hi 7.
My circuit in post #24 is one simple option, if you are able to get the power resistors.
E

 

The reason why a regulator like that heats up is because it's receiving so much power and providing a regulated (lower) power. For a moment lets use some wildly exaggerated numbers that don't exist in the real world: Suppose you have a power supply that supplies 1,200 volts and you want to regulate it down to 12 volts. While you're getting the desired 12 volts you have to do something with the remaining 1,188 volts. You're not just blocking that excess voltage you're converting it into a different form of energy. Energy that exists as 1200 volts is still the same amount of energy whether you keep it as electric pressure (the correct term for voltage) or if you convert it to thermal energy (heat). Since I'm using a wild number as an example, your regulator must convert 1188 volts (assume at 3 amps) (that's 1188 x 3 = 3564 watts) your regulator must do something with that 3500 watts. It's converted into heat. That's the main reason why people are suggesting PWM (Pulse Width Modulation) as a means for reducing the voltage.

The way PWM works (in your real world situation) (24 volts to 12 volts) is that it turns ON and OFF rapidly. The difference between the on time and the off time gives you an average voltage. So if 24 volts is on half the time and off half the time then the average voltage is 12 volts. It's not converting (wasting) extra voltage, it's merely shutting it off for a short period of time. DC to DC buck converters do exactly that - turn on and off to give you an average voltage output.

All things have their limits, so you have to consider three factors: Voltage in and out; Amperage; and Wattage. If you exceed the device's capabilities you will burn it out. No matter what you do with a 7812 it's going to have to dissipate the unused power. It doesn't just go away, it goes away as heat. That's the nature of 7800 regulators.

Hi Tony,

Your explanation is quite good. I searched for PWM first but I didn't find any circuit diagram online. So I decided to go with whichever is easily available and simple to make.
I agree with all your points and I am still searching for PWM. Ping me if you have one.

Regards,
Shan

 

You can reduce the voltage to the capacitor SOME by changing the circuit to a half wave rectifier instead of the full wave bridge. That will charge the capacitor to the peak voltage instead of the peak to peak voltage. But really, the transformer voltage is excessively high for your application. If the transformer has a center tap you have an option to change to a full wave circuit that will provide half the DC voltage that you get now. If it has a dual voltage primary, 120/240, and your line voltage is 120, you could connect the primary for 240 and cut the secondary voltage in half.. OR you may be able to purchase a transformer with a center tapped secondary, or even a 12 volt transformer. One additional idea is to build up a discrete transistor regulator to provide your DC output. You may also be able to put a well insulated resistor in the mains feed to your transformer.

 

One more idea that will work if you only have one load that needs 12 volts, and the current is always the same. such as a 12 volt relay coil. That option is to simply use a resistor with the correct value and adequate wattage. Cheap, simple, and easy. Not very efficient, but I doubt that it matters here in this application.

 

@MisterBill2 is right. Simple solutions sometimes work the best. Can't tell you how many times I've engineered a method for taking a branch out of a tree when all I needed to do was cut the limb and let it fall. Of course, a better understanding of what you're doing would go a long way for us to help you find a good solution.

 

@MisterBill2 is right. Simple solutions sometimes work the best. Can't tell you how many times I've engineered a method for taking a branch out of a tree when all I needed to do was cut the limb and let it fall. Of course, a better understanding of what you're doing would go a long way for us to help you find a good solution.

Indeed, Tony. And I think that I read someplace in this thread that the purpose was to operate a relay, which is a nearly ideal constant load. Even better, if it is indeed a 12 volt relay, no filter is needed. Just two diodes back-to-back across the supply voltage and connect the relay coil across one of them. The positive half cycle drives the relay and on the negative half cycle the collapsing magnetic field keeps the relay energized. It worked very well in stopping the buzz of a big contactor controlling a set of auditorium lights. We used ten amp 400 volt stud mounted diodes because it was a big contactor. A really cheap thick that works well.