# B In a Full-wave rectification plus smoothing

1. Oct 22, 2016

### Potatoishere

So in a full wave rectification, a capacitor is being connected in parallel with a resistor. During the period of capacitor discharging, will the capacitor discharge through the AC generator aswell? And why is the voltage across the resistor is equal to the change of voltage in the capacitor?

Last edited: Oct 22, 2016
2. Oct 22, 2016

### BvU

In a Full-wave Rectifier with Smoothing Capacitor (picture half-way the page) the current can't flow backwards to the generator: the diodes prevent that.
Load resistor and smoothing capacitor are parallel (as you say) and that's the reason both 'see' the same voltage

3. Oct 22, 2016

### Potatoishere

Even though the capacitor is in parallel with the resistor, but there are still current coming from the AC generator. And hence, wouldn't the sum of current through the resistor be the current from the capacitor and the generator? And V=IR, thus the voltage drop across the resistor must be the sum of voltage of capacitor and the generator?

4. Oct 22, 2016

### BvU

The voltage over the resistor/capacitor parallel circuit is one and the same voltage.

In the part of the half-period of the AC that the voltage over the load is lower than what the ac source provides, two of the diodes conduct and there is current flowing into the capacitor and into the load. We see the voltage go up ("C Charges" in the picture)

In the part of the half-period where the purple line is below the black line the diodes block and the current flows from the capacitor into the load resistor. We see the voltage go down ("C Discharges" in the picture)

5. Oct 22, 2016

### Potatoishere

I am sorry, but I still can't see why the capacitor can't discharge through the generator from the diode diagram? I think my main trouble is that I can't seem to comprehend that during the first half cycle, when the AC generator reaches its peak voltage, the capacitor will also be charged up at the peak value. And when the voltage of the AC generator is decreasing in the first half cycle(after reaching its max voltage value) , the capacitors will discharge. But wouldn't the voltage across the load be the effective voltage of the circuit and not just the capacitors?

6. Oct 22, 2016

### CWatters

If the instantaneous voltage from the generator is higher than the voltage on the capacitor the capacitor is being charged. Eg the capacitor current is negative.

The resistor current in that case is the generator current less the capacitor current.

7. Oct 22, 2016

### wud-wurks

A full wave bridge will produce half cycles of the AC all above the zero voltage base line. So the output will look like a series of humps. When you apply a 'smoothing 'capacitor across the output of the bridge the effect is to remove a lot of the humps by charging the cap as rhe AC hump rises from zero, on the up side of the hump, and filling part of the gap at the top of the hump with energy stored in the cap ( current) from the cap) is fed into the load as the AC cycle begins to fall the down part of the hump.

The energy stored in the capacitor CANNOT flow back into the generator as the diodes are in a reverse polarised state and effectively OFF to this direction of current flow.

8. Oct 22, 2016

### Potatoishere

So when the capacitor is discharging, will the there still be current coming from the AC generator?

9. Oct 22, 2016

### davenn

yes, the second 1/2 cycle. BUT the capacitor never fully discharges between cycles. In fact, it is why the capacitor is used after the
bridge rectifier .... to keep the avg voltage up.

AC voltage BEFORE the rectifier ......

Rectified voltage AFTER the rectifier ... NO Capacitor .....

Rectified voltage with smoothing capacitor added ( top diagram)
Note that the voltage no longer drops to zero as in the lower diagram

Dave

10. Oct 23, 2016

### Potatoishere

From the graph when the capacitor is discharging, why is the voltage drop across the load is the equal to change in voltage across the capacitor? Since there is still current coming out from AC generator, hence, shouldnt the voltage drop across the load equal to the sum of voltage drop across the capacitor and the AC generator at that very instant?

11. Oct 23, 2016

### CWatters

The voltage across the load is not equal to the _change_ in capacitor voltage. It is equal to the capacitor voltage because they are in parallel.

12. Oct 23, 2016

### Potatoishere

What about the voltage from the AC generator when capacitor is discharging?

13. Oct 23, 2016

### davenn

as the voltage from the generator starts building up for the second 1/2 cycle then the voltage on the capacitor starts
increasing again. The capacitor only starts discharging for the briefest of time

Is there a specific part in my diagrams you didn't understand ?

Dave

14. Oct 23, 2016

### Potatoishere

Let's assume two batteries and a resistor connected in parallel, and assuming one battery has higher emf than another(), what will be the voltage across the load? I think this scenario is very much similar to full-wave rectifier with smoothing. When the capacitor is discharging, it is because the pd across the capacitor is greater than the resistor and the generator.

Last edited: Oct 23, 2016
15. Oct 23, 2016

### BvU

Dear Potato,
I am afraid
is either wrong, or I am misinterpreting it.

Let me turn it around

When and only when the AC source during the half-period delivers a voltage that is higher than the capacitor voltage the capacitor will be charged - and the parallel load receives current from the AC source as well.

At all other times the AC source voltage is lower than the voltage over load and capacitor, so the diodes will block any current from the AC source and the capacitor discharges over the load. So: while the capacitor is discharging, there is NO current from the AC source

But I find I am repeating myself (#4) and Watters (#6).

There must be someethign in the way of your grasping this and you have to help us find it, if you can.

16. Oct 23, 2016

### Potatoishere

Why would the diode stop conducting when the pd across the capacitor is greater than the AC source?

17. Oct 23, 2016

### BvU

That's what it reason for being. Ideal Diodes only conduct in one direction.

18. Oct 23, 2016

### sophiecentaur

It can help to think in terms of the current flowing, as well as the voltage variations in time. When the supply voltage is higher than the Capacitor volts, a high charge current will flow because the source resistance of the supply (not shown in those diagrams, but very relevant in practice) is (usually) fairly low. During this period, current will also flow from the source through the load R. When the supply volts start to drop, the charging current will rapidly go to zero and the C will take over the current into the load. The lower the source resistance, the higher the peak voltage across R and C (there is a 'voltage divider' circuit, formed by source resistance r and load resistance R; r will always account for some voltage drop. With a poor AC supply (high series r) the peak volts may be significantly short of the nominal supply volts when there's a heavy load (Low R). A massive smoothing C can reduce the Ripple but can't help with the overall voltage drop under load.
The finite voltage drop across the diodes and their own power dissipation will also affect the outcome.

19. Oct 23, 2016

### Potatoishere

When capacitor is discharging, there is no current coming from the AC source at all? Once the AC source has reached a maximum voltage, the current will start to decrease but there will still be current coming out from the AC source into load, am I right?

20. Oct 23, 2016

### BvU

Not correct, see below
You're not very clear on what current you are referring to here.

The current to charge the capacitor is easily calculated: $\ Q = CV \Rightarrow I = C {dV\over dt}\$. The total current through the diode is therefore $I = C {dV\over dt}\ + {V\over R}$. That means it peaks at the moment the diodes start conducting and decreases from that moment onwards. It becomes zero when $\ {dV\over dt} = - {V\over R}\$. Only then The only source for the current to charge the capacitor is the AC source.

So your statement holds for the period from the moment $\ {dV\over dt} = 0\$ until the moment $\ {dV\over dt} = - {V\over R}\$. The capactor discharges and the diode delivers current.

21. Oct 23, 2016

### Potatoishere

During which the capacitor is discharging, will the current passing through the resistor be the sum of current from the capacitor and the AC source?

22. Oct 23, 2016

### sophiecentaur

As I said in post #18, you need to consider a (even very small) source resistance in order to discuss accurately what happens even in a simple circuit. As long as there is a positive PD across the resistor r, current will flow (treat the diodes as ideal). If the C and r large enough (i.e. the rC time constant) , the C volts will never reach peak supply volts and, even on the way down, the supply will be supplying current to the C and the R.
But the mistake in this thread has been to try to describe what happens in the form of 'tell me a story'. If you draw out the circuit and include the resistor r then you have a chance of solving the problem. Use some fairly basic Maths and follow the rules, rather than trying for a verbal description. After all, this process involves solving differential equations and use of exponential functions. They don't go well with simple arm waving.
Edit: I have been searching for a suitable link to show that the waveform during the rising portion of the cycle is actually not part of a sinusoid. There is a lag, due to the rC charge time constant. Anyone who has ever tried to make a conventional power supply, sufficient to supply a high power audio amp will have found out that they cannot get away with anything other than a very beef transformer - because a significant series r will cause the supply rail to dip when there is a big current demand. Nowadays, of course, people use regulators and SMPSs which helps.

Last edited: Oct 23, 2016
23. Oct 23, 2016

### Tom.G

24. Oct 23, 2016

### Potatoishere

Thanks alot guys for the great explanation, I finally got it :)

25. Oct 24, 2016

### BvU

Kirchoff's law says you can always look at it that way. But the current from the through the diodes will be zero from the moment $\ {dV\over dt} = - {V\over R}\$ -- as I tried to explain. Is it still unclear ?