28v to 12v conversion

Hello, I am new to this forum and I would appreciate any help you folks can provide.

I'd like to convert 28v to 12v, and I purchased an LM7824 and a few capacitors, but doing further research, it seems I may need to add a heat sink to this. Does anyone have any experience in this application that can provide some assistance?

Thanks!

LM7824 is 24v fixed voltage regulator. Maximum output current it can deliver is 1A with a fairly large heat sink. At 1A output it will get too hot.So a large heat sink is required.
To convert 28V to 12V you need a 12v regulator. Again Maximum input voltage for 7812(12v fixed voltage regulator) is 27V. So if you connect it to 28v it may burn out immediately.
You other option can be using variable voltage regulator like LM317/ LM338/ LM350.Now how much current you wan the regulator to deliver and what is your input source of 28V, A transformer or a regulated 28v dc power supply?

Hey, thanks for the quick reply to the thread.

I was actually just looking at some information on the LM317. I do need 1.0A along with 12 volts. The 28v are coming from a battery. It does seem that the LM317 would be a better fit.

1. LM7812/uA7812 would be the correct device for this purpose.
2. If a linear voltage regulator is suitable at all, depends on the intended output current, that hasn't been said.
3. A heatsink is most likely required. Without it, the maximum output current will be limited to about 100 mA. Fortunately the device has a thermal shutdown to protect against overtemperature.

Here are links to datasheets you need to download and see it by yourself.
LM7812 Datasheet pdf - 3-TERMINAL 1A POSITIVE VOLTAGE REGULATORS - Fairchild Semiconductor

For LM317 here is the other one.
LM317 Datasheet pdf - 3-Terminal Adjustable Regulator - National Semiconductor

Again Maximum input voltage for 7812(12v fixed voltage regulator) is 27V. So if you connect it to 28v it may burn out immediately.

Click to expand...

If you're refering to LM7800 datasheet, it's a missunderstanding. 27 V is the specified maximum input voltage to achieve 12V/1A output (with respective heatsink) without running into thermal shutdown. The maximum input voltage is 35V nevertheless. But a safety margin should be provided, if an input voltage above 30V must be expected, 7812 isn't safe.

7812 may be used Depending on manufacturer's absolute maximum ratings. There are 1A as as 1.5A version available from different manufacturers. But Keeping in mind the safety margins it may not be a good choice to reach the absolute maximum rating.Exeeding those rating may be worst.

I've looked at the data sheets, but there is so much information that I cannot decipher.

If I need to use a large heat sink, the project is pointless. How big of a heat sink are we talking? Does the LM317 lessen the need for a heat sink? I know it is a large drop in voltage, thus a bunch of heat needs to go somewhere, but I don't know enough about this stuff to understand how much heat it really is.

The required heat sink size will be similar for all linear regulators. You can save a small amount by using regulators with a lower junction to case thermal resistance. But it depends mainly on the package and is identically e.g.between LM317 and LM7812. So the only adavantage of LM317 is a sligthly higher voltage rating.

Can you give me an estimate of the size or a link to a heat sink?

Almost any device will generate heat while it regulates to the required voltage and when a load it applied to it and in respect to the amount voltagge dropped from the input voltage to the output voltage. As you say that your requirement is 1.0A @12 volt, i recomend the use a LM7812 regulator TO220 package whose heatsink may be 1 Inch x .75 Inch with 2 MM thickness. Fins will be an added advantage for better cooling. The one shown here too is agood choice

Ok, thanks! 1x.75 isn't bad at all. I just need to figure out how to put this all together now and some way to encase it.

---------- Post added at 11:15 AM ---------- Previous post was at 11:13 AM ----------

While we are on the topic, would it require an even bigger heat sink to convert down to 5v? That was actually the original plan.

Time for some theory I think !

No matter which linear regulator you use the power it dissipates in Watts is (input voltage - output voltage) / Input current. For simplicity, assume the Input current is the same as the output current flowing into your load.

Now look at the data sheets for heat sinks, you will see them rated in degrees Celsius per Watt. If we ignore the effects of mounting losses, you can assume the temperature will rise the stated number of degrees for each Watt the regulator dissipates. You will see that big heat sinks have lower C/Watt ratings, meaning less temperature rise for a given power.

So for example, dropping 28V to 12V at 1 Amp would dissipate (28-12)/1 = 16Watts. A heat sink rated at 1C/Watt would expect a temperature rise of 16 degrees.
If you dropped to 5V the figures would be (28-5)/1 = 23W so the same heat sink would show a rise of 23 degrees.
It's up to you to decide how hot it should be allowed to get but obviously keep it well within safe limits.

Note that the heat doesn't just vanish, all heat sinks assume some ventilation or conduction to release the heat energy, observe the manufacturers test conditions.

If you want to drop to 5V, (even 12V would benefit) a better solution is to use a switching regulator but they are somewhat more complicated to build.

Brian.

You should look at things like National's "Simple Switcher"
or the newer Vicor DC-DCs. More expensive than a bare
linear IC, but much more efficient and self-contained.

Thanks, Brian.

If I understand correctly, this ( Digi-Key - HS365-ND (Manufacturer - 576802B03900G) ) would rise 27.3C for every watt it had to dissipate?

Betwixt had it right. If you are going from 28V down to 12V with 1A output, then the regulator has to dissipate 16W. That's a huge amount of power to dissipate, especially from a TO220 package. I checked the 7812 datasheet and it has a thermal resistance from junction to tab of around 5C/W. So, whatever the tab temperature is add 16W*5C/W=80C to that for the junction temperature. And still you have to ad the thermal resistance of the Heatsink. If the heatsink has 5C/W thermal resistance, then at room temperature, the junction temperature would be 16W*(5C/W+5C/W)+25C=185C. Max for the 7812 is 150C. I think you would need a liquid cooled heatsink or a very very large heatsink in order to keep the temperature from burning up the device. You need to use a switch mode power supply for this.

Ok, thanks. I'll look into the switch mode power supplies.

So if the input amperage was higher, the temperature would be lower, correct?

I'm asking, because I have a 24v to 5v converter that has little to no heat dissipation. It is an adapter to charge a USB device from a cigarette lighter.

With a linear regulator, such as the 7812, the input current is nearly equal to the output current, which in your case is 1 Amp. So, the only way to reduce the power dissipation in the device is to reduce the input voltage. Power dissipation is (Input voltage - output voltage) divided by input current.

The reason your 24V to 5V converter generates so little heat is because it is most likely a switching supply. Probably a buck converter. Even the most crudely built buck converters have an efficiency of at least 80% whereas in your application the linear regulator would have an efficiency of 12/28=42.9%.

Buck converter and 28v to 5v it is, then. Now to find out how it build it!

Thanks, again!

Hey guys, I just came across LM2575. The data sheet states that it requires a much smaller (or none at all) heat sink than an LM7805. Are any of you familiar with it?