In a Full-wave rectification plus smoothing

[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?

 

[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

 

[Potatoishere]

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

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?

 

[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)

 

[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?

 

[CWatters]

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?

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.

 

[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.

 

[Potatoishere]

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.

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

 

[davenn]

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

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

 

[Potatoishere]

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 ...

View attachment 107870

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

View attachment 107871

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

View attachment 107872
Dave

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, shouldn't the voltage drop across the load equal to the sum of voltage drop across the capacitor and the AC generator at that very instant?

 

[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.

 

[Potatoishere]

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

 

[davenn]

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

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

 

[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.

 

[BvU]

Dear Potato,
I am afraid

yes, the second 1/2 cycle.

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.

 

[Potatoishere]

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

 

[BvU]

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

 

[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.

 

[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?

 

[BvU]

When capacitor is discharging, there is no current coming from the AC source at all?

Not correct, see below

Once the AC source has reached a maximum voltage, the current will start to decrease

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.

 

[Potatoishere]

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.

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?

 

[sophiecentaur]

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?

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.

 

[Tom.G]

@Potatoishere
It sounds like you have some confusion about Diodes. Try these sites to see if they help any:
https://learn.sparkfun.com/tutorials/diodes
http://www.electronics-tutorials.ws/diode/diode_5.html

 

[Potatoishere]

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

 

[BvU]

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?

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 ?

 

[sophiecentaur]

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 ?

Is there a C missing?
Is that not just "another way" of looking at it, though? Plus, it's a bit less accessible than the familiar Kirchoff.

 

[BvU]

There certainly is, thanks sophie. I was trying to pin down the moment where the diodes stop conducting and forgot about the C in front of $dV\over dt$.

So the corrected version of #20 is :

The current to charge the capacitor is easily calculated: $\ Q = CV \Rightarrow I = C {dV\over dt}\$ . The total current through the diodes is therefore $I = C {dV\over dt}\ + {V\over R}$.
[edit] under the condition $V_{\rm AC \ source } > V$

That means it (the current through the diodes) peaks at the moment the diodes start conducting and decreases from that moment onwards. It becomes zero when $\ {dV\over dt} = - {V\over R\bf C}\$. 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\bf C}\$ . During that short period, the capactor discharges and the diode delivers current.​

 

[sophiecentaur]

And this is why I claim that the source r is relevant because, if you include it in the calculation, the period over which diode and Capacitor current both contribute to the current in R will increase as r increases. As the rC time constant increases, there will be an extended period of current from both, flowing into R. The peak Volts on C can fall well short of the peak AC emf. The "V" in your equations will actually vary in time because it isn't the emf.
If we are discussing a Bridge Rectifier circuit then I think we should include the situation of the Power supply. The 'quality' of the transformer (detailed spec) will count much more than the normally quoted waveform diagram suggests. This is not only relevant for large steamy equipment but also to some of those cheap and cheerful wall warts.

 

[BvU]

Agreed, but for the original poster the situation was already so hard to understand.

 

[sophiecentaur]

Agreed, but for the original poster the situation was already so hard to understand.

Agreed - he originally didn't understand the function of Diodes. That's one useful lesson for him.

 

[Mark Harder]

The first question has been answered. Draw a full-wave rectified circuit and you will see that the (positive) current can't flow backward. Kirchoff's voltage law supplies the answer to your second question. The law says that the voltage across a circuit is the sum of the voltages of the series components of the circuit. In the power supply circuit, there are no series connections between the + and - poles of the output of the supply, which applies to the resistor and the capacitor both.. Therefore, the voltage across them will be the same at all times. You said somewhere here that the currents through the capacitor and the resistor will add, and that's absolutely true.
Please note that the load to which the supply is delivering current is also a resistance in parallel to the capacitor and the supply's resistor. So the current out of the capacitor is distributed through both resistances as if they were one. You need to apply the parallel resistance rule to determine the current delivered by the supply to its own resistor and that of the load. That is why resistors added to power supply outputs have high values, in the megohm range commonly, so that most current is directed to where it's needed to do work.
A less formal way to think about it is this: Suppose the voltage across the capacitor is greater than that across the output resistance. Since the current can't flow backward, all of it will flow through the resistance. Then you must consider that Ohm's law says that it will create a voltage drop across the resistor (of the same polarity as the capacitor.) The current flowing out of the capacitor is initially high and the capacitor will lose voltage rapidly. On the other hand, the voltage added to the output resistance by that rush of currrent will be rapidly increasing Thus, the 2 voltages will approach each other rapidly, at which point they will be equal. The circuit will always act to equalize the capacitor and resistance voltages.

 

[BvU]

The first question has been answered. Draw a full-wave rectified circuit and you will see that the (positive) current can't flow backward. Kirchoff's voltage law supplies the answer to your second question. The law says that the voltage across a circuit is the sum of the voltages of the series components of the circuit. In the power supply circuit, there are no series connections between the + and - poles of the output of the supply, which applies to the resistor and the capacitor both.. Therefore, the voltage across them will be the same at all times. You said somewhere here that the currents through the capacitor and the resistor will add, and that's absolutely true.

So far, so good. Yes, one can add currents in one's mind. Kirchoff current laaw is clear in that current from AC source + current ex capacitor = current into resistor load. So if current from AC source = 0 THEN current ex capacitor = current into resistor load

Please note that the load to which the supply is delivering current is also a resistance in parallel to the capacitor and the supply's resistor. So the current out of the capacitor is distributed through both resistances as if they were one.

No. the diodes block current ex capacitor from going into the supply's resistor. I dislike this complication, but yes, real AC sources do have internal resistance.

You need to apply the parallel resistance rule to determine the current delivered by the supply to its own resistor and that of the load. That is why resistors added to power supply outputs have high values, in the megohm range commonly, so that most current is directed to where it's needed to do work.

This is nonsense and it is going to confuse potato no end.

A less formal way to think about it is this: Suppose the voltage across the capacitor is greater than that across the output resistance. Since the current can't flow backward, all of it will flow through the resistance. Then you must consider that Ohm's law says that it will create a voltage drop across the resistor (of the same polarity as the capacitor.) The current flowing out of the capacitor is initially high and the capacitor will lose voltage rapidly. On the other hand, the voltage added to the output resistance by that rush of current will be rapidly increasing Thus, the 2 voltages will approach each other rapidly, at which point they will be equal. The circuit will always act to equalize the capacitor and resistance voltages.

More formally: Q=CV AND V = IR . The wiring between C and R is supposed to have no resistance, so there is no voltage difference over those wires. therefore V_capacitor = V_{load resistor} at all times.

Potato, where are you now ? All clear or is the mist getting worse ?