AutoBiasing Transistors

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Sorry KChriste, I don't understand your formulas

Divider Z = R3//R4 (by R3 & R4 do you mean the PD resistors?) by parallel do you mean 1/ ((1/R3)+(1/R4))?
Transistor base Z = Beta (RE + r'e), do you mean Re? i havent seen this notation?
Input impedance , do you mean 1/ ((1/DividerZ)+(1/TransistorBaseZ))
 
I measured the output impedance of some 2-wire electret mic to be 5k ohms in parallel with a 10k resistor that powers it making 3.3k ohms.
A 3-wire electret mic has a much lower impedance because its FET is a source-follower. But most electret mics are 2-wire.

I selected a voltage gain of 100 for the preamp because then a 5mV signal from the mic is amplified to 500mV. The gain control reduces it to 100mv to 220mV which is the input sensitivity of most power amplifiers.

Um, what is (was?) a tape recorder? Was it something like a record player? A floppy disc drive? A blue-ray DVD burner? An 8GB RAM IC?
 
I now understand the 3.3k bit, but why put this device in parallel with a resistor?

I also understand the reason for having 100gain.

but how did you calculate the required impedance?
 
The 5k ohms 2-wire electret mic is powered by a 10k resistor. The two are in parallel to the signal creating a source impedance of 3.3k ohms.

The input impedance of the amplifier should be 5 times to 10 times higher to avoid signal loss.
The input impedance of the amplifier has the two divider resistors in parallel to the signal and the input of the transistor also in parallel.
 
UTMonkey said:
Sorry KChriste, I don't understand your formulas
Divider Z = R3//R4 (by R3 & R4 do you mean the PD resistors?) by parallel do you mean 1/ ((1/R3)+(1/R4))?
Yes. If you want to know WHY this is so, then read up on Thevenin's Theorem. It can really simplify the analysis of some circuits.
Input impedance , do you mean 1/ ((1/DividerZ)+(1/TransistorBaseZ))
Yes. That's why I wrote (// = parallel). It just makes things clearer in my mind. Sorry to confuse you.
Transistor base Z = Beta (RE + r'e), do you mean Re? i havent seen this notation?
To clarify a typo: Z = Beta (Re + r'e)
Ok, I'll also quote myself from a few posts back in this thread:
Audioguru said:
Um, what is (was?) a tape recorder? Was it something like a record player? A floppy disc drive? A blue-ray DVD burner? An 8GB RAM IC?
Yea... Something like that. I was trying to think of something that was junk that the Monkey could rob a mic from and nobody would care.
 
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I salvaged electret mics from telephone answering machines that got zapped by lightning and from worn out cell phones. I got a huge one from an old wired phone. Digikey actually sells many.
 
Ok, I have taken KChriste's formulas and plugged (nay thrown) in the values from my Amplifier.

DividerZ = R3 \\ R4
1/R3 = 1/16000
1/R3 = 0.0000625

1/R4 = 1/4400
1/R4 = 0.000273

DividerZ = 1 / (0.0000625+0.000273)
DividerZ = 1/ (0.0002898)
DividerZ = 3,450 ohms

Tran BaseZ = Beta(Re + r'e)
Tran BaseZ = Beta(130 +2.5)
Tran BaseZ = 200 * (132.5)
Tran BaseZ = 26,500 ohms

Input impedence = DividerZ \\ Tran BaseZ
1/DividerZ = 1/3450
1/DividerZ = 0.000290

1/Tran BaseZ = 1/26500
1/Tran BaseZ = 0.000038

Input Impedence = 1/(0.000290+0.000038)
Input Impedence = 1/(0.000329)
Input Impedence = 3048 ohms

So what would happen if i attached one of AudioGuru's Microphones (3.3k) to my amplifier?

AudioGuru suggests that the input impedence should be 5-10 times higher (16,500 - 33,000) Where would I start looking to improve this?

Regards
 
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If you connected a microphone that has an impedance of 3.3k to the input of your amplifier that has an input impedance of 3048 ohms then the output of the microphone will be loaded down so it is less than half.
The gain of your amplifier is so low that its output is useless.

The current is high on your amplifier and the output impedance is low at only 400 ohms. Multiply the resistors by 100 then the current is less, the output impedance is high and the input impedance is much higher.
Divide the emitter resistor into 2 resistors and bypass one with a capacitor. Then the gain is much higher but the input impedance is reduced.

Compromise with the resistors 10 times higher than at first and the same ratio of the emitter resistors to keep the increased gain the same then it has a reasonably high input impedance and has gain.
 

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Thanks for this.

I shall pop your figures into my simulator and let you know how I get on and also match your figures with what I come up with (its the only way the penny drops), 3 questions occur to me though:

1. for a microphone of 3.3k, what kind of input impedence would you expect an amplifier of this design to have, and knowing its input impedence what output impedence (for the next stage)?
2. I suppose I should be asking this later on (or in another thread) , the output in your last snapshot equates to 1.5mA @ 6v, is this enough current to drive a speaker, or is this a seperate topic worthy of another thread?
3. Whats the capacitor doing? I can see by your diagram it has contributed to the increased gain with a lower impedence on the input and output, how is it doing that? (pretty cool anyway)

In my previous post I quoted an input impedence of 5-10 times of source impedence (3.3k) which is 16,500 - 33,000 - your last snapshot demonstrates an input impedence of 14.5k, is this close enough?

Once again, many thanks.
 
UTMonkey said:
1. for a microphone of 3.3k, what kind of input impedence would you expect an amplifier of this design to have, and knowing its input impedence what output impedence (for the next stage)?
The input impedance of the mic preamp should be 5 times to 10 times the impedance of the microphone.
The output impedance of the mic preamp should be 1/5th to 1/10th the input impedance of whatever it is driving.

2. I suppose I should be asking this later on (or in another thread) , the output in your last snapshot equates to 1.5mA @ 6v, is this enough current to drive a speaker, or is this a seperate topic worthy of another thread?
A speaker is driven with high current from a power amplifier. The peak current from a 10W amplifier into an 8 ohm speaker is 1.6A. The peak current from a 100W amplifier is 5A.
This is a preamp circuit that drives a power amplifier.
The output impedance of many power amplifiers is only 0.04 ohms for good damping of a speaker's resonances.

3. Whats the capacitor doing? I can see by your diagram it has contributed to the increased gain with a lower impedence on the input and output, how is it doing that? (pretty cool anyway)
The capacitor is a short circuit at AC frequencies so the value of the emitter resistor is reduced to the value of the unbypassed capacitor. Then the gain is increased and the input impedance is reduced. The output impedance is not affected.

In my previous post I quoted an input impedence of 5-10 times of source impedence (3.3k) which is 16,500 - 33,000 - your last snapshot demonstrates an input impedence of 14.5k, is this close enough?
It is close enough.
This lousy circuit is for you to learn about transistors. Its distortion is very high at high levels and its output impoedance is high. Its input impedance is low. It is a circuit that was used in the sixties.

Modern preamps use opamps that have a very high input impedance, a very low output impedance, extremely low distortion and as much gain as you want.
 
audioguru said:
Modern preamps use opamps that have a very high input impedance, a very low output impedance, extremely low distortion and as much gain as you want.

And are MUCH, MUCH, easier to use!.
 
Hi AudioGuru, I have checked your figures and mine come pretty close to yours.

Accept for the last screen shot, as well as reducing the resistors by a factor of 100 you dropped the capacitor value.

Am I right in thinking that if the capacitor short ciruits the resistor with AC signals that the resistor bypassed is not included in the calculation of the TransistorBaseZ?

Regards
p.s. don't worry I feel this thread is coming to an end.

p.p.s. I missed the 60's (by a decade)
 
The capacitor has reactance (AC impedance) that is low at high frequencies and it "shorts" the resistor it bypasses.

At 100Hz, the 10uF capacitor has a reactance of only 16 ohms. At 16Hz the reactance is 100 ohms, the gain is reduced 3dB and the input impedance is a little higher than at higher frequencies.
 
Ahh I see.
So could I calculate the total resistance of Re thus:

Re = 100 ohms + (1.2k \\ reactance of 10uF at 1khz)

1/1.2k = 0.0008333
Reactance of 10uF @ 1khz = 1/(6.28 x 0.000010 x 1000) = 16k
1/16k = 0.0000625
Re = 100 + ( 1/(0.0008333+0.0000625))
Re = 100 + ( 1/0.0008958)
Re = 100 + ( 1116)
Re = 1.2K

Is that right?
 
In my circuit:
1) The transistor has a current of 1.1mA so its internal Re is 23 ohms.
2) The capacitor is a short at most frequencies.
3) The 100 ohms resistor is unchanged.
So the total Re is 123 ohms and the gain is 4k/123= 32.5 times.
The gain is reduced by the impedance of a load which is in parallel with the 4k collector resistor.
 

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