Does a LM317 act like a transformer

[johnmarkp]

This started as a question about how big a transformer I needed for a PS. The final voltage after the LM317 was 12v. I needed 0.9 amps. I used a 18 volt transformer. Does it need to be 0.9 amps or does the LM317 act like a transformer. In this case 12v time 0.9a = 10.8VA. So at 18v I would need only 0.6a. Or do I need the full 0.9a from the transformer. Any help would be appreciated.

John

[DigitalJunkie]

You need the full current.(and maybe a bit more,for headroom) The LM317 is a linear regulator,so the excess voltage is simply 'burned off' as heat. It won't trade voltage for current like a switchmode regulator could.

[imix500]

For a 900mA load I'd do at minimum 2A transformer. 2.5A would be better though.
Remember to heatsink the regulator well.

[infinia]

Hi
Standard series voltage regs need ~3V headroom looking at the data sheet. So the transformer needs to provide Vo plus an extra 3 V. Your load or circuit determines the current.
In your case if the transformer uses full wave rectifiers, then converting backwards from the Vreq out , (15+3) Vdc /SQRT 2 = 12.7 V AC. Also transformers drop some voltage at full load due to regulation so assume small XFMRs are about 85% then use a transformer rated for at least 14-15 Volts AC with at least 15 *0.9 VoltAmp (VA) power rating. Using a XFMR with more AC voltage than 15 will waste excess power across the LM317 and increases the XFMR size and LM317 heatsink accordingly.

[gootee]

This has just about everything you need to know:

Unregulated Power Supply Design

And this thread works through an example, in detail, for a regulated power supply:

Is a 9v transformer large enough?

You will want to remember to account for 10%-low AC mains voltage, and transformer regulation, and then make sure that the troughs (minima) of the ripple voltage don't cause the regulator's input voltage minus output voltage to dip below its dropout voltage spec. That can be pretty ugly. (A "low-dropout" (LDO) regulator can help, if you're ever stuck with a transformer that might not give quite enough voltage under worst-case conditions.)

The first link I gave shows how to calculate the min and max of the ripple voltage, given your smoothing capacitance value and the current draw of your load. After that, it's easy to figure out what transformer will work, given your regulator's dropout voltage spec.

And, as already mentioned, you don't want a transformer with a secondary voltage that's too much higher than you need, because then the regulator might get very hot, and you'll just be wasting power.

Cheers,

Tom Gootee

[janneman]

Originally Posted by infinia

Hi
Standard series voltage regs need ~3V headroom looking at the data sheet. So the transformer needs to provide Vo plus an extra 3 V. Your load or circuit determines the current.
In your case if the transformer uses full wave rectifiers, then converting backwards from the Vreq out , (15+3) Vdc /SQRT 2 = 12.7 V AC. Also transformers drop some voltage at full load due to regulation so assume small XFMRs are about 85% then use a transformer rated for at least 14-15 Volts AC with at least 15 *0.9 VoltAmp (VA) power rating. Using a XFMR with more AC voltage than 15 will waste excess power across the LM317 and increases the XFMR size and LM317 heatsink accordingly.

This is quite correct; I only would like to add that the diodes in the rectifier also drop voltage. For a bridge rectifier there'e always 2 diodes in series which worst case can drop up to 1.5V. So, the Vreq would be 12V output + 1.5V for the rectifiers + 3V for the regulator makes 16.5V. Converted to VAC by dividing by sqrt of 2 gives a minimum of some 12VAC.

jd

[janneman]

Originally Posted by janneman

This is quite correct; I only would like to add that the diodes in the rectifier also drop voltage. For a bridge rectifier there'e always 2 diodes in series which worst case can drop up to 1.5V. So, the Vreq would be 12V output + 1.5V for the rectifiers + 3V for the regulator makes 16.5V. Converted to VAC by dividing by sqrt of 2 gives a minimum of some 12VAC. If you want to be prepared for up to 10% mains undervoltage you're at 13VAC allready.

jd

[johnmarkp]

Thanks to all. This answers my question and the next six I would have had. You have to love this forum.

John

[infinia]

Originally Posted by janneman

This is quite correct; I only would like to add that the diodes in the rectifier also drop voltage. For a bridge rectifier there'e always 2 diodes in series which worst case can drop up to 1.5V. So, the Vreq would be 12V output + 1.5V for the rectifiers + 3V for the regulator makes 16.5V. Converted to VAC by dividing by sqrt of 2 gives a minimum of some 12VAC.

jd

duh oh yea the diode drop. usually 15V Vregs come in pairs for +/-. So using a full wave with a CT XFMR using 2 pairs of diodes, the diode drop is only one Vd per Vreg. As you mentioned the line voltage regulation spec is the real heat waste designing for commercial gear. Being DIY we have the advantage or opportunity to narrow this down considerably.

EDIT
Usually a good guideline for series Vreg's >10V, is to select a transformer Vac equal to the Vdc output.
Also in choosing XFMRs pay particular attention to the output voltage specs> is it at max current and nominal line voltage! this is a gotcha esp. for buying surplus XFMRs

[firechief]

[QUOTE=infinia;2243458]

Usually a good guideline for series Vreg's >10V, is to select a transformer Vac equal to the Vdc output.


This is one of thouse valuable little nuggets that has poped up on the forum a number of times. It's a reasonable guidline.