# Question regarding bridge rectifier.

1. Oct 29, 2010

### Hyperspace2

In the picture 1, the capacitor is grounded
In the picture 2, the capacitor is not grounded.

what are the purpose of this different models.

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2. Oct 29, 2010

### Hyperspace2

I guess in the first pic , there is RC filter
and second one is purely capacitive.
Please describe the detail process what this two different circuits are going to do .

I know that these circuits turn the ac into dc .
I also know that the introduction of capacitors will try to maintain a constant dc voltage.
But my main focus is why in pic 1 capactior is grounded and in pic 2 capacitor is not grounded.

Last edited: Oct 29, 2010
3. Oct 30, 2010

### davenn

in the first the - DC rail is grounded in the second its floating
its nothing special, the PSU is still going to work the same way

In the grounded situation it would be more common to see that GND point connected to the metal chassis of the equip. to which the GDN (earth) lead of the mains cable and the transformer core would also be connected to.

Ina non grounded situation it may be because you dont want a ground the equip is hat is called "double insulated" or that the case of the equip is non-conduction (plastic) so you couldnt have a ground anyway.

Dave

4. Oct 30, 2010

### Hyperspace2

Thanks for the reply.
I didn't understand the logic of the transformer.

I also have a few other question.
Logic 1
I was told that capacitor and resistor in parallel make RC filter network(low pass filter).
That would sense that the capacitor would behave high resistor for ac signal (because it has frequency) an it would allow only non ac signal.
Logic 2
The other logic told to me was that at first cycle of rectifier , the capacitor would get charged and second half of cycle ,it begins disharging to keep constant level of dc.

but I couldnot find the relation of both of these logic anyway.

I also have a question, How we determine the value of capacitor that we need to maintain the dc level. (or we could keep any)

5. Oct 30, 2010

### Hyperspace2

is the purpose of the ground is to pass ac signal to the ground???????

6. Oct 30, 2010

### davenn

ok... in any GOOD and safely built system, the transformer core (which is a huge lump of metal) would also be grounded. This us so that if the transformer develops an insulation fault, from the 120/240 etc mains voltage to the transformer core then that fault current will be taken to ground, causing the protection fuse to blow. (It should have fuse protection on the mains side ! :) ) Note the transformer isnt shown in your ccts but assume that that 12VAC is the output of the transformer secondary

well not in this case an RC filter would have one of the 2 components in parallel and the other in series
What that resistor is doing in this cct is proving a load across the DC output of the supply so that you can connect an oscilloscope across it and see the effect of the rectification and the capacitor smoothing.
If you have an oscilloscope.... for an experiment.... set the cct up as shown see the DC voltage with the small amount of ripple as in the pic you posted above.
Then disconnect one end of the capacitor and look at the difference on the scope.
you will see lots of 1/2 cycles see pic below
and that will answer you next Q about the charging and discharging of the capacitor to produce a smoother DC voltage

The value of capacitor ..... many many moons ago when I was much younger ;)
I was taught that "as a rule of thumb" You should have 1000uF per each amp of DC current

so a 10 amp supply have at least 10,000uF 20 Amp ... 20,000uF etc
Now that holds reasonably true for linear PSU's (ones like your one above that drops mains voltage to something much lower)
I have noticed in switchmode PSU's that there is much less smoothing used on the DC side
basically cuz of the AC to DC conversion and smoothing already occurring on the primary side of the much smaller transformer. And although I'm not an expert, I'm guessing that the higher operating frequency of a switchmode PSU means that the ripple isnt so difficult to get rid of.
( I will wait to be corrected on that assumption ;) )

cheers
Dave

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7. Oct 30, 2010

### davenn

It can be in some situations, but in a PSU more commonly for safety
as commented in the other post, to ground dangerous AC currents under fault conditions to cause the fuse to blow

Dave

8. Oct 30, 2010

### sophiecentaur

As far as the voltage (potential difference) across the output terminals there is no difference. The ground on one circuit is, as yet, uncommitted so you could get either a positive or a negative supply from it. If you wanted a negative out, you'd ground the '+' out and if you wanted a positive out, you'd ground the' -'. This is the same as when you connect a battery up to give a positive or negative Earth.
You would just have to look at the context of each circuit.

9. Oct 30, 2010

### Hyperspace2

Thank you very much for your answer.
But Negative or positive out won't be necessary , or is it we have to choose sometime for our purpose(i don't Know)
An here if we want to connect into oscilloscope ,then both out are needed ,isn't it? and still there is ground?

Aah I got it , I was mistaken .

10. Oct 30, 2010

### sophiecentaur

This circuit would normally be used as a power supply (though no always). Many circuits (amplifiers / processors etc.) need positive and negative supplies and a ground in between. For that, you would use two rectifier circuits, connected appropriately.
An oscilloscope may or may not have one input of a channel connected to ground. The trace will just show you the potential difference between the two input terminals - they could be at 1000V+10V and 1000V, respectively and the scope will tell you 10V.

11. Oct 30, 2010

### Hyperspace2

This example you have given (1000V and 1000+10v), this is the out of two different rectifiers ,isn't it?
The next question, I have not tried connecting two terminals of batteries in oscilloscope .
let suppose 10 v battery is connected to the oscilloscope ,then it shows 10 v in oscilloscope ,isn't it?,( I confused that + and - sign of battery , the potential difference it will show will be 10-(-10)=20 V)
Oh I have a lot of confusion in my mind isn't it?

12. Oct 30, 2010

### davenn

it will just show a 10V level put the crocodile clip lead of the scope probe on the negative
of the battery and the tip of the probe on the positive ... and you will see a hotizontal line at +10V (make sure you have the scope set at at least 2V / division)

if you reverse the scope brobes you will just see a - 10V line on the scope
(because the measured voltage is referenced to the ground lead of the scope probe)

Dave

13. Oct 30, 2010

### sophiecentaur

My 1000V example was just to indicate that an ideal oscilloscope just shows the potential difference between two points - it will (/should) ignore the absolute potential wrt the ground of the scope. Note: there are two terminals on the input of any measuring instrument. If one (say the black terminal) happens to be connected to Ground, then the red one will measure the potential relative to the ground. If it is connected to a different potential then the red one will measure relative to that reference (black) potential.
If you live half way up a mountain then you can travel uphill or downhill i.e. increase or decrease your gravitational potential. The overall trip from bottom to top of the mountain is twice the height of where you live relative to the valley floor. It's the same when you consider Electrical potential and potential difference.

If you put two batteries in series then the + of the 'upper' one will be V above the middle point and 2V above the other end of the lower battery. If you choose to call the centre point Ground then you will have +V and -V from the two batteries. Likewise for two separate, interconnected transformer - rectifier circuits you can obtain a positive and a negative supply.

14. Oct 30, 2010

### Hyperspace2

Thanks for the reply.

I am always confused about the battery. Suppose there is battery 0f 1.5 V.
Then what does that mean, is that mean its + side is +1.5V and -side is -1.5 (but that make potential difference 3V )
or that mean +side is1.5 V and -sighn is actually ground,making potential difference 3V

Can we measure the absolute electric potential of only one side of battery (or we can't) that is by grounding other side of the battery as posted in the following figure I have attached

In following figure pic 1, battery potentail is measured to be V volt by multimeter

In figure 2, + sighn is grounded that multimeter is connected , then I think the multimeter should measure absolute potential of the - sighn of the battery ( I told this Vabs in figure because it would provid ea number with respect to ground ,Vabs=Electical potentail of the -side of the battery -(minus) ground potential = V(potential) - 0 =Vpotential)
Amn't I correct. Will I be able to measure the absolute potential of one side of battery that may be a different number than V

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15. Oct 31, 2010

### sophiecentaur

The concept of Vabs is not meaningful in circuit theory. It may be vaguely relevant when you consider a net unbalanced charge on the Earth, perhaps but current flows only when there is a potential difference between two points on the circuit. PD is the thing.
Where would you connect your multimeter if you wanted to measure its "absolute potential"? It potential wrt ground would depend upon the actual unbalanced charge on it (i.e. how many random electrons had attached themselves to it as you brushed against it) Trying to measure that voltage would give a value of 0V (eventually) because of the input resistance of the multimeter - which would constitute a resistive path to ground.

If a battery has 1.5V marked on it, it will have a PD of 1.5V across it wherever you happen to connect one end of it. It' s just like a staircase - you ascend 4m, whether the staircase is in a valley or up a mountain - and you will hurt yourself exactly the same amount when falling down it, wherever it is sited.