Off Topic: NPN Transistor Amplifier Configuration Help
Off Topic: NPN Transistor Amplifier Configuration Help
I know this topic is quite off-topic for this forum, and there does not seem to exist a similarly named 2n3096.org to ask on, but I can't think of a better group of people with whom to ask the question. Many here have both digital and analog experience in both audio and RF domains.
I've designed common-emitter with emitter feedback, common-base, and common-collector amplifiers before, even making a 60dB, 3-transistor amplifier once (deadbug style) to boost a signal from a crystal radio set. The formulae for these are well understood, and circuit operation is completely explainable by treating the transistor as a variable current sink (or source if common-collector).
However, one transistor amplifier configuration I'm not familiar with is the common emitter with collector feedback and a (potentially) grounded emitter. I've tried researching this configuration on the Internet and am left without any results. My ARRL Handbook lists a few circuits with this configuration, but the section on amplifiers lacks an explanation for how the thing works.
As far as I can see, since the collector always sits at a higher voltage than the base, the current from the collector into the base always guarantees a stable collector voltage. Any signal presented at the base will cause a counter-current which cancels the incoming signal, again leaving the output essentially constant voltage. By my calculations, this happens with both low and high values of collector feedback resistance.
Clearly my understanding of this particular configuration suffers. Worse, it seems as though this transistor configuration is taboo in professional as well as amateur circles.
If anyone can point me to documentation which explains how this configuration actually works, and provides examples of how to design one from scratch, I'd be most appreciative. Thanks.
I've designed common-emitter with emitter feedback, common-base, and common-collector amplifiers before, even making a 60dB, 3-transistor amplifier once (deadbug style) to boost a signal from a crystal radio set. The formulae for these are well understood, and circuit operation is completely explainable by treating the transistor as a variable current sink (or source if common-collector).
However, one transistor amplifier configuration I'm not familiar with is the common emitter with collector feedback and a (potentially) grounded emitter. I've tried researching this configuration on the Internet and am left without any results. My ARRL Handbook lists a few circuits with this configuration, but the section on amplifiers lacks an explanation for how the thing works.
As far as I can see, since the collector always sits at a higher voltage than the base, the current from the collector into the base always guarantees a stable collector voltage. Any signal presented at the base will cause a counter-current which cancels the incoming signal, again leaving the output essentially constant voltage. By my calculations, this happens with both low and high values of collector feedback resistance.
Clearly my understanding of this particular configuration suffers. Worse, it seems as though this transistor configuration is taboo in professional as well as amateur circles.
If anyone can point me to documentation which explains how this configuration actually works, and provides examples of how to design one from scratch, I'd be most appreciative. Thanks.
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ElEctric_EyE
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Thermal run-away is a problem with any silicon device, including FETs. See http://en.wikipedia.org/wiki/Thermal_ru ... er_MOSFETs
NPN and PNP transistors typically use emitter feedback to stabilize the output, which is why emitter feedback is so commonly documented. Still, circuits exist using collector feedback, and I wish to understand those circuits.
Class-D and class-E amplifiers can use (and have used) bipolar as well as FETs.
NPN and PNP transistors typically use emitter feedback to stabilize the output, which is why emitter feedback is so commonly documented. Still, circuits exist using collector feedback, and I wish to understand those circuits.
Class-D and class-E amplifiers can use (and have used) bipolar as well as FETs.
- GARTHWILSON
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The 2N3096 seems to be a pair of 110A SCRs. Is that what you meant, or 2N3906, the small-signal PNP?
Not necessarily. In an NPN, the collector usually sits at a higher voltage than the base, but unless the load is too heavy, the collector can pull that load down nearly to ground, ie, within a few millivolts of ground, while the base is up at .6V or a little higher.
You'll probably figure this out before I finish writing, but here goes.
Without diagrams, it sounds like you want a DC voltage divider resistor pair from your collector to ground with the tap going to the base. The B-E resistor current should be at least a few times as much as the base current so variations in β don't foul up your bias point. Make that resistor about .6V divided by that current. (Base current of course is the target quiescent collector current divided by β.) The C-B resistor will have that much current plus the base curent, so its value will be the target quiescent collector voltage minus .6V for the base, divided by the current.
If the collector voltage gets too low because it's conducting too much, the voltage-lowering at the base tends to shut it down. If the collector voltage is too high, the base voltage increases and gives the base more current and turns the transistor on harder, bringing the collector voltage back down.
The input signal will usually be coupled capacitively, and there will be a very low input impedance. The base voltage won't budge much at all. You can use this configuration kind of like an op amp where you bring the signal into the inverting input via a resistor. Ideally the signal voltage at the inverting input is zero, held there by infinite gain and an output which tries to put the same amount of current through the feedback resistor that we find coming in the input resistor in order to hold the inverting input at the same voltage as the non-inverting input. In a couple of our products, I've used this configuration of single-transistor amplifier (with an extremely high-gain NPN) without the input resistor, where the signal source was essentially a current source. Ideally the output signal voltage would be equal to the feedback (C-B) resistor (or should I say complex impedance if you also want a capacitor across that resistor) times the input signal current. Realistically it will be slightly lower.
When I worked in VHF and UHF power transistor applications engineering in the mid-1980's, our bipolars couldn't handle more than a 3:1 VSWR in (hopefully non-destructive) testing, because of the thermal runaway problem. As they got hotter, the emitter-base voltage would drop, inviting more current thus increasing the collector current (there was no DC load to limit collector current, only a choke to the power supply), which in turn increased the temperature more, making it spiral out of control and making it produce a flash and a high-pitched pop as it destroyed itself. I saw it too many times to count. Production testing of our MOSFETs included a 30:1 VSWR, all phase, at full rated power.
Quote:
since the collector always sits at a higher voltage than the base
You'll probably figure this out before I finish writing, but here goes.
Without diagrams, it sounds like you want a DC voltage divider resistor pair from your collector to ground with the tap going to the base. The B-E resistor current should be at least a few times as much as the base current so variations in β don't foul up your bias point. Make that resistor about .6V divided by that current. (Base current of course is the target quiescent collector current divided by β.) The C-B resistor will have that much current plus the base curent, so its value will be the target quiescent collector voltage minus .6V for the base, divided by the current.
If the collector voltage gets too low because it's conducting too much, the voltage-lowering at the base tends to shut it down. If the collector voltage is too high, the base voltage increases and gives the base more current and turns the transistor on harder, bringing the collector voltage back down.
The input signal will usually be coupled capacitively, and there will be a very low input impedance. The base voltage won't budge much at all. You can use this configuration kind of like an op amp where you bring the signal into the inverting input via a resistor. Ideally the signal voltage at the inverting input is zero, held there by infinite gain and an output which tries to put the same amount of current through the feedback resistor that we find coming in the input resistor in order to hold the inverting input at the same voltage as the non-inverting input. In a couple of our products, I've used this configuration of single-transistor amplifier (with an extremely high-gain NPN) without the input resistor, where the signal source was essentially a current source. Ideally the output signal voltage would be equal to the feedback (C-B) resistor (or should I say complex impedance if you also want a capacitor across that resistor) times the input signal current. Realistically it will be slightly lower.
When I worked in VHF and UHF power transistor applications engineering in the mid-1980's, our bipolars couldn't handle more than a 3:1 VSWR in (hopefully non-destructive) testing, because of the thermal runaway problem. As they got hotter, the emitter-base voltage would drop, inviting more current thus increasing the collector current (there was no DC load to limit collector current, only a choke to the power supply), which in turn increased the temperature more, making it spiral out of control and making it produce a flash and a high-pitched pop as it destroyed itself. I saw it too many times to count. Production testing of our MOSFETs included a 30:1 VSWR, all phase, at full rated power.
GARTHWILSON wrote:
The 2N3096 seems to be a pair of 110A SCRs. Is that what you meant, or 2N3906, the small-signal PNP?
Here's the schematic I'm talking about:
Code: Select all
o
|
<
> Rc
<
Rf |
+--^v^----*----> out
| |
| |/
o-||--*-------|
|\
|
|
---
///
- GARTHWILSON
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Ok, well, it'll be pretty unstable without another resistor from base to ground. Since the impedance to the signal is so low at the base anyway, the resistor to ground will not have much effect on the signal-- only on keeping control of the circuit biaswise. Make that resistor to ground take several times as much current as the base, so a small change in DC collector voltage will result in a large change in base current because of its non-linear curve.
To control the gain, you'll either need a resistor in series with that input capacitor, or have a source that already looks more-or-less like a current source.
To control the gain, you'll either need a resistor in series with that input capacitor, or have a source that already looks more-or-less like a current source.
The input impedance is around 58 ohms (0.7V divided by the current through the collector), which makes it a good match for RF work coupling to coax. (Yes, this is why I'm curious about how these circuits work -- they appear in radio schematics with sufficient frequency that I feel I need to know about them.)
Rf typically is in the tens of kilohms, while Rc typically is less than 500 ohms. For example, Rf = 70K while Rc = 270.
Input sources tend to be current-mode (e.g., photodiode) anyway.
But, my question basically boils down to, why would Rf = 70K? I can figure out Rc easily enough, but it's Rf that confuses me, and the fact that it taps the collector output means you can not think of the transistor as a variable current source anymore (with or without a resistor from base to ground), for now your output voltage is the result of a simultaneous equation, thanks to the fact that your transistor's bias (that which should be most stable) is now predicated on the amplified signal (that which is most variable).
Rf typically is in the tens of kilohms, while Rc typically is less than 500 ohms. For example, Rf = 70K while Rc = 270.
Input sources tend to be current-mode (e.g., photodiode) anyway.
But, my question basically boils down to, why would Rf = 70K? I can figure out Rc easily enough, but it's Rf that confuses me, and the fact that it taps the collector output means you can not think of the transistor as a variable current source anymore (with or without a resistor from base to ground), for now your output voltage is the result of a simultaneous equation, thanks to the fact that your transistor's bias (that which should be most stable) is now predicated on the amplified signal (that which is most variable).
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Quote:
The input impedance is around 58 ohms (0.7V divided by the current through the collector), which makes it a good match for RF work coupling to coax.
Ideally you'd have a network analyzer, directional couplers, attenuators, RF power meters, and all kinds of very expensive equipment if you're going to get serious about RF design; but it really is beyond the hobbyist's budget to go to that extent. The lab I worked in had about $100,000 of equipment per engineer, and that was almost 25 years ago. If you're only at HF (heaven knows a pull-up resistor in the collector circuit will sure limit the frequency!) and really know what you're doing, you could probably jerry-rig a significant part of it while getting a few things from the electronics swap meets or eBay. Still, it would be good to sneak it into a decent lab to at least get some calibration data on your many pieces of equipment.
Quote:
why would Rf = 70K?
Quote:
and the fact that it taps the collector output means you can not think of the transistor as a variable current source anymore (with or without a resistor from base to ground), for now your output voltage is the result of a simultaneous equation, thanks to the fact that your transistor's bias (that which should be most stable) is now predicated on the amplified signal (that which is most variable).
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Nightmaretony
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- GARTHWILSON
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Quote:
The current gain from the transistor, fed back through Rf, will negate any input signal you provide it, which means your output voltage remains more or less fixed.
In other words, the transimpedance gain of this amplifier is quite close to zero.
In other words, the transimpedance gain of this amplifier is quite close to zero.
This transistor configuration may seem inoperable, but the truth is that we've sold thousands of communications microphone pre-amplifiers made in this configuration, and their performance is very predictable and dependable. It is, in essence like a low-quality op amp whose non-inverting input pin is hidden.
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Rick Cortese
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Yes, I am that kind 'special' to the point where they wanted me to ride the short bus to school. Problem is I can't remember things like my phone number or year the Treaty of Trent was signed. Circuit looked familiar with a Forrest Mims article or two I read ~40 years ago so I googled with his name and
http://www.creative-science.org.uk/lightbeam.html
He capacitively couples the input and output to block DC voltage which makes a lot of sense.
I'm pretty sure he used it more then a few times so you can probably find one where he explains it. He is still around and active on the internet so you may want to email him.
Rick
http://www.creative-science.org.uk/lightbeam.html
He capacitively couples the input and output to block DC voltage which makes a lot of sense.
I'm pretty sure he used it more then a few times so you can probably find one where he explains it. He is still around and active on the internet so you may want to email him.
Rick
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leeeeee
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Nightmaretony
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