Voltage shedding question

A few diodes between the reservoir cap and the regulator input will drop the volts. 1N400x diodes are good for 1A and wil drop about 1V per diode. Don't forget thatthey still need to radiate that heat somewhere.

Graham

Hi, Torben. If you've got a couple of ohms worth of power resistor you can put between the transformer and the regulator, you'll be pleased with the results. Look around in your junkbox. A couple of 1N5401 diodes in series between the cap and the regulator will work well, too.

Especially when you are dissipating higher wattage, a good heat sink for the IC is important. If you're doing a home brew, make sure the surface of the heat sink is perfectly flat where it meets the back of the semiconductor. Also, get some heat sink compound (the white zinc-oxide-based stuff is sold by Radio Shack in small tubes), and place a thin coating on the back of the IC before you bolt it to the heat sink.

Good luck Chris

--- You could do it by putting a suitably sized capacitor in series with the primary. That would bring the transformer's output voltage down to something the regulator could handle, and the capacitor wouldn't dissipate any power except that due to its ESR.

Here's the circuit: (view in Courier)

To find the value of C1, we'll work backwards from the output.

Since you want 12volts out at 1 amp, the worst-case load will look like a 12 ohm resistor, and it'll dissipate 12 watts.

The 7812 needs at least 2 volts of headdroom, and if the 4700µF cap is at the low end of its tolerance it'll be generating:

I dt 1A * 0.0083s dV = ------ = ------------- ~ 2.2V C 3.76E-3F

of ripple and that, added to the 2V headroom spec for the 7812 means that the cap needs to charge up to at least 16.2V on peaks.

On the other side of the cap we've got a couple of diode drops that need to be added in, so that brings the peak output voltage needed from the transformer to 16.2V + 1.4V, which is 17.6V. Divide that by 1.414 and you get 12.44V, (say 12.5) which is the RMS voltage you need to get out of your transformer.

Right now your transformer puts out 18VRMS for 120V in, so to get the output voltage down to 12.5V you'll need to get the input voltage down to:

Now, you'll need an amp from the transformer to feed the 12 ohm load, plus, since you have a capacitive filter, you'll need another about 0.8 amp to charge the cap on peaks, for a total out of the secondary of 1.8A

Since the transformer's turns ratio (for the windings you're using) is:

Vp 120 N = ---- = ----- = 6.67 Vs 18

the current in the primary will be:

Is 1.8A Ip = ---- = ------ ~ 0.27A N 6.67

And the impedance looking into the transformer will be:

Vp 83.3V Z = ---- = ------- = 308.5 ohms Ip 0.27A

Since the voltage into the transformer needs to be 83.3V and we have

Since current in a series circuit is everywhere constant, if we simplify the circuit to:

we'll have 120VRMS supplying 0.27A into the circuit, so the impedance of the whole circuit must be:

E 120V Z = --- = ------- ~ 444 ohms I 0.27A

In order to determine what part of that impedance is due to the capacitor's reactace we can write:

Z = sqrt (R² + Xc²)

and then rearranging to solve for Xc we have:

Xc = sqrt (Z² - R²) = sqrt(444² - 309²) ~ 319 ohms

Now, since the reactance of a capacitor can be found from:

and since we know the reactance and frequency we can solve for the capacitance by rearranging:

So, unless I made a mistake somewhere, it looks like if you put about 8µF in series with the transformer's primary it'll get you where you need to be without dissipating any power (other than about

If you decide to try it, use good quality mains-rated caps and make sure the transformer and the cap(s) aren't resonant at 60Hz!

-- John Fields Professional Circuit Designer

If 12 V is what you want coming out, I'd get a transformer that outputs less than 25 V peak in the first place.

But to answer your question, the standard technique is to put another regulator upstream from the 7812. Maybe an LM317 set to 18 volts.

The ultimate answer is to go with a switching regulator (not as easy to design yourself) since they do not turn the excess volts into heat.

That's going to be a bit pricey though !

I'd suggest also having a high value bleed resistor across that cap to discharge it when disconnected from the mains.

Graham

It is ? Doesn't sound at all 'standard' to me at all ! Talk about overkill.

Graham

A 10 ohm, 10 watt resistor - actually, you should probably use a

Hope This Helps! Rich

Agreed. A simple series power resistor can be used to shed the excess voltage. Let's presume you want 3V of headroom for the regulator. Also for nice round numbers let's presume you want to be able to pull 1A @ 15V. You need to drop the other 10V across a series power resistor. Pulling those nubers into V=IR you get

R = 10 ohms.

Note that the input voltage to the regulator will be higher the less lightly loaded the circuit. For example using the same equation you find that at 300 mA the voltage drop across the resistor will be V=(0.3A)10ohms = 3V. So in that instance the regulator will have 22V presented to it and will have to dissapate P=VI P=(22V-12V)*0.3A = 3W of power. But that's the same amount of power that would need to be dissapated when fully loaded too.

Now onto the wattage rating of the resistor. P=(10V)*1A = 10W. If you use a

Hope this helps,

BAJ

Not only overkill, it doesn't reduce the area of heatsink needed, so he's still at square 1.

Ed

If you have a 120:12V transformer, you could wire its 12V secondary to buck the primary side of the 120:36, in a pinch.

Ed

Thanks! Most of the answers here are things I was thinking might work...glad to find that everyone has a slightly different way they'd suggest. I love electronics for that.

I guess I'd want to throw in about 7 of these guys were I to go this way, at the far end of the enclosure from the regulator perhaps (since the diode would form a 7 watt heater, wouldn't they?).

Thanks again,

Torben

Hi John,

Thanks for the detailed explanation! It's actually better than a lot of the websites around which try to explain power supply design. Or at the very least it'd make a great appendix. :)

The method you describe looks interesting and gives me a lot to think about and read about, such as learning how I would go about checking the resonance of the AC cap/transformer combo. Until then I think I'll stick with a 20W series resistor or so.

Thanks for taking the time to explain all this!

Torben

discharge

Funny how bandaids always seem to snowball, eh? OK, so 2 more parts isn't exaclty an avalanche, but... :)

What about a bleed lamp? Any reason that wouldn't work?

Torben

Yes. So I'd use something like 1N540x - 3 amp rectifiers that have a Vf typ. closer to .8 at 1A. And a 2X safety margin for forward current. :-)

Although, personally, I still like my answer best - 10 ohm, 20W resistor, assuming you can't change the transformer. :-)

Cheers! Rich

Hm. I've actually got a small pile of those over there. :)

I think I like that solution best too...at least, it's the simplest and easiest to do the math for. I do have the option to change the transformer but hey---love to have some ideas for the case which will one day arise when I cannot.

Thanks again!

Torben

My Pleasure! :-) :-) :-)

Cheers! Rich

It's advisable to expect the reservoir cap voltage to droop a little on load, and allow for some variation in line voltage too. That would suggest a slightly smaller value R. When doing this, it's a good idea to be exactly sure of the unregulated voltage too.

Graham

They would. You're wisely aware of the effect of proximity. Also be aware that especially in a non-vented enclosure, things can get a lot hotter in 24/7 use than you might expect from leaving it for a few hours on the bench.

Graham

discharge

Oh, in parallel across the ac input. Yes that would do the job. I was just concerned about electric shock hazard.

Graham