• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Rebuilding My Power Supply

Status
Not open for further replies.

MrAl

Well-Known Member
Most Helpful Member
The data sheet states that Adjustment Pin Current is typically 50 and max 100 micro Amps. 1.25 / 100 micro amps = 12.5K, so 500 is well within the operating area. 120 and 240 ohms are convenient standards, but not the rule, IMO. I don't suppose you would believe I have a very rare zener LED would you? No? OK you're right, It's drawn in backwards, Thanks.
EDIT I see you strongly believe in 120 ohms for R1, Think you'll ever change your mind and accept that it can be something else? http://www.electro-tech-online.com/threads/lm317-calculating.119697/

Yes, I did a little work with 400Hz inverters, the theory behind 400 Hz is quite interesting, if I remember right, the ideal frequency for motors is 380 Hz, so they rounded it up to 400 Hz, a motor running at 400 Hz can produce 3 time the HP of the same size and weight motor running at 60 Hz.

The negative side won't get used much and just reducing the ripple will be ok for what I do. I will probably just set the voltage around 16 and forget about it.

Hi,

Just so you know, the minimum resistance on the output is NOT based on the adjustment pin current. It is based on the minimum OUTPUT current that should be drawn from the device so that it maintains regulation. Below a certain OUTPUT current (not adjustment pin current) the output could go up somewhat and that means the output voltage is no longer regulated. That's why they always show a somewhat low value like 220 ohm at the output of the LM317 regulators. But 100 ohms is better because it covers more devices under more operating conditions, so that's why such a low value is recommended.

There is one time when this is not necessary, but you have to be careful here. This is when the output will ALWAYS be loaded by the following circuit at more than something like 10ma. If your circuit always draws 10ma then you can use other resistor values that are somewhat higher like you wanted to do, but you have to be very careful here because the following circuit may NORMALLY draw 10ma but during startup it may only draw 1ma for example, and if that happens then during startup the following circuit will receive an over voltage condition which could burn it out and then it wont work at all.

The 100 ohmish resistor is chosen to always draw around 10ma or so and that keeps the min output current spec happy regardless of the following circuit current draw or how that current changes over the operating time of the product.

About LED's and zeners...

LED's do act like crummy zeners because their operating curve looks a little like zeners. I used an LED as a zener in a circuit long time ago when i realized that powering an LED from an unknown voltage source that can vary quite a bit would change the brightness of the LED quite a bit when using just a series resistor alone.

The LED acts a lot like a zener as it does 'regulate' the voltage to some degree when the current is limited to something the LED can easily handle. But when done it is not connected backwards, it is connected as a normal LED.
 
Last edited:

dr pepper

Well-Known Member
Most Helpful Member
Yes of course leds make naff zeners, this was indeed a crummy supply, and it was part of the I limit not the voltage regulation, like I said a pair of trannys acting as a current source would be a lot better.

I have a comercial psu that blows up a pic micro if you connect it to it on power up even with the o/p set to 5v, I wonder if a load resistor will cure that, I allways thought it was due to the pass trans being a PNP and the base being driven on by being pulled to ground through a resistor, my thinking was that it took a while for the circuit to remove the base drive and regulate the o/p causing a spike, but maybe its just the fact there isnt a load, theres just a cap on the o/p.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
And I thought aircraft used 400hz because it was the speed at which the engine turned the alty!
As far as I'm aware aircraft used 400Hz because it makes everything electrical smaller and lighter - such as smaller transformers and smoothing capacitors - presumably the same holds for electrical motors? (for the same reasons).
 

dr pepper

Well-Known Member
Most Helpful Member
Makes sense.
Smaller soothing capacitors.
 

audioguru

Well-Known Member
Most Helpful Member
The data sheet states that Adjustment Pin Current is typically 50 and max 100 micro Amps. 1.25 / 100 micro amps = 12.5K, so 500 is well within the operating area.

I am not talking about the tiny ADJ pin current. I am talking about the current for the entire IC that flows through the output in order to make the ADJ pin current very small. The datasheet spec's the minimum load current at 10mA (5mA for the more expensive LM150) which can be provided by using 1.25V/10mA= 125 ohms or less from the output to the ADJ pin.
The datasheet says, "If there is insufficient load on the output, the output (voltage) will rise."
So you should do it correctly.
 

kinarfi

Well-Known Member
I tried to check my info about 380 Hz being ideal, but couldn't find anything, but did find that the speed of the motor has a huge effect of its output. google 400hz.
another nice thing about using LEDs like zeners is they can handle a lot more current and you can see if they're working by how much light they're putting out.
 

audioguru

Well-Known Member
Most Helpful Member
The power dissipation (heating) of an LED bugs me.
A white 1W LED has a voltage of about 3.5V and a current of 285mA. Then it heats with 3.5V x 285mA= 0.998W correct?
But since it is releasing A LOT of light then isn't its heating dissipation less? Light is energy isn't it?

If it didn't get hot and didn't even get warm then wouldn't it get cooler when it lights up?
 

kinarfi

Well-Known Member
If
I am not talking about the tiny ADJ pin current. I am talking about the current for the entire IC that flows through the output in order to make the ADJ pin current very small. The datasheet spec's the minimum load current at 10mA (5mA for the more expensive LM150) which can be provided by using 1.25V/10mA= 125 ohms or less from the output to the ADJ pin.
The datasheet says, "If there is insufficient load on the output, the output (voltage) will rise."
So you should do it correctly.
If you are NOT going to have of a load that draws the necessary 10ma current, then yes, use bias current to get the necessary 10ma current draw via R1 & R2, but if your load is drawing ,say 1400ma, then the extra 10 ma through R1 & R2 only pushes you closer to the max current for the LM317 and does nothing extra for the regulation bias set point where a 1K R1 and 7K R2 will work just fine for a 10 v output, plus if you have a 30 volt supply and 10 v out the 10.4 ma using R1=120 & R2 = 840 use 104 mw in the resistors and 208 mw is added to the heat of the LM317, where the use of 1K & 8k produce 11.25 mw in the resistor and 22.5 mw in the LM317 while still providing very adequate voltage regulation. I believe your confusing bias current with load current.
 

audioguru

Well-Known Member
Most Helpful Member
The LM3xx adjustable voltage regulator IC needs a minimum load current of 10mA for it to regulate the output voltage. With a load current of less than 10mA then the output voltage of some of them will rise and not be regulated.
You have 500 ohms from the output to the ADJ pin that has 1.25V across it so it draws a current of only 1.25V/500 ohms= 2.5mA so the output voltage of your project will rise if it has no load or if it is powering a low current Cmos circuit.

It is best to use 10mA from the output to the ADJ pin (then through the pot to ground) so that a very high power low resistance is not needed at the output to ground of the project.
 

kinarfi

Well-Known Member
The LM3xx adjustable voltage regulator IC needs a minimum load current of 10mA for it to regulate the output voltage. With a load current of less than 10mA then the output voltage of some of them will rise and not be regulated.
You have 500 ohms from the output to the ADJ pin that has 1.25V across it so it draws a current of only 1.25V/500 ohms= 2.5mA so the output voltage of your project will rise if it has no load or if it is powering a low current Cmos circuit.

It is best to use 10mA from the output to the ADJ pin (then through the pot to ground) so that a very high power low resistance is not needed at the output to ground of the project.
I agree to disagree with you and shall leave it at that.
 

MrAl

Well-Known Member
Most Helpful Member
The power dissipation (heating) of an LED bugs me.
A white 1W LED has a voltage of about 3.5V and a current of 285mA. Then it heats with 3.5V x 285mA= 0.998W correct?
But since it is releasing A LOT of light then isn't its heating dissipation less? Light is energy isn't it?

If it didn't get hot and didn't even get warm then wouldn't it get cooler when it lights up?

Hi,

In today's world there is always some loss of power associated with producing light. The LED is more efficient then some other forms of lighting, but it still has it's internal resistive part and that dissipates energy as heat. So there is some radiant visible light energy being released but there is still some heat because of that resistive part.

If it was 100 percent efficient in converting electrical energy to light energy then it would not get warmer, but it would never get cooler because that would require absorbing heat energy from the surroundings and converting that to light too. There have been experiments that show that this can be done (absorb heat, convert to light), but it's on a very small scale and the LED's were not 100 percent efficient anyway so they end up absorbing some of their own heat. That makes them a little more efficient but they still produce heat along with the light.
 

MrAl

Well-Known Member
Most Helpful Member
I agree to disagree with you and shall leave it at that.
Hi,

I think what audioguru is saying is that if you do not use that 100 Ohm ish resistor then you risk not being able to turn the voltage down low when you need too if the load resistance is some fixed value like 500 Ohms. Lets look at what happens with a 500 Ohm load and a high value upper 'feedback' resistor for the LM317...

With 500 ohm load and the power supply set for 10 volts, the output current is 10/500=20ma and 20ma is enough current so the LM317 is doing just fine at regulating the output at 10 volts.
Now turn the voltage down to 5 volts. The 500 ohm resistor then draws 5/500=10ma and the LM317 is still doing fine because the current is around 10ma.
But now turn the voltage down to 2 volts. That means with a 500 ohm load the output current is now only 2/500=4ma, which is considered too low to be sure we can get adequate regulation. This means that the voltage may NOT ACTUALLY GO DOWN TO THAT 2 VOLTS, but may stay up higher like 2.5v or maybe even higher. So what this means is that we can not adjust the power supply down to the normal lower limit of 1.25v anymore.
Now change the feedback resistor to 100 ohms, and suddenly everything works normally again. We can easily adjust the output down to 1.25v and we get 1.25/500=2.5ma output current.

Also, the reason we were talking about this is because of your reply that i quoted in post #21 where you talked about the bias current not the load current. The effect of the bias current is another issue. That parameter is needed to calculate the approximate output voltage fluctuation with temperature. If the 'feedback' resistor is too large then the temperature effects are more severe. But again this is a different issue not about the voltage regulation vs output load current.
 

audioguru

Well-Known Member
Most Helpful Member
When the resistor from the output to the ADJ pin is the only load (through the pot to ground) then a 120 ohm resistor draws 1.25V/120 ohms= 10.4mA at ANY output voltage setting so the voltage will always be regulated and the resistor stays cool, dissipating only 13mW.

Imagine using a 120 ohm resistor at the output to ground as the minimum load. When the output voltage is set to 1.25V then the resistor draws 10.4mA and regulation is fine.
Then turn up the output voltage to maybe 30V then the current in the resistor is 300mA and it is frying at 9W.
 

kinarfi

Well-Known Member
Here's my finished design, seems to be working very well, but I haven't had it finished long enough to make possible mistakes that could damage it yet. Very simple to build, if you have the heat sink and drill and tap to mount the components, started with 20 amp, 13.6v powers supply and made it into a + and - 20 volt supply that can handle a combined 10 amps with out getting a ripple on the out put.
 

Attachments

MrAl

Well-Known Member
Most Helpful Member
Hi,

Well are you saying that you will always have a load of 1 ohm at the output?

Because of the extra pass transistor you should still check the current out of the LM317. If there is less than 10ma it may not work right at low voltages, less than 5ma is even worse although many LM317's do work with only 5ma output.
 

kinarfi

Well-Known Member
Hi,

Well are you saying that you will always have a load of 1 ohm at the output?

Because of the extra pass transistor you should still check the current out of the LM317. If there is less than 10ma it may not work right at low voltages, less than 5ma is even worse although many LM317's do work with only 5ma output.
NO, I just forgot to remove it from my spice file before I copied it and posted it. I'm well aware of every ones concerns about lightly loading the LM317 & LM337, so I will be watching it closely to see if it causes any problems. Later to day I'll hook up my Fluke to the output, set it to 3 volts and have it record min max for an hour or so. got any suggestions?
 

audioguru

Well-Known Member
Most Helpful Member
The LM317 should ALWAYS have a load of 10mA because if you need to replace it the replacement might NEED 10mA even if the one you are using now doesn't need this as its load.
The resistor from the output to the ADJ pin should be 120 ohms for an LM317 and can be 240 ohms for a more expensive LM117. Then this resistor with 10.4mA in it stays cool even if the output voltage is increased. Of course the voltage setting pot must have its value work with the 120 ohm resistor.
 

large_ghostman

Well-Known Member
Most Helpful Member
couldnt you just use an indicator led? that would give you 10-20mA
 

MrAl

Well-Known Member
Most Helpful Member
NO, I just forgot to remove it from my spice file before I copied it and posted it. I'm well aware of every ones concerns about lightly loading the LM317 & LM337, so I will be watching it closely to see if it causes any problems. Later to day I'll hook up my Fluke to the output, set it to 3 volts and have it record min max for an hour or so. got any suggestions?
Hi,

kinarfi:
Yes i have a suggestion, use a 120 ohm resistor <smile>.
Anything else and you really have to check the output over the full temperature range that you intend to use this at, and that could be difficult.
What you could test more easily though is try to find out what an even higher value does, and maybe find the limit at your ambient temperature. This would be interesting to all of us. Try 120, then 240, then 500, then 1k, then 2k, etc. See what happens to the output when you try to adjust all the way down to 1.25 volts. That will show you first hand what happens when the device does not have enough output current all the time.
But please also read below to see the secondary detrimental effect of using a higher value for the normal 120 ohm resistor.

LG:
Yes that's a very good idea. There is a catch though, that is if the output voltage is adjusted to a value that is too close to the characteristic voltage of the LED, the LED stops conducting, and that means we're back to zero output current. If the LED is connected in series with the 120 ohm resistor, then the LED will never conduct or will conduct very little if it is a standard red one. I dont think there is any easy way around this problem.

The reason why the 120 ohm resistor can always cause a 10ma output current is because of exactly where it is placed in the circuit. In the place where it is connected the LM317 maintains approximately 1.25 volts across it, and so it always draws 10 ma or thereabouts. That's why this connection is the best way to do it. If it is connected to the output, then it draws too much current when we raise the output voltage, or at the least it wastes too much of the available current.

TEMPERATURE CONSIDERATIONS
When the output is constant that is a different story, but then there is another issue that comes up, the issue about the temperature dependence of the output voltage. With a constant 120 ohm load we do in fact satisfy the output current requirement assuming we are using the simple typical connections for this device. But we still have to have another resistor in place of where the 120 ohm resistor normally goes. So say we use a 1200 ohm resistor, which allows us to use a larger lower value resistor too. The output should regulate just fine this way, but there's another catch here. The adjustment pin current varies over temperature, and that variation causes a voltage drop in the lower placed resistor (not the 1200 ohm one). With the adjustment pin current changing with temperature, that means the voltage divider voltage changes with temperature. And because any fixed change in current causes a LARGER change of voltage in a higher value resistor, the higher the resistor value the MORE the output varies over temperature. So what we end up with roughly is for a 10 times change in the feedback resistor (120 to 1200 ohms) we'll get an output voltage temperature sensitivity of roughly 10 times, which is much worse.
So although that 120 ohm load with 1200 ohm feedback resistor and maybe 12k lower resistor (or whatever we need) still regulates, we now see the output fluctuate much more over any given temperature range.
So that's two reasons why it is better to use a 120 ohm resistor as the 'feedback' resistor. Of course lower is better, but we dont want to eat up too much current if we dont have to.
 

audioguru

Well-Known Member
Most Helpful Member
couldnt you just use an indicator led? that would give you 10-20mA
Where would you connect the LED?
Not at the output because the output voltage can be turned down lower than the voltage of LEDs. Also it would need a power current regulator.
Not between the output and the ADJ pin because only a resistor is needed there to help set the output voltage.
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top