Well, unfortunately this is not true. Re helps a lot.But the VBE voltage still remains: Vth * ln (IC / Is).
So, its DC operating point's dependence on the temperature remained the same.
Ups... yes,you are absolutely right. I forgot about that.Jony130, you have apparently ignored the fact that the impedance looking into the collector is not simply Hoe, but is substantially higher due to the feedback provided by the external emitter resistance. Have a look here for an analysis.
Wow, lots of good information! It is interesting to see how temperature affects a given transistor circuit.
Austin
That resistor is the value equal to around (R1 / Av.) AC gain not DC gain.Since I don't know how Ron derived the correct value for R5, I am going to stay put with just a single Re resistor.
As for calculating the value of the cap across the emitter resistor, the equation below is necessary:
\(X_c=\frac{1}{2 \pi f C\)
What would be an adequate reactance for Xc?
Austin
I see so Xc=(R1 / Av) - for AC gain.That resistor is the value equal to around (R1 / Av.) AC gain not DC gain.
DC gain is R1 / R2) AC signal gain is R1 / R5..
So that resistor in series with the cap ensures AC feedback with a higher input impedance?You want some AC feedback too, if it were plain cap. then input impedance would be VERY low, to signal source, with some AC feedback it allows some better high frequency response as well.
As well as kepping a stiff gain so as to not be transistor parameter dependant.
So that resistor in series with the cap ensures AC feedback with a higher input impedance?
Austin
My design proceeded as follows:Since I don't know how Ron derived the correct value for R5, I am going to stay put with just a single Re resistor.
As for calculating the value of the cap across the emitter resistor, the equation below is necessary:
\(X_c=\frac{1}{2 \pi f C\)
What would be an adequate reactance for Xc?
Austin
Thanks Ron, glad to have that post. I did notice the approximations, but I'll do as you advised and begin with the precise calculations.My design proceeded as follows:
1. I arbitrarily decided to set Ic≈Ie≈1mA.
2. I chose Ve≈1V, which is large enough to make the nominal ΔVbe≈2mV/°C relatively insignificant, but small enough that the signal swing at the collector can be fairly large. This requires that R2=1k.
3. I chose Vcc=20V for large signal swing, and Vc≈10V so that limiting on positive and negative peaks will occur at approximately the same input voltage amplitude. This requires that R2=10k.
4. I chose the bias network so that I(R3) and I(R4)≈Ic/10, for stability with changes in β due to temperature and/or individual device characteristics.
5. In order to get a gain of 100, the total emitter resistance re+(R2||R5) needs to be approximately equal to R1/100 (remember re≈26Ω @ Ie=1mA). This assumes that the impedance of C2<<R5 at midband. The calculated value of R5 is 82Ω. I chose the next lowest standard value (75Ω) because I knew that there would be a little gain loss due to β and Hoe.
6. I arbitrarily chose F(-3dB)≈20Hz. For C2, F(-3dB)=1/(C2*2*∏*(re||R1)+R5). 100μF is the next highest standard value.
7. I have to admit I sorta pulled the value of C1=1μF out of a hat, knowing from experience that I would get about the same -3dB point for the input coupling. It might be better if C1 a little larger, in order to get the 20Hz cutoff point.
Note that I used a lot of approximations here. The precise calculations are good for understanding where the approximations can be made. Component tolerances are such that the approximations I made are justified, and make life a lot simpler.
The whole process of designng, drawing, and simulating the circuit took a lot less time than it did for me to write this up.
Thanks loosewire. Once I ideally master Class-A amps, I'll post another Attempt #X that will give a practical approach on how to bias for Class-A amps for everyone's benefit.Are you being advised about putting your work out that can be copied
anywhere in the world. I admire your Intelligence real or seeking to be
a great contributor.
Am I to simply measure the typical input impedance of the microphone and then subtract that from what impedance I need? Then I can use the difference for that resistor value?Yes it does...
It is difficult to measure the impedance of an electret mic because it has a FET transistor inside. Its impedance is about 3.3k ohms and it is powered from a 10k ohms resistor. Then the preamp input impedance should be 15k ohms or more to avoid the mic output from being loaded down.Am I to simply measure the typical input impedance of the microphone and then subtract that from what impedance I need? Then I can use the difference for that resistor value?
Austin
Hi Audioguru,It is difficult to measure the impedance of an electret mic because it has a FET transistor inside. Its impedance is about 3.3k ohms and it is powered from a 10k ohms resistor. Then the preamp input impedance should be 15k ohms or more to avoid the mic output from being loaded down.
But your transistor will have an input impedance of only 4k ohms which is too low. Don't add a series resistor to increase the input impedance because it will just reduce the level.
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