Rc coupled amplifier

[earjun]

Hello guys can someone help me out with the working of an RC coupled amplifier?

 

[colin55]

The "C" prevents the voltage from one stage interfering or upsetting or (controlling unnecessarily), the next stage. It allows the AC component to pass from one stage to the next and blocks the DC component.
For instance, one stage may be working on 5v and the next stage may be working on 25v. The capacitor will completely separate the two stages but allow any signal to pass from the first to the second.
That’s what we mean by RC (resistor-capacitor) coupling.
We can also have transformer coupling, direct coupling or DC coupling where a resistor is placed between the two stages.
RC coupling is also called “AC” coupling as DC will not pass from one stage to the next.
Do you want more?

 

[earjun]

That was good 1but could you explain me the design as well.I understan that it is a voltage divider biasing used for the transistor.
could u xplain me that in detail?

 

[colin55]

You asked about RC coupling. The item you are now wanting to be explained is: "Biasing." Namely: "stage biasing." This is a competely different topic.

 

[earjun]

SO could you please help me on that the stage biasing?
I do not understand this biasing well how do I learn more about that?

 

[colin55]

Post a circuit of two stages that are directly coupled and I will explain how the circuit works.

 

[earjun]

could u pls tell me how to post a circuit here?

 

[colin55]

Are you getting the circuit from your computer or from the internet?

 

[xiongyw]

The "C" prevents the voltage from one stage interfering or upsetting or (controlling unnecessarily), the next stage. It allows the AC component to pass from one stage to the next and blocks the DC component.
For instance, one stage may be working on 5v and the next stage may be working on 25v. The capacitor will completely separate the two stages but allow any signal to pass from the first to the second.
That’s what we mean by RC (resistor-capacitor) coupling.
We can also have transformer coupling, direct coupling or DC coupling where a resistor is placed between the two stages.
RC coupling is also called “AC” coupling as DC will not pass from one stage to the next.
Do you want more?

Dear Colin55,

Thanks for the information. I just have a question about the AC coupling, in terms of how it works. Say in a typical BJT circuit, the single positive power supply is 15V, and the base is biased at about 7.5V by two divider resistors. The input AC signal, which is small in its voltage amplitude, say, 0.1Vpp, and it's AC coupled to the base by a capacitor (suppose the capacitor allows the AC signal pass in its frequency range). So the result would be that whenever there are fluctuations in the AC signal, it will directly "added" to the 7.5V bais voltage, just like the AC signal is "riding" on the DC signal. I am puzzled on the question of how this voltage "riding" actually works, i.e., how it possible. Can you please give some explanation and/or point to some source of more info?

Thanks a lot.

 

[colin55]

Firstly, are you talking about a common emitter stage (NPN) with the emitter connected to the 0v rail?
If so, the base will not rise above 0.7v when two resistors are connected to the base.
It does not matter what value resistors are placed on the base, the base will be 0.7v. We are assuming, of course that the resistors are such that they are turning on the transistor and are in the acceptable range of sensibility.
Different value resistors will simply allow more current to enter the base (or less current to enter the base) and if you use a 1k load resistor on the collector, the actual collector voltage will be high or low, depending on the value of the base resistors you have chosen.
If you connect a capacitor to the base and apply a 100mV signal, the signal will pass through the capacitor and at the same time the capacitor will charge.
The reason why the capacitor will charge is due to the following:
The base has a resistor connected to 0v rail and a “voltage divider” is formed between the signal and the base. When the signal rises 100mV the base may rise only 80mV. (When the signal falls 100mV, the base may fall only 80mV.)
This increase in voltage on the base will cause additional current to flow into the base and the transistor will amplify this current and cause an increased current to flow in the collector-emitter circuit.

 

[xiongyw]

Thank you very much for your prompt reply!

I guess I still need some time to fully understand your comments. The circuit I was refering to is one from a book I am reading, which is about the emitter-follower configuration. see attached picture. Since Emitter is not directly connected to 0V rail, the base bias is, according to the book, 7.2V, slightly below 7.5V. Why it's below 7.5V, according to the book, is because that the input impedance of the transistor is parallelly connected to R2, so it divides less voltage than R1. Actually i have another question here: why the input impedance of the transistor is paralelly connected to R2, not R1? what "the input impedance of the transistor" actually mean -- i mean it has 3 terminals, from which two terminals we "see" its impedance?

Also, in this case, it will be helpful if you can also explain why the signal voltage can be added to the base voltage. the confuse i have is that the base has a much higher voltage (7.2V) than the signal voltage (0.1v, e.g.), why the small voltage can still directly imposed on the big voltage?

As you have found out, I am a newbie in this. I know I need to study basic stuff in details to sort all these out, but it may takes too much time and may also boring than jumping into real circuits/topics and get my doubts cleaned bit by bit, especially given that i do not have enough spare time/energy for a systematical study in this field

Thank you again!

 

[andy257]

well it looks like an emitter follower (common collector amp) only ive never seen the RE resitor shown on the circuit. Usually the resitor is placed between +Ve and Collector to control the amount of current that can flow.

I am asuming the output will be identical to the input minus 0.7V. The current gain is limited by RE (usually Rc???)

Be interested to hear other peoples comments.

 

[Nigel Goodwin]

well it looks like an emitter follower (common collector amp) only ive never seen the RE resitor shown on the circuit. Usually the resitor is placed between +Ve and Collector to control the amount of current that can flow.

I am asuming the output will be identical to the input minus 0.7V. The current gain is limited by RE (usually Rc???)

Be interested to hear other peoples comments.

An emitter follower has a resistor in the emitter, NOT in the collector, it's a perfectly normal and common circuit.

It provides no voltage gain, but high current gain, and a high input impedance.

 

[audioguru]

The transistor has a small amount of base current that reduces the +7.5V from the two 10k resistors to about +7.2V. The input signal causes the +7.2V to increase and descrease by 0.1V peak-to-peak.
The emitter voltage averages +6.5VDC which increases and decreases 0.1V peak-to-peak.

 

[andy257]

An emitter follower has a resistor in the emitter, NOT in the collector, it's a perfectly normal and common circuit.

It provides no voltage gain, but high current gain, and a high input impedance.

Hi Nigel,

Little bit confussed with my thoughts on this. Is that Emitter resistor an actual physical resitor or is that the INTERNAL emitter resistor?

Also ive seen this circuit before used as a voltage regulator and there is always a resitor in the collector???

Thanks

 

[audioguru]

The emitter resistor in the emitter-follower circuit is an actual resistor. The internal emitter resistance of a transistor has a value that is much lower.

 

[Nigel Goodwin]

Hi Nigel,

Little bit confussed with my thoughts on this. Is that Emitter resistor an actual physical resitor or is that the INTERNAL emitter resistor?

An external resistor.

Also ive seen this circuit before used as a voltage regulator and there is always a resitor in the collector???

No, you don't have a resistor in the collector, the circuit is complete and correct, and is an absolutely standard common circuit.

 

[xiongyw]

The transistor has a small amount of base current that reduces the +7.5V from the two 10k resistors to about +7.2V. The input signal causes the +7.2V to increase and descrease by 0.1V peak-to-peak.
The emitter voltage averages +6.5VDC which increases and decreases 0.1V peak-to-peak.

thanks for the explanation, audioguru! now i am a little bit clear on the value of the bias voltage. so it seems, since the base current is flowing throught the transistor to the emitter, not the collector, then the author can say that the input impedance is parallelly connected to R2, not R1. it also explains another point of the author about the choice of R1 & R2: choose R1 & R2 to get a current (through them) which is bigggger enough (i.e., more than 10 times) than the base current. this is because, i guess, that the effect of the base current can be ignored w.r.t to the bias voltage imposed on the base terminal (i.e., the change from 7.5V to 7.2V can be ignored here, for the purpose of the circuit).

this is a digress and let's go back to the original AC coupling question. In this configuration, the author also present the scope pictures of the voltage at each point of the circuit, and from the pictures i can see exactly what you said: "The input signal causes the +7.2V to increase and descrease by 0.1V peak-to-peak.
The emitter voltage averages +6.5VDC which increases and decreases 0.1V peak-to-peak." actually the input signal in the book is not as that small, but is 5Vp-p, and the behavior is the same: the input signal causes base and emitter voltage change accordingly with the same amplitude. my question is how it can? I guess this is a too simple question, but it puzzles me when i want to understand the details. so let's discard the transistor in the circuit, we just keep R1, R2, +15V power supply, the ground, and the coupling capacitor C1 and the input AC signal Vi. In this simplified configuraiton, will Vi has the same effect on the voltage drop on R2 (and R1)? Can Vi be treated as another voltage source, and how is it connected to the DC power supply (parallel or in series)? Will the DC power supply affects the input signal instead of the opposite?... Sorry, I have a lot of doubts

Thanks again for your help,
/bruin

 

[audioguru]

so it seems, since the base current is flowing throught the transistor to the emitter, not the collector, then the author can say that the input impedance is parallelly connected to R2, not R1.

The input impedance of the circuit is the input impedance of the transistor in parallel with R1 and also in parallel with R2.

it also explains another point of the author about the choice of R1 & R2: choose R1 & R2 to get a current (through them) which is bigggger enough (i.e., more than 10 times) than the base current. this is because, i guess, that the effect of the base current can be ignored w.r.t to the bias voltage imposed on the base terminal (i.e., the change from 7.5V to 7.2V can be ignored here, for the purpose of the circuit).

Correct.

this is a digress and let's go back to the original AC coupling question. In this configuration, the author also present the scope pictures of the voltage at each point of the circuit, and from the pictures i can see exactly what you said: "The input signal causes the +7.2V to increase and descrease by 0.1V peak-to-peak.
The emitter voltage averages +6.5VDC which increases and decreases 0.1V peak-to-peak." actually the input signal in the book is not as that small, but is 5Vp-p, and the behavior is the same: the input signal causes base and emitter voltage change accordingly with the same amplitude. my question is how it can?

The input signal is strong enough to drive the approx 3.5k ohms input impedance of this circuit without much signal voltage loss.

let's discard the transistor in the circuit, we just keep R1, R2, +15V power supply, the ground, and the coupling capacitor C1 and the input AC signal Vi. In this simplified configuraiton, will Vi has the same effect on the voltage drop on R2 (and R1)? Can Vi be treated as another voltage source

Yes, Vi is a valtage source, an AC signal voltage source to C1.

how is it connected to the DC power supply (parallel or in series)?

It is connected to the input of the circuit and to ground.

Will the DC power supply affects the input signal instead of the opposite?

The DC power supply has nothing to do with the input signal source.

 

[xiongyw]

Thanks, audioguru!

The input impedance of the circuit is the input impedance of the transistor in parallel with R1 and also in parallel with R2.

yes, you are right about the input impedance of the "ciruit". i was refering/implying to the input impedance of the "transistor" as seen by the DC power supply--in the context of the author's explanation of voltage drop from 7.5 to 7.2 on R2.

The input signal is strong enough to drive the approx 3.5k ohms input impedance of this circuit without much signal voltage loss.

what does "strong" mean here? Is "3.5k" big enough or small enough in your context? I guess, by "strong", you meant the signal source is some sort of constant "voltage" source, that it can impose a constant voltage drop on the load disregard the load's impedance. right? if this input signal is coming from a CD player or MP3 player or sound card lineout, can these signal be regarded
as "strong" enough? and why?

It is connected to the input of the circuit and to ground.
The DC power supply has nothing to do with the input signal source.

Trying to understand this (simplified case), i am imagine a kind of hydraulic analogy of this: say we have a vertical pipe in 20M height, and the water in side is about 15M height (leaving 5M empty in the top).
this 15M water head is an analogy of the DC power; then in the middle of the pipe, there is a small hole, and a horizontal pipe is connected to the vertical pipe by the hole. these two pipes are not "conducted", but separated by a rubber plate, which is an analogy of the coupling cap (I borrow this cap analogy from http://amasci.com/emotor/cap1.html); and finally there is a pump at the end of the horizontal pipe, simulating the AC signal input.
Is this a valid analogy for our topic (AC coupling)?
I am imagining that in the begining, the pump is not activated. the pressure at the hole is about 7.5M water head, and this pressue is transfered to the cap, and also to the pump throught the cap and the horizontal pipe (suppose the horizontal pipe is also full of water). so can i say that the DC power supply has effects to the signal (pump) in this case?

Now imaging the pump is activated, then the rubber is vibrating back and forth, which will in turn introduce vibration of the 15M water head in the vertical pipe (suppose the rubber plate is big enough to introduce apparent water head changes in the vertical pipe). so can i say that the DC power supply is affected by the signal via the cap? may be i went to far with this analogy...


Thanks again,
/bruin