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Basic Power Supply Questions

by mishima
Tags: basic, power, supply
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mishima
#1
Dec30-13, 05:35 PM
P: 303
Hi, I have two questions about power supplies. I am just curious of what is practical and possible.

i) I have been using a 30V wall wart regulated to 18V for a radio telescope project (using LM317). The load does not really need a set 18V, and could actually swing between 13V and 18V. The current on the other hand should stay close to 200 mA. Is there a way to limit current to this value while allowing voltage to swing through or be within that range?

ii) In the same project another part of the circuit requires a regulated 5V supply. I was curious what might be the best way to merge the two so that I only have one single plug. The way I understand things, simply using the 30V for both regulators would cause a huge power dissipation on the 5V regulator especially.
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Okefenokee
#2
Dec30-13, 06:22 PM
P: 218
i) The LM317 can be configured as a current regulator. Do you need to clamp the voltage between 13V and 18V to prevent damage to the device? If not, just set the LM317 to deliver 200mA and you're done. If you need clamping then we can come up with something.

ii) Yes, 30V - 5V is a large voltage drop. I don't know you're current requirements for the 5V supply but the watts will add up quickly.

The LM317 is an LDO (Linear drop-out regulator). Those types of regulators do internal switching. That means the power dissipation is internal too.

You can get a switching controller to make a power supply. They use an external transistor, a coil, and a capacitor to do the bucking. The coil will do most of the dissipation and it can handle it. Here's one you can look at.

If your current requirements for the 5V are low enough you can still use a simple LDO. Your data sheet can tell you how to calculate the limit.

Texas Instruments makes an LM317. I'll give you a hint, always go with TI if you can. Their documentation is head and shoulders above other brands. The data sheets have clear explanations and plenty of design examples. They're written in proper english too.
mishima
#3
Dec30-13, 07:27 PM
P: 303
I tried measuring current from both power supplies with a brand new load (suspected the older one was damaged). Now I am thinking maybe (i) is not a concern. It seems that with everything setup and a strong signal source directly in front of the dish the current being drawn from the regulated 18V is only 180 mA. This matches what I was originally expecting just by looking online about LNBs (most require ~200 mA). The fact I was getting 700mA to a full 1A earlier probably just meant that something in the old device was damaged.

As for the the regulated 5V supply and the chopper circuit, with everything setup and functioning properly I get 5mA drawn. With a strong signal source directly in front of the dish it comes up to 6mA.

So, its
30V->regulated to 18V needs 180mA and
8V-> regulated to 5V needs 6 mA

I want to turn into a single supply. Thanks for bearing with me as I plod through this. I will check the datasheet to see if the 5V current is low enough, I am pretty sure I know what you mean. Those calculations are what I need to do right now.

tfr000
#4
Dec30-13, 07:33 PM
tfr000's Avatar
P: 119
Basic Power Supply Questions

Daisy chain the regulators - pull your 5V from your 18V. Still a big drop, but better than 30 to 5.
Okefenokee
#5
Dec30-13, 08:03 PM
P: 218
6mA @ (18V-5V) = 0.078W

Even if the thermal resistance of your LM317 is high (~ 50C/W) you'll only get a temperature increase of 3.9C. Seems like you'll be ok. Just need to find the thermal resistance and maximum temperature in your datasheet to be sure.

Your formula should be something like:

(power dissipation)x(thermal resistance to air) + ( ambient temperature )< (maximum junction temperature)

For ambient temperature, military grade is 70C. That means it can work in extremely hot environments. You can go lower for industrial or commercial grade.

The maximum junction temperature is probably either 125C or 150C. It's a result of the manufacturing process.

Thermal resistance to air may be split up in your datasheet. It could be (die-to-case)+(case-to-air) or something along those lines. I hope that helps because not all datasheets explain the thermal calculations.
mishima
#6
Dec30-13, 09:34 PM
P: 303
Well from the TI datasheet it is:

(thermal resistance junction to ambient) must be less than [(125-ambient temperature) / power dissipation]

From the STMicroelectronis datasheet, my values are

50 < (125-40) / 0.078 (seems unlikely I will be using at above 40 C)
50 < ~1090 (true)

For my 18V regulator on the other hand, its

(125-40) / 2.16 = ~40 which is not greater than 50. So its a good thing I've been using that heat sink there. :)

I think for simplicity's sake I may just break down and buy an actual regulated 18V wall wart. I already have a 5V regulator circuit soldered together anyways, and it doesn't need a heatsink. Then I could just totally eliminate the other regulator circuit (its a handheld type of thing, so less bulk is good).

I also think I found out why when I was using batteries they were draining so fast, yet no one else ever mentioned this problem. The design suggests 1.5 AA batteries, but I was using 2 9Vs in series. This website claims that at 200mA a 9V can only maintain 80% voltage for 1 hour. 1.5 AA batteries are eight times that! I had no idea there was that much of a difference. If I had used AAs, I could go from fresh batteries to 13V (still operational) in 35 hours (compared to a lousy 2 hours for 9Vs!). So that's what I get for not following instructions it seems.


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