25-45 DC in, I need 12V DC out.

Firstly, msimon  is correct the input voltages of a 7812 will not withstand prolonged high voltage. There is a high voltage equivalent but it may be better to consider a transistor series regulator.

I have attached a simple circuit for reference. Select a transistor that has enough voltage Vce rating and current.

Filter capacitors do not need to be huge. Input filter is the most critical. Make it perhaps 470Uf 63V. Zener diode is 12 volt.

Thanks for the help! I'll see what I can come up with. Can you perhaps suggest some components I should use, like the transistor and the resistor? I went on the local hardware store's website and there so many different kinds that it's overwhelming. Once again, I apologize for asking such questions. Also, about the capacitors, the ones apart from the input, can I get away with using ceramic capacitors?

A little update, I found this: DC/DC Понижаващ модул с LM2596HV, LM2576HV | Захранване Ком
It's pretty cheap and the store is in my country. It's based on the LM2576HV and I've seen some reviews on the internet and people seem happy with it. On the page there's a "drawbacks" section. It says that when the difference between the input and the output voltage is greater 15-20V, the board can't reach its output of 2 amps without overheating, in order to do it, it needs a different inductor. I don't think a bunch of LEDs can draw 2 amps (though as I said I'm no expert so I could be entirely wrong) what do you guys think?

It would depend upon the final current draw of the indicators - I would suggest perhaps a TIP3055 on a heatsink and a 2k2 resistor (1 watt). Make the zener a 1 watt as well (softer regulation and more robust). Output filter capacitor can be 100UF 25V the one across the zener can be the same.

Very reliable circuit and extremely robust - but you should always have a fuse in circuit for protection (input side).

I'll source the components and build it. Thanks a lot for the help!

Actually I have a little 40mm (maybe smaller) fan with a 2mm thick heatsink attached to it. I think it should work, what do you think? Is it overkill?

@ WarrenW - I don't think your values will work very well::

Suppose the load current is 1A, the TIP3055 isn't accurately specified for current gain at Ic  = 1A but interpolating from the ST and On Semi data sheets we'll assume a gain of 40. So we need a base current of 25mA. To regulate reasonably the 1W zener current should never be less than 5mA and never more than 80mA so the resistor value should be about 475R, use 510 which gives a resistor current of about 75mA at a supply of 50V. 25mA will go into the base of the transistor and the rest into the zener. The circuit will start to lose regulation when the resistor current drops to 30mA, at which point the input voltage will be 12 + (0.03 * 510) = 27.3V. This may be just about acceptable.

It's not  a nice design because the resistor needs to be rated at 5W (and could have to handle 3.3W continuously - the problem is that the TIP3055 doesn't have enough gain.

Lets try a BDX33C, cheaper than the TIP3055, gain  minimum 750 at 3A, rated at 100V and 10A max, use a 1N4743A 13V zener.

The zener has a test current of 19mA so we can expect the voltage to be below 13V for lower currents. The darlington transistor will only need a base current of 1.3mA so, if we aim for lowest working input voltage of 20V, and 5mA zener current the resistor will need to be (20 - 13)/(.005 + .0013) = 1076R, lets use 1k. And at max input voltage of 50V the resistor current will be 37mA, and it will dissipate 1.7W.

So my advice is use R = 1k, 3W, zener 1N4743A, transistor BDX33c, all parts available from Farnell.

The output capacitor and zener capacitors can be small - 47uF at 25V will be fine, the input capacitor depends rather on the frequency of the AC, which we don't know, but I would go for 1000uF (will drop 5V in 5mS at 1A) to start with. It ought to be a decent part and rated at 60V or more.

MK

Thanks a lot for the input! I already got the parts Warren suggested. I talked to my digital/analog schematics teacher at school and she said that we can test this circuit tomorrow at her lab. We will be using a power supply for the input so it might be the best case scenario, but I guess it's better to try it before investing more time into it. If it doesn't work, I'll try what you suggested. Once again big thanks to all.

Thanks a lot for the input! I already got the parts Warren suggested. I talked to my digital/analog schematics teacher at school and she said that we can test this circuit tomorrow at her lab. We will be using a power supply for the input so it might be the best case scenario, but I guess it's better to try it before investing more time into it. If it doesn't work, I'll try what you suggested. Once again big thanks to all.

So today we tested the circuit on a project board and on the output we got about 7V with 12V in, so we put more voltage on the input and at 30V in (the maximum on the DC power supply we had on hand) we got 8.6V output. We put a 2K load resistor and the values remained the same. We measured the voltage across the Zener and it's the same as the output. Before I try to build the other scheme I'd like to know what went wrong so I know in the future. Thanks for the replies!

Try removing the transistor and measuring the voltage across the zener for input = 10V ..... 30V in 5V steps.

Check your caps are all the right way round.

Check it is a 12V zener !

MK

Thank you for the suggestion! Thing is, that I don't have an adjustable DC power supply at home and the teacher doesn't have access to the lab at all times, so I'm not sure how to test it out. I've got a 1A 18V transformer with a bridge rectifier on it that my dad built long before I was born, but I'm not sure whether it will work for testing. I tested the little fan I was talking about (12V, 0.06A) and it caused the voltage from the transformer to drop to 15V, so I'm really not sure how to test it. The zener is "PL12Z" and a quick google search revealed that it is indeed a 12V zener (at least from what I managed to gather). About the capacitors, I'm pretty sure they're the right way around because we ran it for quite a while so my guess is that if they were flipped, they would've popped. Regardless I'll make sure to check them when testing. I apologize if I'm being a pain in the bottom, I feel too uneducated to be in this community, but I'm willing to learn. Once again, thanks to everyone for helping.

I have to go home now but I'll make some suggestions first thing tomorrow.

Have you got a reasonable voltmeter   ?

MK

I've got a "Pro's Kit MT-1210" multimeter, it beeps and it's also got a back light. I think it'll get the job done. Thanks for the help, have a nice evening.

Its a pretty poor zener specced for a test current of 50mA. With 12V and 2k resistor your current would have been about 2.5mA and with the 30V in about 10mA. The spec of the zener doesn't say what it will do at such low (compared with the spec) currents but 8.6V is not impossible.

The zener I suggested is better (test current 19mA) but it will still vary as the supply voltage varies.

To get good results you need to keep the current in the zener closer to the test current but this is very difficult without using more parts.

To check the zener you need a range of resistors or a variable power supply. Your home supply needs a capacitor across the output (1000 uF would do) if you are going to use it. So long as you measure the current and voltage through the zener for different resistor values you can still work out the characteristics of the zener.

(It's just a bit easier if you have all the gear.)

MK

So if I opt for a different zener, the values are going to be closer to 12V correct? If so, I'm going to go to the hardware store and ask them for the specific part you specified. If it's not available, I should look for a replacement one with low current spec, correct? Thanks for the advice, everything becomes more and more clear with every reply I get!

Hi,

What you want to do is make sure that the zener is regulating - as suggested remove the transistor. Your 18V transformer with a bridge should be fine for confirming the zener is indeed regulating. This simple circuit is effectively known as an emitter follower series regulator. The base of the transistor is held at the zener voltage and the output (emitter) will be approx. 0.6 volts lower than the zener voltage (with a small load on it - less than 0.5 amp). The regulation will be pretty stable as the current into the load increases. If you are using a BDX33 this is a darlington arrangement (effectively 2 transistors in one package), it requires less drive current into the base as you have 2 transistors to 'amplify' the current. With 2 transistors you have 3 base emitter junctions each at 0.6v so the output voltage will be 1.2volts lower than the zener voltage, but would be VERY stable under even small loads.

Assuming the Hfe (current gain) is 100 and you want 1 amp of output. 1 amp divided by the gain would be the required input current to the base. Or in this case 10mA. For good regulation the zener current should be approx. 5 times the base current so let us assume we will want a total of 60mA (10 to the transistor and 50 to the zener). The zener voltage is 12 volts and when running at 50mA (P=I*V) the power dissipated would be 600mW so a 1 Watt zener is the smallest you would go for.

The real problem you have is the very wide range of input voltage. If we say you have 45 volts of input, the voltage dropped across the resistor is going to be 45 -12 = 33 volts. We have to supply the zener / transistor with 60mA so that makes the resistor value 33/.06 = 550 ohms (closet is 560) and the wattage will be 33*.06 or 2 watts.

As the input voltage decreases the amount of voltage across the resistor drops and therefore the current into the zener /transistor drops as well. At 24 Volts input we have only 12 volts across a 560 ohm resistor = 20mA  if your load is 1 amp and the transistor is 100 gain then you have 10mA going into the base and 10mA into the zener. It will still regulate but is now dependant upon the characteristics of the zener.

Best option is to find a higher gain transistor as this will allow you to drop the current through the zener / transistor combination. A TIP142 may be a better option with a VCE of 50V allowing you input voltage to get close to 60V. It also has a current gain of 1000. This means for a 1A load you only need to supply 1mA into the base! you can run with 5 -10mA zener current and this means your zener can go down to a 400mW version if needed for better regulation. The resistor value would now be 3300 ohms to give you a 10mA current flow into the zener /transistor at the 45V input voltage.

When the input voltage drops to 24V the current in the zener / transistor would then be approx. 4mA. All good!

The resistor wattage also reduces to a 1/2 watt version.

Hopefully this long drawn out response will be of some assistance.

Warren.

Allright, just got done testing the zener with the transformer I got, it was reading around 13V across it while radiating a considerable amount of heat. I think I baked it back at the lab because there was some smoke coming out when my teacher tested the zener, I guess the PSU there supplied too many amps. Flipping it around gave me around 0.03V while attached to my transformer but it was buzzing a lot louder. Warren, just to confirm, you're suggesting I build a new circuit with the values you stated above? I once again apologize for my newbie questions, it's all a little complicated to me.

What was the resistor value you were using?

The Zener sounds like it was working overtime and dissipating a bit of heat!

I would really suggest you look at the components I indicated and the values. The calculations are all based on Ohms law for power and voltages etc. P=V x I , V=IxR, I=V/R, R=V/I.

Have a look at this for a better explanation Ohms Law Tutorial and Power in Electrical Circuits

With a TIP142 transistor the power dissipations go right down due to the gain of the device.

I would recommend you build the circuit. Sounds like the zener may now be fried. Replace it to be safe.

Cheers

Warren.