Digital power supply

Agent Cobalt

I don't know if anyone has done this, but I'm trying to make a dual digital power supply. The power supply needs to have outputs from 3-20V and anywhere from 1-5A. I know that there are two Nuts & Volts articles that touch on this, but the design presented had limitations. Those limitations being it only had a range of 3-10V and output 1A. A plus for the power supply would be to measure current draw and to be able to limit the current draw to a specific amount (i.e. 750mA). Any thoughts as to assistance on this would be appreciated as I'm not the most experienced electronics hobbyist and would like to build this for use in my semi-existant lab.

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PJAllen

The LM-1084ADJ works the same as an LM-317 (and has the same pin-out), but it's a Low Drop-Out type.· The lowest voltage it will do is 1.25V.· It's rated for 5A, but you'll need a heavy-duty heat-sink for that.

If you need an output of less than +1.25V, you can reference the adjust circuit to -1.25V, instead of to circuit Ground, and that'll give you adjustability all the way down to 0V.

Agent Cobalt

I looked at the datasheet for the LM1084ADJ and it seems that it would be adequate for what I need. I looked at the calculations for the resistor values that I would need to get my voltage range and the results came out very small. Like it said that to get 20V out, I needed a 1.8k ohm variable resistor for my 3-20V yet in the datasheet, a 5k ohm variable resistor is used to get 1.25-15V. To check my math, I plugged in the values specified in the example (1.25-15V) and got that the output voltage would be 52V. So if anyone knows how to calculate what resistors I would need to get 3-20V, please let me know. Thank you.

P.S. I've included a picture of the example circuit provided in the datasheet for reference.

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MikeK

Are you going to be able to draw the current you want at low output voltage? The spec sheet says that Ilimit is 600ma (typical) when Vin-Vout is 25V. For your 3-20V range, you need 22V in (21.5 actually), which gives you a maximum delta V of 19V. I didn't see a graph for this function, but it doesn't appear you'll be able to draw much more than 2A at 3V (assuming Vin<22V).

Unless I totally misinterpreted the data sheet....

Mike

Agent Cobalt

Yeah. That portion of the datasheet confused me. The column header though says, "Conditions" and it says that for V_in-V_out = 25V the typical current limit is 0.6A. What I got from that is that when the difference between the input and output is 25V that the current limit is about 0.6A. So if I'm understanding and reading that right, I will only be able to draw 0.6A when the voltage is around 3V but I can draw the full amount at 20V. I don't know though. This is my first time messing with any voltage regulators and datasheets are usually cryptic to me because half of the stuff I have no clue what it means. *Shrugs*

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Agent Cobalt

I decided to use what Lon Glazner used in Nuts & Volts columns 49 and 50, the MAX726.

Lon Glazner (in column #50) said...

There are a few points that must be stated here. The digital potentiometer (DS1267-010)
was very effective in controlling the output voltage of the MAX726 over a short range of
voltages. Specifically, I had luck controlling the MAX726 between 3Vdc-10Vdc. Over
10Vdc, there was not enough resolution available in the DS1267-010 to allow accurate
voltage control. So at this point, I reduced the power supply output to 10Vdc maximum,
and modified the user interface code to reflect this change.

Now what I'm wondering, is if there is a digital potentiometer out there that would have better resolution?
I've looked on a couple different manufacturers' sites and couldn't find any that might fit. Though upon reading further,
it seems that the max resistance used in the DS1267 is around 5k (not sure if this is correct or not). So I figured that if I could find a digital pot that only
went to 5k ohms and had the same number of steps (256), that the resolution would be better. But looking on the sites (again),
I found that it seems that 10k ohms is the lowest digital pot made.

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pjv

Hello Agent;

A better precision potentiometer will not solve your problem, and that is heat.

In order for your supply to provide the 20 V out you need, then the input to the regulator needs a voltage around 23 volts or higher. And it had better be a beefy source to be able to deal with that 5 Amps you stated.....lots of concern over ripple at that current.

Then, when your wish is to have the unit set to a low output yet high current, 3 Volts at 5 Amps, the thing must dissipate 20 Volts (23 Volts in minus 3 Volts out) at 5 Amps, and that's a nice tidy 100 watts. No way you're going to do that with ANY integrated circuit regulator, regardless of the precision of your resistors!

What you need for those ratings is a switcher, or something that will lower the input voltage when you wish a low output voltage. Look at the device dissipation ratings; probably the best is 25 watts with a GOOD heatsink and blower.

For what you're looking for (your ratings) from an IC regulator, you are on the wrong track.

Cheers,

Peter (pjv)

Agent Cobalt

Well, the original design in the Nuts & Volts article only had the supply output max of 1A. Maybe if I stick with that, then the heat won't be as big of an issue? Just wondering, what calculation are you using to get how much heat would be dissipated? Because it seems that it must be pretty common (at least when dealing with voltage regulators) and that I might want to remember it for future reference.

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Agent Cobalt

Well I found something interesting in the Nuts & Volts article (Column #50).

Lon Glazner (in column 50) said...

In my initial design, I was shooting for a digitally-controlled dual power supply with on
output range of 3-20V and about 2A current source capability. A linear regulator, such as
the National Semiconductor LM317, could be used. But if I used a linear regulator,
cooling and heatsinking would be mandatory. For example, an 8Vdc output linear
regulator, providing 500mA of current, with an input voltage of 24Vdc would have to
dissipate (24Vdc-8Vdc)*0.5A = 8 watts. Even with external pass elements, I would have
to account for considerable power dissipation.

On the other hand, a switching power supply would provide me with the efficiency
necessary to minimize power dissipation concerns. It was primarily due to power
dissipation considerations that I selected the MAX726 (Maxim Integrated Products: 1-
800-998-8800 for samples) step-down, PWM, switch-mode DC-DC regulator, as a power
supply. You can approximate the power dissipation requirements for the MAX726 by
multiplying the load current by 1.1Vdc. This is described in the MAX726 data sheet.
Therefore, by using a MAX726 instead of a linear regulator, your 8Vdc supply would
only have to dissipate 1.1Vdc*0.5A = 0.55 watts; which is a significant improvement
over the linear regulator.

I guess that with that little bit of information, I'll stay on track with using the MAX726 instead of the LM317 or the LM1084.
I am new to all of this so I have to go by what people say or what is written in the Nuts & Volts articles.

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pjv

Hi Agent;

The formula for heat (watts) is the voltage drop (Vin minus Vout) times the current passed. So in my example from your requirements, (23 V in - 3 V out)·times 5Amps is 20 * 5 = 100 watts.....quite a bit of heat. (Ever put your hand on a 100 Watt light bulb?)

I encourage you to keep reading and then experimenting on circuitry about what you have read in order to gain a full appreciation of the words. For me, words alone don't seem to complete the understanding process; the experimentation really helps nail it.

Do expect to smoke some components, but that is normal.

Above all, have FUN.

Cheers,

Peter (pjv)

Post Edited (pjv) : 11/20/2005 4:27:55 PM GMT

John Bond

Hi Agent

My very old variable power supply needs to be retired. I've been looking at the LM317/337 datasheets and application notes the last month or so.·- What applies to these would probably apply to other regulators.

• you can use a negative voltage on your control pin to enable an output from 0 volts. This negative voltage will be the same as the regulator's offset voltage.
• With the LM337, you're not limited to a maximum of 15/30/45 volts. its the difference between the output and the control pin. (usefull if you need say 60 volts)
• You can control dissipation by switching in additional windings on your transformer as the voltage increases. Use a multi-tap transformer, a couple of opamps and relays. Remember that the difference between input and output voltages has a direct relationship to dissipation so by doing this, you can up your Amp output.
• One of the application notes have a·good schematic for a PSU that controls both Volts and Amps. I always like to limit amps when I first power up a new circuit.
• I'm·planning to use·an 8 output I2C chip and resistor ladder to give me 255 steps. Im·building a 0..25 volt regulator so the individual steps will be roughly 0.1 Volts. Most digital pots give 100 steps or a 0.25 volt resolution.

I found the applications quite a lot of fun and the LM117/317/337 are simple enough for me to understand what's going on.

Whatever you choose, have fun... and let us know how things work out.

Kind regards from Kwa Dukuza

John Bond