I agree with all sugestions.

One question though...

How about using bigger electrolitic caps ( ex. 1000uF )?
Not recommended or just overkill ?

.

 

Not in this application. It would be over-kill since he is using an AC adapter outputing 12Vdc. There won't be that much ripple. The larger caps would be used if powering from an AC source, post-rectification.

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Hi ruijc,

using an electrolytic of 1,000µF per expected current flow of 1A is no overkill.
ICs tend to work most reliably with smooth DC for supply.

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I certainly agree that ICs work best with smooth DC, but when he asks if they are over-kill, I think not necessary (over-kill). The AC adapter filters out most of the ripple anyway. A 100uF cap on the input and output is sufficient. But, ruijc, if you want to spend the extra money and real estate and use a larger cap, it won't hurt anything.

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Hi

sorry to dig up an old thread, but i just have a quick question and this is very relevent - I am building a relay box/power supply/square wave generator.

I plan on having 230v IEC mains plug on the back, through a fuse, transformer to 15v, bridge rectifier (with the filter caps earlier in the thread).

Now on to the question, I want to have a 7805 drawing 200mA, a 7805 capable of drawing 1A, and a 7812 capable of drawing 1A.

What size of capacitor would be suitable to use as the storage cap (the one that goes after the rectifier before the regulator. I have a few 2200uF and a few 4700uF. would any of these (or a combination) be suitable for this, if not what would?

The load on 1 of the 7805's and the 7812 may be switched on and off regularly.

I have all the smaller caps for use with the regulators.

Thanks

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a single 2200uF should work fine, but a single 4700uF would be OK also, any more would be over kill and just take up more space. All the input filter needs to do is to keep the lowest peak of the input ripple voltage above the minimum input DC voltage specification for the regulators.

Your biggest problem will be the size of heat sink you will need, certainly for the 7805 that will be supplying 1 amp. That one has to dissipate 10 watts of heat which is a whole lot of heat. You might want to put a series power resistor in front of that 7805 such that the input voltage is closer to the 9-10volt range when supplying a one amp load.

Lefty

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I guess the idea is to use the Power Resistor is used to dissipate some of the power, helping the 7805 by not having to dump as much to regulate to 5v.

I have a few boxes of 10watt power resistors in various values, can you suggest a value that may be suitable.

I doubt that I will ever draw the full current from them, but as its a 'tool' im building, i dont really know what i'll be plugging into it - but it might as well be able to handle itss full rated capacity.

It is going in a rather large project box (A4 roughly, by about 100mm tall) so there is plenty space for heatsink and for a fan if neccisary.

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Yes the idea is to spread the heat dissipation between the two components, the regulator and the voltage dropping resistor.

5 or 6 ohms should work. Just use ohms law, E= how much voltage you wish to drop and I = the max current to be drawn.

There are newer regulator chips that use switching methods for voltage regulation that dissipate a lot less heat at the same power levels. They cost somewhat more but sometimes that is worth the extra efficiency, especially if one is using batteries as the power source. For a bench supply old 78XX regulators work fine if you watch out for there heat sinking needs.
Lefty

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Measurement changes behavior

 

The input voltage should stay about 2v above the output voltage. If the DC is unfiltered it goes all the way down to zero so the regulator would shut off every 8.3 mS.
Imagine troubleshooting that with a DC voltmeter.

12vdc equals a peak value of PI/2 x 12 = 19v. Your lowest desired value is 7V. Approximately, if your current is 0.5A you need a minimum of I ΔT/ΔV = 0.5 (.008)/(19-7) = 330 µF.
An exact solution is kind of messy, and caps have a wide tolerance anyway.

I wouldn't go with very large cap values; you could exceed the I squared T rating of the diodes in your adaptor, thereby giving you more troubleshooting tasks.

Do you have a scope?