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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.
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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.
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???
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.
The input impedance of the circuit is the input impedance of the transistor in parallel with R1 and also in parallel with R2.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.
Correct.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).
The input signal is strong enough to drive the approx 3.5k ohms input impedance of this circuit without much signal voltage loss.this is a digressand 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?
Yes, Vi is a valtage source, an AC signal voltage source to C1.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
It is connected to the input of the circuit and to ground.how is it connected to the DC power supply (parallel or in series)?
The DC power supply has nothing to do with the input signal source.Will the DC power supply affects the input signal instead of the opposite?
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 impedance of the circuit is the input impedance of the transistor in parallel with R1 and also in parallel with R2.
audioguru said:The input signal is strong enough to drive the approx 3.5k ohms input impedance of this circuit without much signal voltage loss.
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).audioguru said: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.