Simplest DC 5V power supply design

Hello experts!

I want to design the simplest DC power supply of 5V variable.

Here is the block diagram.


I want to design first unregulated power supply.

OK here is my idea for designing,
I want to design it for 5V and 1A so my transformer must be rated at 5V and 1A too? right?

First tell me this then I will ask further.

Thank you.

so my transformer must be rated at 5V and 1A too?

Click to expand...

Who will pay for the rectifier and linear voltage regulator losses? A typical design would use at least a 7V/1.5A transformer, better 8 - 9V.

Ohhh I see...
We should take transformer above than the required voltages and current in order to compensate the losses. Right?
Nice point thank you sir.

OK so I am taking transformer 8v and 1.5A rating.

OK now my second step is to design the rectifier. I am choosing bridge rectifier. Since my requirement for supply is 5v and 1A, so rectifier's output equation would be,
\[{V }_{L(pk) }\]=\[{V }_{S(pk) }\]-1.4
\[{V }_{L(pk) }\]=\[\frac{{ V}_{rms }}{0.707 }\]-1.4
\[{V }_{L(pk) }\]=\[\frac{8 }{0.707 }\]-1.4
\[{V }_{L(pk) }\]=11.315-1.4
\[{V }_{L(pk) }\]=9.915V

It is the output of rectifier. Is it correct?

And which diodes should I use for bridge rectifier? If I use 1N400X series as shown in the below datasheet table, then its forward current is just 1A while I am using transformer of at least 1.5A of rating. But diode's forward voltage is OK.


So how to choose diodes? In order to choose diodes, anyone must consider forward current, forward voltages and power dissipition of that diode. Right sir?

Thank you.

Diode maximum peak current must be considered according capacitor placed on bridge output.


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I would not use a diode rated at 1A in a 1A power supply. I would use a diode rated at about 1.5A instead. In fact, I would use a 1.5A bridge rectifier module instead of separate diodes.

Texas Instruments recommends a minimum input of 7V for their uA7805C voltage regulator IC.
The bridge rectifier might have a forward voltage drop of 1.6V and the filter capacitor might have ripple of 1V. Then the peak of the sinewave from the transformer must be a minimum of 7V + 1.6V + 1V= 9.6V and its minimum RMS voltage is 6.8V. An 8V/1.5A transformer will be fine.

Eshal said:

9.915V

It is the output of rectifier. Is it correct?

Click to expand...

Yes, you calculated it correctly.
In real life, the transformer won't give exactly 8V rms and the voltage drop across each diode won't be exactly 0.7V, so you're unlikely to get exactly 9.915V. Maybe it will be 9V or maybe it will be 11V. Good enough.

Eshal said:

And which diodes should I use for bridge rectifier?

Click to expand...

It's easier to just buy a bridge instead of making one out of four diodes.

- - - Updated - - -

P.S It's probably easier to find a 9V transformer than an 8V one.

All replies helped me a lot. Thank you experts.
Now I have stepped ahead in designing DC power supply. OK so I have chosen transformer of 8V and 1.5A rated.

Yes experts, I know bridge rectifiers modules are available there. OK I will use it with these ratings,
output voltage = 9V minimum and 11V maximum
output current = 1.5A nearly.
Just these two parameters are to be considered in choosing bridge rectifier IC or anything else?

Thank you sir.

Help Please

Eshal said:

Just these two parameters are to be considered in choosing bridge rectifier IC or anything else?

Click to expand...

Nothing else. Only forward current and reverse voltage.
In this case reverse voltage isn't a problem because the voltage is so low.

OK thank you sir for you reply.
Now, next step is to design the filter capacitor. I have made the calculation of the capacitor for the capacitive filter, here is that calculation.

\[{V }_{L(dc) }\]=\[{ V}_{L(pk) }\]-\[\frac{{ V}_{r } }{2 }\]
\[\frac{2{V }_{L(pk) } }{\pi }\]=\[{ V}_{L(pk) }\]-\[\frac{{ V}_{r }}{2 }\]
\[\frac{2(9.915) }{\pi }\]=9.915-\[\frac{{ V}_{r } }{2 }\]
\[\frac{ 19.83}{ \pi}\]=9.915-\[\frac{{ V}_{r } }{2 }\]
6.312=9.915-\[\frac{{ V}_{r } }{2 }\]
6.312-9.915=-\[\frac{{ V}_{r } }{2 }\]
-3.603=-\[\frac{{ V}_{r } }{2 }\]
3.603=\[\frac{{ V}_{r } }{2 }\]
3.603*2=\[{V }_{r }\]
7.206=\[{V }_{r }\]
\[{V }_{r }\]=7.206V

Above I calculated the ripple voltage for the capacitor from the equation you can see in the first line of the calculation.

Now from this ripple voltage I calculated the capacitor value, show below,

C=\[\frac{{I }_{L } }{{V}_{r } }\]
C=\[\frac{1.5 }{7.206 }\]
C=0.208
C=208mF

Did I do all calculations correct? Is my calculated capacitor value correct for the the designing of the dc power supply for 5V and 1A?

I used IL=1.5A as discussed in the previous post in order to compensate the losses.

I didn´t understand what the parameters your model was based.

why don´t you perform a simulation at some tool such as PSPICE or some other else in order to take a fine tune ?



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Eshal said:

Did I do all calculations correct?

Click to expand...

No. Here is an example of how to work it out:

If we assume that the regulator requires at least 7V input and we assume that 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.

It looks like you got lost trying to solve equations and forgot about the goal, which is to make sure the output voltage from the unregulated supply is always high enough for the regulator. I'm surprised you didn't notice that your answers are obviously wrong.

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?

Important:
1) I did not say the regulator requires at least 7V input. That depends on what regulator you use.
2) I did not say that the peak output voltage from the rectifier may be as low as 9V. That depends on what transformer you use and what your actual mains voltage is.

You need to know these things before you try to calculate the capacitor value.

Eshal said:

I used IL=1.5A as discussed in the previous post in order to compensate the losses.

Click to expand...

Using components rated for higher current is not to "compensate the losses", it is to provide a safety margin.

godfreyl

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.

Click to expand...

I am getting what are you trying to teach me. But I didn't get how do you come to use capacitor of 4700uF!!!!
Is it calculated value, guess or your experience?

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?

Click to expand...

Obviously sir, you are right. You raise a very good point and I agree with you that I did mistake.

But now I am confuse with one thing now. The rectifier output is VL(dc) or VL(pk)? And what should be the unregulated input to the regulator, is it VL(dc) or VL(pk)?

As you said,

(because 9V - 2V = 7V).

Click to expand...

Here, \[{V }_{L(pk) }\]=9v right?
and \[{ V}_{r }\]=2v right?
But what is 7v?

In previous post, we discussed and I also mentioned the output of the rectifier is \[{V }_{L(pk) }\]=9.915v. So it is obvious then if it is output of the rectifier then it will act as an input for the filter capacitor, right sir? 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?

Sir, are you getting what is confusion in my mind and what I am trying to ask you sir?

Thank you.

PS: sorry for late rep because my internet connection was disabled.

andre_teprom
Hello, yes sure, I can perform simulation but I will not. I want to learn mathematics behind electronics. What engineers could do when there were no electronics software. Obviously they perform mathematics and implement. So, I want to be like as those engineers were.

Thank you sir.

Eshal said:

But I didn't get how do you come to use capacitor of 4700uF!!!!
Is it calculated value, guess or your experience?

Click to expand...

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.

Eshal said:

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?

Click to expand...

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.

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.

Click to expand...

I have gotten all things sir.

So, I decided to choose 6800uF because 1A is the current which is my requirement as the output of the supply. So obviously, at least I need 1.5A and out of this 1.5A some will destroy in the supply's component during there working in the form of heat. So I decided to use 1.5A and 6800uF capacitor. This thing is also discussed in the post#1 by FvM.

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.

Click to expand...

Yes, sir you are right again. I have seen datasheets of regulators like 7805, LM117, LM317 etc.
But I want to design my own regulator, i.e series or shunt regulator with pass transistor and I hope you are going to help me through all this tutorial because you have great knowledge and you are raising very good point and I am understanding all your saying sir.

Now, I have decided my filter capacitor value i.e. 6800uF. Now, next step is to design regulator. Can you tell me sir, which regulator is good, either series or shunt pass transistor regulator? I think shunt.

Thank you.

Eshal said:

which regulator is good, either series or shunt pass transistor regulator? I think shunt.

Click to expand...

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?

godfreyl said:

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.

Click to expand...

According to following chart, 4700uF seems to be reasonable result for above calculation ( considering Vpp=2v and Io=1A ) :




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godfreyl

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?

Click to expand...

Hello sir, thank you for your reply. And I choose it because it is a shunt. Means, uses a transistor in parallel with load. And we all know that anything in parallel holds all voltage across it unlike series components in which voltages are divided. :-/
May be I get it right. But you know better.

By the way I have simple design for regulator. May show you sir?

andre_teprom
Hello sir, thank you for your reply to me. 4700uF lies between 2000uF and 8000uF it means in between 0.1Vpp and 0.03vpp at 1amp. But my vpp of Vr is 2V. ????????????? Is my point wrong sir?

Thank you all
Princess

Eshal said:

andre_teprom
Hello sir, thank you for your reply to me. 4700uF lies between 2000uF and 8000uF

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I presume the curves at above chart was drawn on a logarithm scale, what means that once the point [ 1A , 2Vpp ] is located approximately at 1/4 distance between 2.000uF and 8.000uF, is somewhat around 2x factor on that scale ( ~2x 2000uF ~ 4.000uF ). Note that this chart don´t provide a way to determine precisely some result, but allow us to check if calculated result fit on expected magnitude.


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Eshal said:

I have simple design for regulator. May show you

Click to expand...

Yes, of course.

Eshal said:

4700uF lies between 2000uF and 8000uF it means in between 0.1Vpp and 0.03vpp at 1amp.

Click to expand...

No, look at the scale at the bottom of the graph to see where 1 amp is.