title. --Wtshymanski (talk) 02:18, 14 August 2022 (UTC)Reply