Eshal
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so my transformer must be rated at 5V and 1A too?
Yes, you calculated it correctly.9.915V
It is the output of rectifier. Is it correct?
It's easier to just buy a bridge instead of making one out of four diodes.And which diodes should I use for bridge rectifier?
Nothing else. Only forward current and reverse voltage.Just these two parameters are to be considered in choosing bridge rectifier IC or anything else?
No. Here is an example of how to work it out:Did I do all calculations correct?
Using components rated for higher current is not to "compensate the losses", it is to provide a safety margin.I used IL=1.5A as discussed in the previous post in order to compensate the losses.
I am getting what are you trying to teach me. But I didn't get how do you come to use capacitor of 4700uF!!!!No. Here is an example of how to work it out:
If we assume that the regulator requires at least 7V input and we assumethat the peak output voltage from the rectifier may be as low as 9V, then the ripple voltage must be no more than 2V pk-pk (because 9V - 2V = 7V).
If we want ripple less than 2V pk-pk when 1A of current is drawn, then the smoothing capacitor should be at least about 4700uF.
Obviously sir, you are right. You raise a very good point and I agree with you that I did mistake.For example, you assumed that the peak voltage from the rectifier is 9.9V and you calculated that ripple voltage is 7.2V.
But if peak voltage is 9.9V and ripple voltage is 7.2V, then the minimum output voltage from the unregulated supply = 9.9V - 7.2V = 2.7V.
How do you expect a linear regulator to give 5V output when the input is only 2.7V?
Here, \[{V }_{L(pk) }\]=9v right?(because 9V - 2V = 7V).
Rough calculation. Here is a simple, easy to remember formula to calculate the capacitance you need:But I didn't get how do you come to use capacitor of 4700uF!!!!
Is it calculated value, guess or your experience?
The input to the regulator must obviously be higher than 5V, but how much higher depends on which regulator you choose. Some regulators will work with 6V or less. Some may need 7V or more. If you look in a regulator's datasheet, you will see what it's minimum input voltage is.But what is 7v?
And out of these 9.915v, how did you say that 7v is the least output of the filter capacitor. why not 6v, although 6v are also greater than 5v?
I have gotten all things sir.Rough calculation. Here is a simple, easy to remember formula to calculate the capacitance you need:
C = 10 000uF * current (in amps) / ripple voltage (in volts)
It's not exactly accurate, it always gives a value a bit larger than what you need.
Using that formula, C = 10 000uF * 1 / 2 = 5000uF, but we know that's a bit more than needed, 4700uF should be OK, and it's a standard value. Just to be safe, a higher value like 6800uF could be used, in case current is sometimes a bit higher than 1 amp.
Why the formula works:
When a capacitor is discharged at a constant current, the rate of change of voltage (in volts/second) = current / capacitance.
So the voltage across a 10 000uF capacitor will reduce by 100 Volts per second when it is discharged at 1amp.
However, in a supply like this, the capacitor gets recharged 100 times per second. So the discharge time is a bit less than 0.01 second and the voltage can only reduce by a bit less than 1 volt before the capacitor is recharged again.
Yes, sir you are right again. I have seen datasheets of regulators like 7805, LM117, LM317 etc.The input to the regulator must obviously be higher than 5V, but how much higher depends on which regulator you choose. Some regulators will work with 6V or less. Some may need 7V or more. If you look in a regulator's datasheet, you will see what it's minimum input voltage is.
Why do you choose shunt?which regulator is good, either series or shunt pass transistor regulator? I think shunt.
If we want ripple less than 2V pk-pk when 1A of current is drawn, then the smoothing capacitor should be at least about 4700uF.
Why do you choose shunt?
In what way do you think it will be better than a series regulator:
a) Lower cost
b) Better load regulation
c) Better input ripple rejection
d) Higher efficiency
e) Easier to design and build
f) Something else?
andre_teprom
Hello sir, thank you for your reply to me. 4700uF lies between 2000uF and 8000uF
Yes, of course.I have simple design for regulator. May show you
No, look at the scale at the bottom of the graph to see where 1 amp is.4700uF lies between 2000uF and 8000uF it means in between 0.1Vpp and 0.03vpp at 1amp.
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