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TRANSISTOR ATTENUATOR?


prateeksikka
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HI ALL!
LET US CONSIDER A TRANSISTOR OPERATING IN ACTIVE REGION AS AN AMPLFIER.IT IS IN COMMON EMITTER CONFIG.IT HAS A GAIN OF SAY 50.I APPLY 1mV signal at i/p and i get 50mV at the collector.what would happen if i now apply a 50 mV signal at the collector?will it attenuate the signal?what do you think will it give 1mV at the base?Now the main thing .I have noticed that it does give 1mV at the base .it is surprising.I too was surprised that we cant have a reversed transistor operation.But when i carefully monitored the signal,the frequency of the signal was not the same as the one i had applied at the collector.tomorrow i am going to simulate it on ORCAD .any explanations? ::)
prateek

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Feeding a signal to the collector varies the collector and emitter current. Therefore the emitter's resistance, Re, has a varying voltage across it. Since it is also in series with the base-emitter junction, the base voltage varies a little amount. The signal at the base is not the same frequency because of severe distortion caused by the exponential voltage change requirement at a base to make a linear change at a collector.
Two resistors connected in series to make a voltage-divider would make a much better attenuator. ;D 

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I think the gain/loss is a lot more complicated:
1) Much of the input current through the 47k resistor is dumped in Rbias.
2) The input impedance of a transistor without an emitter resistance is fairly low so also attenuates the current through the 47k input resistor.
3) In addition to current gain, the transistor has some voltage gain (same thing?).
4) The feedback resistor feeds current into the 47k input resistor to cancel some of it, into Rbias, and into the low input impedance of the transistor.

I just finished doing my daughter's income tax so I don't want to calculate anything more.
A perfect application for a smart sim program or actual bench testing! ;D

An inverting opamp circuit is much easier to figure its gain/loss. 

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Yes audioguru an op-amp would be much better.

The the current though Rf cancels out the low emitter impedance and Rbias this is an example of how negative feedback increases the input impedance. I've done a quick simulation and the loss is just under Rin/Rf, if you increased the open-loop gain by using a darlington or constant current collector load or even both then the loss would be damn near Rin/Rf.

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this is an example of how negative feedback increases the input impedance.

Hi Alun,
You mean the negative feedback decreases the input impedance, don't you?
The input of the transistor is approaching being a virtural ground. I think that the input impedance is very close to the value of the 47k input resistor.

Is your sim program any good? Frequently they screw-up on complicated issues like this one.  ;D
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Guest Alun

audioguru,
At DC the current through Rf if added to the input current therefore increasing the input impedance this is why Rbias makes very little difference since the base has a very low impedence anyway and 700mV across 2K issn't going to cause much current to flow, and yes you're right the input impedance will be about the value of Rin. This circuit is a lot better than a potential divider but nowhere neard as good as an ap-amp.

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Hi Alun,
Never mind the DC current, the thing is cap-coupled.
The junction of the input resistor and the feedback resistor is a low impedance point at AC because as the input swings positive, the output swings negative and vice-versa. Therefore the junction is a virtural-ground like the inverting input pin on an inverting opamp amp. ;D

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Guest Alun

The op-amp has a very high gain - some op-amps have a gain > 1 million.

Because the gain is so high, the output is always 1 million times greater than the input the input can be considered to be 0. For example, if the outputs 1V, and the input (Vd on my circuit) would be 1uV which for all intents and purposes is virtually 0V.

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A little confusing.
In Alun's inverting opamp circuit, if the output is -1V, then the input of the opamp is +1uV and the circuit's input to the 47k resistor is +0.212767V.
Due to the opamp's extremely high gain and extemely high input impedance, Rf and Rin simply form a voltage divider to determine the gain of the circuit.  ;D

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Guest Alun

Oh sorry of course -1uV with 1V input because it's inverting  silly me.;D Rf is providing feedback, some of the output is fed back to the input and as it's an inverting amplifier the output will be opposite to the input and cancel it out to some extent, hence the feedback is negative, the ratio of Rf and Rin deturmines the amout of negative feedback.

The principle is the same with my transistors circuit, the base is always near ground at 700mV and the transistor has a gain of about 100, Rf and Rin reduce it to 10/47, the transisto has a lower gain than the op-amp so in practice the gain will be a bit below 10/47.

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Guest Alun

prateeksikka,
It isn't considered to be exactly 0V, but near 0V, that's why it's called a virtual ground, and not just ground. If the input to the circuit is 5V and Vd is only 5mV, 5mV is nothing compared to 5V in pracitice Vd will be a lot smaller than 5mV because the gain of a cheap op-amp is a lot higher than 1000.

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  • 2 months later...

I have seen where a signal can originate at the emitter of a transistor, or at the base. Assuming a base resistor, it seemed possible that the signal applied to the emitter would be larger at the base because of the base current and the resistor. And in fact there is a signal at the base, but it is smalller not larger. Scoff if you will, but it seems logical that the signal could be larger dependent on the base resistor.

It's hard to see, but if you plug in some numbers, here is what happens.




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Kevin,
I don't know what your numbers are supposed to be about.
Aren't you making a Cascode Amplifier?
It doesn't need a collector resistor for its lower transistor.
The voltage gain of the lower transistor is only about 1 becase its high impedance collector is severely loaded down by the very low input impedance of the emitter of the upper transistor. Therefore the circuit performs about the same as a single common emitter transistor at low frequencies.

A cascode amplifier is used at high frequencies because the C-B capacitance of each transistor doesn't cause negative feedback at high frequencies as it does in a common emitter amplifier.

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Hi Audioguru, I'm glad you were able to identify the type of cirucit I was describing along with the low impedance observation. I think I used a resistor to increase the impedance. Other than that, here is an acceptable gain circuit. The nice thing about the common base is that the emitter resistor drops most of the voltage creating a large change in current. You know that sometimes you gain a signal only to realize loss somewhere due to the resistances. Aren't we lucky that we can use a resistor here to increase the impedance and at the same time realize a two stage gain amplifier without loss. Amazing.

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