Hello,
I'm having some problems designing a power supply for two voltage regulators. They are needed to power a lamp at 12V and a HT12D chip at 5V.

I am told that the peak voltage of the transformer is 22v @50hz , with an output resistance 0.9ohm. So I then full wave rectify this and am left with 20.6V. I then tried to calculate my value for the smoothing capactior using the formula below to get a 10% ripple:


C= 5xIo
Vsxf

Using this calculation I got a value of 77.27mF ( this seems like a big capacitor but I've never done anything like this before so it could be correct )

I calculated the Io by using ohm's law ( Transformer RMS / 0.9 ohm ) giving 17A

This is where my problems start to begin, I think
I'm using the 7805 and the 7815. Each are rated P<15W. Am I right in assuming that the voltage after the diode rectifier is 22-1.4= 20.6V.

I know that I'll need a resistor to limit the current inputted to the chips to ensure P<15W. I assumed the current inputted to the regulator was 0.6A with a voltage drop of 3 across the resistor ( 17.6 x .6 = 10.56W )
Now I'm pretty sure that this is wrong but as I said I haven't done anything like this before really.

Any help would be much appreciated, I think I've layed this out as clearly as I can, and thanks to anyone who replies!

Damien

 

You are doing this a bit wrong.. I didn't check out your equation and am taking your word for it. You seem to be calculating for a 10% ripple under a 17A load. I don't think your lamp and IC will be drawing nearly this much current!!

Io is the nominal current that you wish to supply, not maximum. Secondly, if your voltage regulators were to supply this amount of current, they'd be destroyed very quickly anyways

Steve

 

Thanks for reply, I got the equation from http://www.kpsec.freeuk.com/powersup.htm#smoothing and it says that Io is the output from the supply? I drew a diagram anyway of what I'm thinking of doing

 

Here the diagram with the resistor before the regulators
omitted ( I forgot about it )

 

Your cart is ahead of the horse. The load determines the current flow at a given voltage. The 78xx regulators can't pass over 1 amp of current, even with heat sinking.

Just as a practical matter, capacitors are cheap. Unless there is some overriding consideration, do not figure for the absolute minimum needed - get a nice large value. Also get about double the rating for the voltage it will be handling. The difference in cost between 100 uF @ 25 volts and 2200 uF @ 50 volts is $1.20.

A large capacitor on a rectifier should charge to peak - look for most of those 22 volts to show up.

Using a resistor to limit current before the regulator wrecks the regulation - make sure you have enough regulator to pass the current the load needs. There are ways to parallel fixed regulators or to use large transistors to pass more than the 1 amp limit.

 

Thanks for the reply,
I understand now to use a smaller value capacitor, 2200uF with the 50v rating as you suggested .

I'm still a little confused as to the current issue. The regulator's are rated at 15W or less, since the voltage entering both of these will be roughly 20V the current will have to be less than 0.666.A. Are you saying to place a resistor after the regulator ( Rload ) and use this to control the current i.e. use a 20/.666 = 30ohm resistor?

 

Power supplies don't just hose power off into thin air. With no load attached, they will use a couple of milliamps of current inside the regulator, but nothing else will go on.

Now let's say you hook up the 12 volt regulator to a load, like a 12 volt motor. Let's say it draws 100 ma when running. The regulator will supply that 100 ma while maintaining the voltage at 12. The step-down transformer may be capable of supplying 100 amps, but the load only need 100 ma, and that is all the current that will flow.

You design the power supply to be able to source current to a known or expected load. No matter the capacity of anything else, though, the load on the supply determines current flow, and therefore the power.

 

Thanks for the reply,
I understand now to use a smaller value capacitor, 2200uF with the 50v rating as you suggested .

I'm still a little confused as to the current issue. The regulator's are rated at 15W or less, since the voltage entering both of these will be roughly 20V the current will have to be less than 0.666.A. Are you saying to place a resistor after the regulator ( Rload ) and use this to control the current i.e. use a 20/.666 = 30ohm resistor?

The power dissipated by the regulator is the current through it times the voltage across it. The current is the current of the load and the voltage across it is the input voltage of the regulator minus the output voltage of the regulator.

 

Thanks for the replies, that's exactly what I needed to know!

Greatly appreciated