Voltage Drop of Regulator Output

Greeetings:
I am trying to build a +5V regulator using a LM350K. I am using a 28V 5A transformer and a two diode (GI750) full wave rectifier. I get
about 10.4V No Load voltage at the filter cap (10,000uf) I am using 120 Ohm resistor for R1 and a 180 Ohm resistor plus a 500 Ohm Pot. for R2. Those are the only components used in the regulator.
I have noticed that the regulator handles variations on the input side of it well. However, the output voltage varies significantly according to the resistance of the load. If I adjust the regulator to provide ~5.25V at no load, the voltage will then drop to 4.66 when the load is a 1.4 Ohm resistor that results in 3.4 amps of output current.
The voltage drop apparently is inversely proportional in a virtually linear fashion to the output current. Here are links to a jpg's of the ciruit and performance:
www.geocities.com/cybervagrant/misc/circuit.jpg www.geocities.com/cybervagrant/misc/performance.jpg
Does anybody know how to stablize the output voltage with regard to output current?
BTW I thought this device might be malfunctioning so I substutited a LM150K and got the same results. So, I concluded this is just the way the devices functions in this circuit.
Additionally, the heat sink I am using is rated at 5.5 degrees per watt. So the thermal resistance of the whole assembly is about 8 degrees per watt.
Best Regards,
CV
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I assume you've connected the 120 ohm resistor directly to the case of the LM350K.
The spec sheet I have suggests that the output current limit could be as low as 3 amps for Vin - Vout less than or equal to 10 volts. What kind of load regulation do you get with, say, 2 amps output?
Chuck
snipped-for-privacy@yahoo.com wrote:

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I used a solder lug provided in the TO-3 mounting kit I bought to connect R1 to the case. It's like a washer on one of the case bolts. I'm using a ring terminal on the other bolt to connect the output.
CV
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A differential voltage of 10V with a 3A output current would result in 30 watts of dissapation for the device. That would activate the thermal protection features very quickly. With my setup at room temprature 12W will max out the dissipation of the device. That means less than 4V differential at 3A of output.
According to the chart I posted the link for at 2A out the voltage is approximately 4.92. The variation in differential voltage is less linear. So, I would have to make a measurement to get an accrurate figure, but I figure around 3.85-3.90. The voltage regulation would be (5.25-4.92)/4.92 or about 7 percent.
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substitute an appropriately sized zener diode in place of r2 and r3.

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By r2 and r3 I presume you mean the combination of the resistor, r2a and the pot., r2b, I used for r2 in the typical circuit provided by NSC.
That would be easy enough. If the internal voltage reference is working the way it's supposed to, then there should be 1.25V +/- 0.05 across r1 between the adjment pin and the case, so a 3.75V zener would seem to be appropriate. I could adjust the value of r1 to provide a current suitable for the zener.
It would be convient, if it works.
Regards,
CV
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snipped-for-privacy@yahoo.com wrote:

Either the measurements are wrong, or there is a bad part somewhere. With a 28 volt center-tapped transformer, something has to be greatly in error for Vin to measure 10.4 volts. You should see about 20 volts across C1. (For reliability you should make that a 35 volt cap.) The delta across the 350 should be about 15 volts, no load. You need to solve that problem first.
Next, you mention drawing 3.4 amps from the circuit. The max for the LM350 is 3 amps. To go over 3 amps, you need a pass transistor. You also need to increase C1 for 3 amps, say about 3 times the value.
With your present circuit, the LM340 will need to dissipate close to 45 watts when 3 amps is drawn. Way too much. So this is what you need to do,*after* solving the 10.4 volt problem:
PNP Vcc ~+20 ---+---------- ----------------+ | e\ /c | | --- | | | ---------- | +---[R1]-----+---|in 350 out|---+--- Vout 5V | Adj | | ---------- [R3] 120 R1 = 1 ohm, 5 watt | | +---------+ | [R4] 180 | [POT] 500 | Gnd ---------------------------------+-------------- Gnd
The pass transistor will begin to take on some of work when the load draws about 700 mA. You need to heat sink both the regulator and the pass transistor, and your existing sink isn't big enough for both, but will work for the 350.
Ed
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With a 28 volt center-tapped transformer, something has to be greatly in error for Vin to measure 10.4 volts.<<<
Cheers, I was waiting for some one to catch my ommission there:). I was hesitant to triple post. I'm putting 115 into the 230 terminals of the transformer's primary.
According to NSC the 3A figure is garrenteed minimum. The data sheet allows for up to a 4.5A max.
The data sheet also shows a good plateau in performance with a differential voltage between ~4V and ~12V. A differential of 15V would cause too much dissipation. My goal was to be able to hold 5V at 3A indefinately, although I will now settle for 2.5A.
The LM323A is the regulator I wanted to use for this application, but I would have had to order it. Anyway, the data sheet for that one states a mininum differential voltage of 2.5V is required. So, in my case here, 3.5V seems sufficient.
The output current at the moment is actually mimimally sufficient. It's the load regulation that is the problem. I will require 1.0A to 2.6A from the regulator with at least 2% regulation and a 4.75V minimum. This setup will actually serve the purpose at hand, if I keep the load between 1 and ~2.2 amps. It just falls way short of the voltage regulation with respect to load variations that I expected from such a device.
The voltage regulation with respect to input variations seems to be outstanding.
Regards,
CV
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snipped-for-privacy@yahoo.com wrote:

Ok - so you are using a 7 volt AC input to get 5 volts DC out. At 3+ amps. With 10000 uf. That rings the alarm bells for me. Your C1 computes to ~8290 using the values you measured. Vripple (or sag if you prefer) into the 350 will be (ImA*8.3mS)/Cuf; at 3 amps it's 3000*8.3/829 or 3 volts. You'll drop at least .6v in the rectifiers, so the solid (ripple free) dc level at the 350 input will be 6.8 volts (10.4 - .6v diode drop - 3v ripple). Not enough overhead for solid regulation at 3 amps. Triple the value of C1.

But you are going beyond the 3 amp maximum guaranteed. After 3 amps, you can't expect the 350 to maintain its regulation specs. It might, or it might not. The performance curve does not show values over 3 amps. In addition, the performance degrades with heat, and the more current drawn, the hotter it gets.

You can get 5V indefinately at 3A - just keep the Vin 3 volts or more than Vout, and keep the chip cool. But you are running at max, so regulation will not be as good as it would be at lower current. It still should be adequate, but a nice pass transistor circuit - 1 transistor, 1 resistor, and a heatsink - puts the LM350 in a better place: lower heat and lower current, therefore better regulation.
Ed

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My design goal was to be able to draw 3A, but I'll settle for 2.5A, which this regulator will do, but then no-load voltage rises to 5.27V, and that far exceeds the claimed load regulation ability of the device. So, there is obviously a problem or mistake somewhere.
--
Your C1 computes to ~8290 using the values you measured.
Vripple (or sag if you prefer) into the 350 will be
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snipped-for-privacy@yahoo.com wrote:

That's close to the computed 6.8 volts at 3 amps. Your graph shows 8.22 volts average vin with the 3.4 amp load - already 2.18 volts below the 10.4 no load level. With the 1.6v sag, it drops to 6.62 v during the sag. That sag occurs twice per cycle, so the 350 drops out of regulation twice each cycle. Total sag by computation is V = 3400ma * 8.3ms/8290uf which equals 3.4 volts. Total sag by measurement = 2.18 + 1.6 or 3.78 volts. At 3 amps or more, the 350 needs over 2.5 volts headroom.

Great. I think you'll see a lot better regulation, once the Vin is stiffened with 30,000uf (or more).
Ed

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I was also thinking of adding a second transformer in parallel with the first :) I put a shunt between the diodes and the center tap, then scoped that. It appears that I'm getting about 13.5A peak current in the charging pulses. Two transformers and four caps, that ought to work well, huh?
CV
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snipped-for-privacy@yahoo.com wrote:

No - you are shorting the transformer winding through the diode and the shunt. Nothing to do with the charging current.
--(||)----[D1]-----+ (||) | (||)---[SHUNT]---+ (||) | --(||)----[D2]-----+
Two transformers and four caps, that ought to

You have a 5 amp transformer - you're drawing only 3.4 amps. Why would you need to add another transformer for more current? Put your meter on the output of the transformer and measure the no load and full load (3.4 amps drawn from the 5 V supply) voltage. What readings do you get?
Ed
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My bad, the shunt went between the center tap of the secondary and the star ground on the chassis, where everything is grounded: No-load there was a flat line; at 3.4 amps out there were 120Hz negative only pulses of about 130mV to 140mV peaks, which equates to 13A to 14A peak currents with my 50mV per 5A shunt. I attribute the 10mV variations to differences in the diodes and in the seperate halves of the secondary windings.
After reading about Schade's work it occured to me that parallelling two transformers would half the series resistance of the secondary windings. Adding more caps in parallel would similarly reduce the ESR factor for the caps. Both of these action would increase the peak instanteous current in the charging pulses, which is only limited by the series resistance of the rectifier circuit. So, instead of 13.5A peak I might get 20A+ with two transformers. I also have some low forward drop schottky diodes (1N6096) that I could try. More peak current, means more energy transfered to the caps each cycle.
But, I must say that I doubt just adding more caps or transformers is going to flaten the performance graph of my regulator. As far as that goes, for a regulator that is doing it's job I would expect a flat line until the point where either the theremal protection features are activated or as you have suggeseted the instanteous voltage at the input drops below a required minimum.
In any case, the performance of my regulator drops linearly right from the start, way before either one of those conditions would apply. This leads me to believe that at the least different external circuitry is required to get the performance I want. I might just toss these regulators out with the trash, and use discrete components instead. At least then I could control all aspects of the regulation.
This circuit is so simple, the data sheet claims something like 0.1% load regulation, and I am surprised that everybody is clueless about why I getting such a voltage drop. It's like I'm the only person who ever built one of these things and ran in to this result.
OK, I feel better now, I've ranted enough :)
Regards,
CV
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I am very pleased to announce that I have identified and solved the problem!!!
It was the circuit breaker on the output side of the regulator that was causing up to a 0.5V drop in the output voltage. I removed the breaker... et voila, il fonctionne bien!
Here's the address for the chart of the successful performance:
www.geocities.com/cybervagrant/misc/success.jpg
Now that's what I call regulation :)
Note: Even after I removed the old Airpax circuit breaker, I still had 43mV drop between the regulators case and the output terminal. So, I moved the ring termainal from under the perf board to between the top of the regulator and the head of the attachent bolt to let it have direct contact with the case. There is now zero mV drop between the case and the output terminal of my regulator. I'm going to do the same for the solder lug that connects R1 to the case.
I love the enjoyment of sweet success. Although, I still have yet to learn how to bang my head against the wall less hard in the process of achieving it. Where is that bottle of ibuprofen anyway :)
Best regards,
CV
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