Hi Friends!

According to text…due to zener diode, Q1 is supplied with constant base voltage. Any change in output voltage is sensed at at the emitter of Q1 which brings change in forward bias of the transistor. Thus transistor changes its resistance at CE junction and compensates for the voltage change at output or input. Here transistor is acting as a variable resistor. Also total current pass through transistor hence it is also called as PASS transistor.

Q1.What all difference would it make, if I replace this zener diode with a capacitor and how would this circuit behave with a 1 Vp-p ripple is superimposed on 5V DC input signal??

I am having multiple queries regarding capacitance multiplier, will take them successively.

Thanks!

Source article- https://www.tpub.com/neets/book7/27k.htm refer fig 4.34, 4.35, 4.36.

 

With Zener diode, you have a voltage regulator. And the output voltage is equal to Vout = Vz - Vbe. But if you remove the Zener diode from a circuit and replace it with a capacitor you no longer have a voltage regulator.
And now you have a "capacitor multiplet". And this circuit will reduce the supply ripples due to the fact that now the capacitor discharging current is (β +1) smaller than the load current due to the "emitter follower" actions Ib = Ie/(β +1).

 

Why do we need to call it a "capacitance multiplier"? It's a low-pass RC filter with an emitter follower on the output.

 

And now you have a "capacitor multiplet". And this circuit will reduce the supply ripples due to the fact that now the capacitor discharging current is (β +1) smaller than the load current due to the "emitter follower" actions Ib = Ie/(β +1).

Thanks Jony130 for your speedy reply.

I have watched the video which you mentioned about and trying to get to some conclusion.

To increase current capability BJT is used. That means all major amount of load current flows through the transistor due to its current gain.
The path way of capacitor discharge is through the BE junction and then through load to minimise ripple.

That means that 2 load current flowing through load :
1. A high current that passes through the transistor, containing maximum ripples and

2. The a discharge current of the capacitor.

Q: I wonder what exactly makes the capacitor discharge current to go low B+1 times since physically the cap size is same.
I don’t think it is the emitter follower since emitter follower copies what exist at base and just bypass all the current which would have initially passed through R1, thus helps with lowering corner frequency with practically low size capacitors.(however that's another concept.

 

Why do we need to call it a "capacitance multiplier"? It's a low-pass RC filter with an emitter follower on the output.

Thats exactly what I am wondering about that how an emitter follower multiplies the capacitance. Actually what the video explains is that it "APPEARS" to get multiplied, actually its the same on physical terms. I think so that the catch is somewhere when the cap discharge current gets lowered down which Jony130 has mentioned in his post.

 

The cynic in my thinks it is a way of making a simple circuit sound complicated.
I think that the idea is that, without the transistor, with just an RC filter on the power supply, R would have to be much smaller - say 10 ohms (because all the power supply current has to go through it), and that would make C=10000uF.
Buffer the RC filter with a transistor, and the filter can be 1k/100uF, so somehow it has "multiplied" the capacitance.
The ripple is attenuated by the ratio of the resistor to the capacitor's impedance at 100Hz.
So, if you replace the capacitor by a zener, the ripple is attenuated by the ratio of the resistor to the zener's slope resistance.
With the RC the output voltage follows the input voltage (minus Vbe).
With the zener, the output voltage is the zener voltage minus Vbe.

There is a snag - it doesn't work if the peak-to-peak ripple voltage is bigger than Vbe. For that situation, it needs a resistor across C1 to make the voltage drop across R1 greater than peak-to-peak ripple voltage - or use a MOSFET which would have a larger Vgs than 0.6V, which could be larger than the ripple voltage.

 

Why do we need to call it a "capacitance multiplier"? It's a low-pass RC filter with an emitter follower on the output.

Because that is what it is called.

 

Buffer the RC filter with a transistor, and the filter can be 1k/100uF, so somehow it has "multiplied" the capacitance.

I would like to raise a doubt...however I agree to you at some level.
With the absence of transistor, the entire current has to go through R. So, not to attenuate Vout more by causing heavy vol drop at R, we have to select a smaller R value. According to formula R = 1/2pi.f.C, if R decrease, capacitance increases. Increased capacitance comes up to be very high values which are not physically realisable. Also it would shift the corner frequency to an upper limit, which hampers filtering action of RC filter. A solution to this was made, that with the addition of transistor, maximum current would flow across it thus we can increase the value of R so that corner frequency is lowered and capacitance decreases(acc to formula), so how buffering an RC filter with transistor "somehow it has "multiplied" the capacitance"

There is a snag - it doesn't work if the peak-to-peak ripple voltage is bigger than Vbe. For that situation, it needs a resistor across C1 to make the voltage drop across R1 greater than peak-to-peak ripple voltage....

Does ripple voltage when less than 0.7 V drops at CE junction of transistor?? If yes than can we consider BJT to be having a role in ripple reduction.?

 

Does ripple voltage when less than 0.7 V drops at CE junction of transistor?? If yes than can we consider BJT to be having a role in ripple reduction.?

Yes.

so how buffering an RC filter with transistor "somehow it has "multiplied" the capacitance"

You get the same amount of ripple reduction for a lower value of C (because you can use a higher value of R)