ac & dc isolation in amplifier circuit

[Disha Karnataki]

hi everyone,
in amplifier circuits usually we pass small ac signals and then pass through various transistor biasing circuits and finally get an amplified o/p.
let us take an example: consider an i/p signal of 150mv(p-p) so it's nothing but 75mv peak. As v know that the base bias voltage for transistor under operation should be about 0.7V=700mV. so considering positive half cycle of i/p signal:base voltage will be .7+.075=.775v.
considering negative half cycle: base voltage =.7-.075=.625v which will make the si transistor to go in cut-off region due to which the o/p will be a distorted dc. these are my analysis & i donot know how far are they true. I am unable to understand how do ac signals can be amplified for larger signals like 2v,3v etc.... because due to these signals though we set the operating point the operating point will keep on deviating and will not be fixed. please someone tell me properly what is actually HAPPENING INSIDE THE TRANSISTOR BY PHYSICAL ANALYSIS & PLEASE TRY TO AVOID TELLING BY MATHEMATICAL FORMULAES..

 

[pinout]

Good question.

The key point that may help your understanding is the typical bias arrangements get a transistor to operate in a linear region, i.e. away from turning off or being saturated.
Taking a common emitter circuit, the input voltage is typically converted to a current which modifies the base current and this change in current is amplified by the transistor gain or more likely the stage gain into a larger (amplified) current change in say the collector circuit.
Again at the output it is the collector load resistor that changes this current back into a voltage.

The above is basic description, but the key point is a transistor is a current amplifier.

hope that helps, with your thoughts and leads to a more complete understanding.

 

[crutschow]

You can not apply that much AC voltage to the base unless there is an un-bypassed emitter resistor to provide negative feedback (which is often done to reduce and stabilize gain, increase input impedance, and reduce distortion).

In general the maximum peak signal at each stage must be small enough so that it does not cause that stage to go into either cutoff or saturation. Otherwise you will get significant distortion, as you noted.

 

[KerimF]

In case of an input voltage of 2V, 3V etc..., the voltage is not applied directly to the base-emitter junction of a bipolar transistor. There is always a resistance in series at the base and/or emitter side. And the ac current passing the base-emitter junction should develop on it a voltage less than +/- 2.6 mV if linearity is important.

 

[Disha Karnataki]

Re: ac & dc isolation in amplifier circuit

hey pinout,
when we input an ac signal to transistor at it's base as we know that transistor is operated at dc level i.e we give +vcc voltage but as we are also giving ac signal won't this ac signal mix with dc supply?? how are they actually isolated??

- - - Updated - - -

i donot know much about power amplifiers but just about to start it know...
i have read the first line of the chapter they have mentioned that power amplifiers are used for large signal amplification.
But a power amplifier consists of nothing but a cascaded small signal amplifier stages...

 

[KerimF]

The analysis could be made much clearer if you have a simple practical example to study.
AC and DC signals (as current or voltage) could be mixed or isolated whenever it is necessary.

 

[Disha Karnataki]

as stated by cruscthow that we can't use it for large signals.
i donot know much about power amplifiers but just about to start it know...
i have read the first line of the chapter they have mentioned that power amplifiers are used for large signal amplification.
But a power amplifier consists of nothing but a cascaded small signal amplifier stages...

 

[pinout]

Hi Disha

hey pinout, when we input an ac signal to transistor at it's base as we know that transistor is operated at dc level i.e we give +vcc voltage but as we are also giving ac signal won't this ac signal mix with dc supply?? how are they actually isolated??

You obviously have a good analytical brain and you are correct. Considering a common emitter amplifier such as this
https://4.bp.blogspot.com/_3b1TEwzFtQg/SgEyNoF6qWI/AAAAAAAAAIM/8u5108ATupQ/s400/Emmiter.bmp
RB sets the DC bias point and hence the voltage at the collector.
This circuit is a bad example to use in an application. The DC conditions are affected by the transistor gain and temperature.
Now your AC signal changes these DC conditions and that is what give you your AC gain.

The normal way of thinking about this is that the DC bias sets up the transistor in a stable way and the AC is modifying this to vary the transistor around this point.

If you want to look at some formula "h parameters" define the transistor from an AC point of view assuming the DC conditions are taken care of.

You last point where you say DC supply, I have answered the above assuming you mean DC bias or set point. From a beginners perspective the actual supply is considered perfect and stable.
#hope this helps

 

[Disha Karnataki]

thanks a lot as this was all about amplifiers i was just going trough the oscillators chapter & found that colpitts ,hartley oscillators have RFC coil connected between collector of npn transisot & vcc. And the answer given was to seperate ac signals from dc signals.
why wasn't the same thing even carried out to the amplifiers?? I have gone through all the amplifiers but didnot find this choke coil placed in amlifiers circuits why is this ??? When we are dealing with high frequency signals even in the amplifier....

 

[pinout]

You don't need or want the inductor in the amplifier circuit. This is only required in the oscillator as part of the tuned circuit at which it oscillates.

In the amplifier we are typically trying to keep the amplitude versus frequency flat over the pass band.
In a tuned amplifier such as found in the high frequency part of radios you will use inductors and transformers to give the required passband.

So it still comes down to, do a DC analysis to set the bias and then the AC to get the frequency response/gain/phase.....

 

[Disha Karnataki]

In a tuned amplifier such as found in the high frequency part of radios you will use inductors and transformers to give the required passband

sorry sorry i didnot get this point.... oscillators are meant to resonate at only and only one frequency right?? then why do we require to use again inductors or transformers to give the required frequency o/p?? am i right?? please explain a little more on this...

 

[pinout]

Sorry I'm confusing you. Yes you are correct reagrding the oscillators and inductor.

My other point was trying to answer

why wasn't the same thing even carried out to the amplifiers?? I have gone through all the amplifiers but didnot find this choke coil placed in amlifiers circuits why is this ??? When we are dealing with high frequency signals even in the amplifier....

Inductors are not normally required in say audio amplifiers as they are not 'tuned'.
But they are used in high frequency amplifiers etc.

 

[Disha Karnataki]

ok fine now i got the point ,
i just wanted some more information about the oscillators i.e as per barkhousen criteria(may be wrong spelling) oscillators are the ones with positive feedback.
As i have read that the feedback network will cause 180deg shift from the o/p due to which there will be 360deg /0deg shift from OR WITH RESPECT TO the original input signal(which is actually absent in oscillators i.e external i/p signal is absent)
MY QUESTION IS WHY IS THERE NEED TO HAVE AGAIN 180DEG SHIFT I CAN JUST INTRODUCE SOME SHIFT OR NO SHIFT AT ALL WHY IS THAT SHIFT REQUIRED WHEN THE EXTERNAL I/P SIGNAL HAS NO EXISTENCE AT ALL IN THE NETWORK???

 

[BradtheRad]

ok fine now i got the point ,
i just wanted some more information about the oscillators i.e as per barkhousen criteria(may be wrong spelling) oscillators are the ones with positive feedback.
As i have read that the feedback network will cause 180deg shift from the o/p due to which there will be 360deg /0deg shift from OR WITH RESPECT TO the original input signal(which is actually absent in oscillators i.e external i/p signal is absent)
MY QUESTION IS WHY IS THERE NEED TO HAVE AGAIN 180DEG SHIFT I CAN JUST INTRODUCE SOME SHIFT OR NO SHIFT AT ALL WHY IS THAT SHIFT REQUIRED WHEN THE EXTERNAL I/P SIGNAL HAS NO EXISTENCE AT ALL IN THE NETWORK???

We have the idea of time-shifting an AC signal 180 degrees, which is something that could be done by any amount, including 179 or 181 degrees, etc.

However our nomenclature also has us say it is 180 degrees out of phase when we invert the polarity of the signal.

Although the results look similar, one action is different from the other.

 

[Disha Karnataki]

yes right,actions are different because i.e in the feedback network we do not have an amplifier circuit but just it provides phase shift... and resonates at a peculiar frequency we want to feed as i/p in the feedback...
my question is why did barkhosen criteria specify that 180deg phase shift is very crucial for oscillations to take place??? please elaborate a little..please..

 

[LvW]

my question is why did barkhosen criteria specify that 180deg phase shift is very crucial for oscillations to take place??? please elaborate a little..please..

It is not true that Barkhausen did "specify 180 deg phase shift" for oscillations.
The Barkhausen criterion requires a loop gain of unity - which means: Overall gain within the feedback loop of unity and a phase shift of 0 deg. (resp. 360 deg).
In some cases, there is a pure signal inversion within the loop. Then, the remaining blocks must provide an additional phase shift of 180 deg at the desired oscillation frequency.
Counter example: In the WIEN oscillator, the frequency dependent RC network provides a phase shift of 0 deg - and the amplifier must be non-inverting to meet the above condition.

 

[Disha Karnataki]

yes he didnot mention that but, when will u get 0deg /360deg only when the feedback loop produces an extra of 180deg so,total rotation in angle=angle rotation of amplifier +angle rotation of feedback loop=360deg/0deg..
so he INDIRECTLY SPECIFIED THAT THE FEEDBACK LOOP SHOULD HAVE AN 180DEG PHASE SHIFT....
SO I A NOT GETTING THAT POINT WHY IS THAT THE LOOP SHOULD HAVE 0/360DEG SHIFT WHYYYY???

 

[LvW]

SO I A NOT GETTING THAT POINT WHY IS THAT THE LOOP SHOULD HAVE 0/360DEG SHIFT WHYYYY???

If I understand right, you are asking WHY the oscillation condition (Barkhausen) requires a loop gain with a phase shift of 0 resp. 360 deg, correct?
The answer is as follows: An oscillator consists of an amplifier that produces its own input signal.
This is identical to a gain around the complete loop of unity. And unity means: Magnitude of "1" and phase of 0 deg.

 

[Disha Karnataki]

ok this is what i have understood after sumarizing all your points let me explain & den plz correct me if i go wrong:
initially before oscillations start there will be noise in the transistor due to the material it is built of. So that noise is being amplified by the transistor,(let me name this stage as amplifier) now i want a peculiar frequency that the oscillator should continue it's oscillations, i.e say XHz so this frequency is being selected by the feedback circuit and is fed into the input side such that what we want as o/p is retained and the oscillations are kick started. As i know that in amplifier circuit the o/p will always be outphase to the i/p signal so... if i want that X frequency to oscillate then i will have to give an i/p exactly out of phase with the prevaling o/p.
So that is what barkhousen stated...

 

[LvW]

As i know that in amplifier circuit the o/p will always be outphase to the i/p signal so... if i want that X frequency to oscillate then i will have to give an i/p exactly out of phase with the prevaling o/p.

Out of phase? What do you mean? 180 deg phase shift?
There are two basic types of amplifier circuits that can be used for oscillators: Inverting (180 deg phase shift) and non-inverting (0 deg phase shift).