Getting more current from power supply

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grant fair
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Getting more current from power supply

Post by grant fair »

I built the power supply with adjustable voltage and current limiting from TJ Byers' January column. It is basically the same as the circuit on the National Semi website datasheet for the LM317, at:<p>http://www.national.com/ds/LM/LM117.pdf<p>The LM317 and pass transistor I am using are both in TO-3 cases on a big heatsink. The pass transistor (I have substituted a BDX66, rated at 16 amps, instead of the TIP125 in the original circuit, which was rated for 8 amps). But I can only get 4 amps out before current limiting sets in. (The transformer I am using is rated at 8 amps).<p>How can I increase the output current? I would like at least 6 amps to be available.<p>Grant
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grant fair
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Re: Getting more current from power supply

Post by grant fair »

PS: The National Semi circuit is at the top of page 14 of the datasheet at the website above.<p>Grant
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gadgeteer
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Re: Getting more current from power supply

Post by gadgeteer »

Page 13 might be a better circuit. REMEMBER---you can't just PARALLEL transistors---you hafta put a couple tenths of an ohm in EACH emitter lead, to prevent "current hogging".<p>I built a power supply based on a zener reference---simple zener diode and resistor---which is amplified, and sent to a potentiometer---amplified again (with common collector, darlington config). But there's a 0.12 ohm resistor on the OUTPUT, and a small transistor with its base and emitter tied right across the 0.12 resistor (at 5 amps the 0.12 develops the 0.6 volts to turn on the limit-transistor); and the limit-transistor shorts the base of the darlington drive to ground, turning it off. IOW, at 5 amps, the voltage falls off ---current limited. Idiot proof. (Hasta be if I'm gonna be using it...)<p>The problem with linear supplies, is HEAT (that's what is hurting you). I bought a supposedly 6-amp supply from MPJA, but it got way too hot at just over an amp. I removed the transistor, re-mounted it on a thick plate---and on the same plate placed a black finned heat sink. I then drilled holes every 1/4 inch in between all of the fins---and internally mounted a FAN. Unit provides 4-6 amps at 12 volts just fine.<p>ON a variable supply, your HEAT varies with the CURRENT, and the VOLTAGE. If you have 20volts in, and 15 volts regulated out, at ONE AMP, it dissipates only 5 watts of heat. BUT---if you're outputting 5 volts @ 1 amp, then you're dissipating 15 watts of heat!!! Power on your drive transistor is equal to its CURRENT, multiplied by its VOLTAGE DROP...
grant fair
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Re: Getting more current from power supply

Post by grant fair »

I know there are potential problems with heat. But that doesn't lead to current limiting in the pass transistor, does it? It does lead to thermal runaway and smoke release when the transistor fries. In fact the pass transistor does not get too warm to touch at 4 amps, and I am running the voltage at close to max to get the current that high. <p>I first made the thing with a LM317 and TIP125 in TO-220 cases - the pass transistor definitely got too hot at about 2.5 amps and heat sinking did not move enough heat fast enough. So I upgraded to the TO-3 versions and a much bigger heatsink.<p>So I'm not convinced heat is the problem. What am I missing? <p>Grant
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russlk
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Re: Getting more current from power supply

Post by russlk »

The circuit is not current limiting, it is just running out of voltage when the drop across R1 brings the LM317 input voltage down to 1.5 volts above the output voltage. Increasing the source voltage or decreasing R1 should solve the problem.
gadgeteer
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Re: Getting more current from power supply

Post by gadgeteer »

Hmmm---Russ may be right. <p>Also---do you have an oscilloscope? It's nice to view the output---sometimes it is discovered that instead of a supply, we've built an OSCILLATOR...<p>I've never trusted the designed "current limit" in the IC. I once bought a high-power regulator (was it an LM339?); it lasted about 11 seconds. Ten bucks (or more) shot.<p>If I were you I would absolutely use the limit configuration I described. (You can also use the voltage developed across your "limit resistor" to drive a low-current meter---and your supply will show AMPS...)
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Re: Getting more current from power supply

Post by Ron H »

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by Russ Kincaid:
The circuit is not current limiting, it is just running out of voltage when the drop across R1 brings the LM317 input voltage down to 1.5 volts above the output voltage. Increasing the source voltage or decreasing R1 should solve the problem.<hr></blockquote><p>The voltage across R1 is limited by the Vbe of the BDX66 Darlington, which will be about 1.4 volts. This happens at (1.4/33)=42ma, which will be the operating current of the LM317. All additional output current flows through the BDX66. The use of the Darlington booster means you need an extra 1.4v of headroom between Vunreg and Vout, a minimum of about 4.5v.
The current will limit when the voltage across R3 exceeds the voltage across R2. If you go through the math (as I did), you will find that <p>Ilimit=1.25*R2/(R3*R5)<p>With the values shown, maximum Ilimit=4.735A, which is very close to the 5 amp limit stated in the datasheet.<p>What value pot did you use for R2? For a given value of R2, you can pick R5 to give you the 6 amp limit you want. Pots are typically +/-20%. You should measure the actual value.
Remember that R3 will dissipate 7.2 watts at 6 amps. You should use a 10 watt resistor.<p>Ron
grant fair
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Re: Getting more current from power supply

Post by grant fair »

Thanks for both the explanation and the information. R2 is a nominal 250K, measures at 236K. R5=336K; I will look for something > 407K to replace it. I don't have a 10 watt .2 ohm resistor in my junkbox, so its off to the surplus stores tomorrow.<p>I do appreciate the help members have given, this is a very positive and helpful list.<p>Grant
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grant fair
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Re: Getting more current from power supply

Post by grant fair »

Oops - I miscalculated, R5 should be around 240K.<p>I am wondering whether there is any reason not to use R3=.1 (it's just a little easier physically to parallel another .2 ohm R with the first .2R than to use a larger .2R 10 watt).<p>Grant
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Ron H
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Re: Getting more current from power supply

Post by Ron H »

0.1 ohm should work well. For 6 amps max, make R2/R5 = 6/12.5, maximum. If R2=236k, then R5=491k. Use either 470k or 510k if you only have access to 5% resistors. Of course, the max current is (inversely) proportional to R3, so if it is 5%, high, your max current will be about 5% low, and vice-versa. Also, the current through the LM317 is about 50-60ma, which adds to the maximum current.<p>Ron
grant fair
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Re: Getting more current from power supply

Post by grant fair »

I put a 490K (measured) resistor in for R5, but I can only get 5.13 amps out. Interestingly enough, the voltage has decreased to about 20.5 and current limiting has started. The transformer is rated for 21 volts, 8 amps. The rectified DC (unloaded) is about 30 volts before the regulator, ignoring ripple. I assumed that I could count on about a 24 volt regulated maximum out of the supply.<p>When I am getting 5.13 amps out, the unregulated DC is down at 25.5. Maybe the transformer power rating is optimistic? <p>I disconnected the unregulated rectified DC from the supply and ran it into a 4 ohm (1%) load and the voltage was 24.9 with a current of 6.2 amps. (So maybe that explains why the transformer was surplus). With the power losses in the diode bridge, transistor, LM317 and R3 then I guess that 5 amps is as good as it gets with this transformer.<p>I should also mention that in the National Semi datasheet, R1 is 33 ohms, but I used the 10 ohms which TJ Byers has for R1 in his schematic for the circuit. <p>One other point - this transformer has two primary windings, 120 and 105 volts. I have been using the 120 volt one. I can switch to the 105 volt winding for a nominal output of 24 VAC. (In fact I may as well make the two windings switch selectable to reduce power dissipation for use at lower output voltages.) That would give me higher output volts if I wanted them; assuming a dropout of 2 volts for the LM317 I might get 23.5 regulated volts. <p>If I increase the input voltage as above, would I need to make other changes in the circuit? <p>Grant
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grant fair
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Re: Getting more current from power supply

Post by grant fair »

PS: I tried the increase by using the 105 volt primary leads and could get 5.8 amps at 23.5 volts out. I am using 7200 uF in the supply, I need to double this to maintain <=10% ripple, so I will. I think that's about the best I can do with this transformer.<p>Grant
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Ron H
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Re: Getting more current from power supply

Post by Ron H »

Grant, you've pinpointed the component that can probably give you the most improvement. Ripple on the unregulated input will cause regulator dropout on the valleys of that voltage waveform. In theory, you can make that cap arbitrarily large (if you have the space and the bucks). In practice, there are at several non-ideal aspects of the circuit that can cause problems. When you make the cap larger, the rectifier peak current increases, because the conduction angle is smaller. This can destroy the rectifier. This fact also increases voltage drops across the winding resistances of the transformer, and also increases capacitor dissipation due to the equivalent series resistance (ESR). This can damage the capacitor. You can make the cap bigger, but you need to ascertain that all affected components can tolerate the change.<p>Ron
grant fair
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Re: Getting more current from power supply

Post by grant fair »

I'm using a BR-252, a 25 amp bridge rated for 300 amps surge current. The worst case Vpeak (using the 105 volt primary) would be about 46 volts. I don't know the resistance of the transformer secondary. Assuming it is about .2 ohms, then the surge current would be 230.<p>At present I am using a Sprague Powerlytic 7200 uF capacitor for filtering. To ensure a ripple voltage of no more than 10% would require about 12,500uF, but it would be practical to just add another cap to get 14,400 uF. Again assuming a transformer secondary resistance of .2 ohms, the capacitor time constant would be .2 x 0.0144=0.0028, or less than 1 ms. <p>I understand a rule of thumb is that if the surge current is less than the rated max for the bridge, and the capacitor time constant is less than 8.33 ms then the bridge will likely be ok. So I guess I am ok there.<p>I could not find a web site for Sprague (they have been bought out) and the cap is likely 20 years old anyway, so I don't know the rated ESR and have not made the ESR meter TJ Byers outlined in his column a year or so ago - still looking for the right size log slider pot. I might have to trade off less ripple for less current - I can just try adding the second cap and see how the current output holds up, no?<p>Thanks for helping me get this far with the power supply. Its been working fine for one of its intended uses (testing steppers).<p>Grant
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Ron H
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Re: Getting more current from power supply

Post by Ron H »

Hi Grant,
I think you misunderstand the reason for repetitive peak current. Each rectifier diode only conducts when it is forward biased. Looking at the two diodes that connect to the unregulated output, this is when the sine wave on the anode is more positive than the output, i.e., during the positive slope of the ripple. All the charge that drains out of the cap must be replaced during this time.If the peak current had a rectangular waveshape, the value would be<p>Ipeak=Iload*T/Tcond<p>where T=8.33ms in the USA, and Tcond is the conduction time. As the cap is increased, the ripple decreases, so Tcond decreases, increasing peak current. In practice, the peak will higher than the value calculated by the above equation, possibly as much as twice as high, because the waveshape is far from rectangular.
Watch out for old capacitors. 20 years may be OK. If it's working, I guess that's the proof of the pudding, as they say. ESR may go up with age.<p>Good luck-
Ron
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