[tiger762]

Being the benevolent tiger that I am, I felt the need to share some of my experience in the art of DC power supply design, on a budget.

A friend of mine who works in an industrial setting brought home some 3-phase motors, transformers, and meters that were being scrapped. He asked if I had any use for them. I asked him if the bear takes a dump in the woods. Among the spoils were some 277/120 and 480/120 transformers. What these do is take 480 volt (typical delta connected 3-phase voltage) or 277 volt (typical wye-connected 3-phase voltage) and steps it down to a safer-to-handle 120 volts. The secondary of these transformers had been connected to some Gould/Crompton meters to passively monitor the voltages present throughout the plant.

When you see 277-to-120, just think of it as a 2.31:1 step-down transformer. Some of the transformers he gave me had big "288" numerals on the sides of them. These are 2.4:1 step-down reduction.

I set about to test the transformers this morning by crimping a couple of screw lug terminals to the end of an old appliance cord. I attached the 120 VAC cord to the PRIMARY of the transformer (which normally would be receiving 277 or 480 volts) and measured the output voltage. The 480-to-120 transformers were indeed generating 30.1 VAC at the secondary, as expected.

The 277-to-120 transformers were generating 56 VAC at their secondaries. This seemed a few volts too high, as if the ratio of windings was off a bit. The 288-to-120's were generating 50.6VAC.

With these available AC voltages, I figured that using them to make rectified DC yields a DC voltage approximately 1.4142 times higher. That is the square root of 2.

Here's a picture of one of the low-power (25 VA) 277-to-120 transformers, whose secondary is connected to the "AC" leads of a 200V/2amp full-wave bridge rectifier, which in turn has its "+" and "-" leads screwed down to the corresponding leads of a 26,000uF 80WVDC electrolytic capacitor. This here is the simplest linear power supply you can make. A transformer, rectifier, and capacitor. The measured secondary AC voltage was 56VAC. The measured DC voltage at the capacitor was 76VDC. As stated above, 25 volt-amps isn't enough to do anything really useful with. It's just to illustrate the principle.

For my setup, I have 80V Geckos and 75V servos. A 72VDC power supply would be perfect. To achieve that, I would use one of the 288-to-120 step-down transformers, to get 2.4:1 reduction in AC voltage. Wall voltage being about 121 volts at my house, divided by 2.4, yields a secondary voltage of 50.4VAC, which once rectified and filtered, should give about 71 volts DC. Perfect!

Check out the picture here. The machinery it is sitting on is my air compressor in my garage. It was the only available free space at the moment That's sad, isn't it?

Enjoy!!!
http://i7.tinypic.com/525vmna.jpg

[gar]

070528-2103 EST USA

Your transformer looks quite small. Thus, in normal use it has a low VA rating. The current rating remains the same no matter what the input voltage is up to saturation. When you use this at a lower input voltage than its rating, then VA rating at reduced voltage = Original VA * Lower Voltage / Original Voltage rating.

These transformers are probably called potential transformers.

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[tiger762]

Yes, I have no plans to drive a servo (or anything) with this transformer. I have a 277-to-120 2.0KVA on order to replace the transformer I am using now. The one I'm using now is a microwave oven transformer with a hand wound secondary winding to give 50VAC.

You're right, on some of the transformers they indeed have "potential transformer" on them. I have five 288-to-120 transformers that have 150VA capacity, and three 480-to-120 that have 500VA capacity.

070528-2103 EST USA

Your transformer looks quite small. Thus, in normal use it has a low VA rating. The current rating remains the same no matter what the input voltage is up to saturation. When you use this at a lower input voltage than its rating, then VA rating at reduced voltage = Original VA * Lower Voltage / Original Voltage rating.

These transformers are probably called potential transformers.

.

[gar]

070529-0656 EST USA

Tiger:

Sorry I did not read your original post closely. However, do note your microwave transformer with a 277 V primary at 2 KVA will be good for about 0.86 KVA when connected to 120 instead of 277. If there is a magnetic shunt on the transformer it should be removed.

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[tiger762]

Actually this is a GE transformer, catalog number 9T51BO192, that I just "won" on ebay for $10. It weighs 50lbs

All the microwave oven transformers I've messed with indeed had a magnetic shunt, for safety reasons I assume, and yes I punched those out with a hammer and drive pin punch.

OK, so what you're saying is that there will be a 7.2 amp draw from the primary, no matter what the primary voltage is? I took 2000 VA divided by 277 V to get a rough idea of what the transformer will pull when running at 277 V. So multiplying by 120 gave me 866 VA, which is what you posted.

So the transformer's capacity comes from how good a magnetic conductor it is, and not as much on its windings' resistance.

0.86KVA will still get me almost 17 amps at the secondary. More than enough for my four motors.

070529-0656 EST USA

Tiger:

Sorry I did not read your original post closely. However, do note your microwave transformer with a 277 V primary at 2 KVA will be good for about 0.86 KVA when connected to 120 instead of 277. If there is a magnetic shunt on the transformer it should be removed.

.

[gar]

070529-0805 EST USA

Tiger:

The input current to a transformer is a function of three things. First, one component is a direct reflection of the load current but modified by the turns ratio. Second, is the reactive magnetizing current (current thru the shunt inductance in the equivalent circuit). And, third, is all current relating to power losses in the transformer.

Normally the magnetizing and power loss currents are small, under 10%, of full rated current.

The rating of a transformer is based on temperature within the transformer. Much of the temperature rise at full load is from power loss in the resistance of the wire from the current flow. Thus, for the most part the rating of the transformer is current rather than VA.

Roughly your transformer is good for 2000/277 = 7.2 A in the primary for any input voltage to the primary up to its rated voltage of 277 V. You will not draw 7.2 A unless you apply an appropriate load on the secondary(ies).

The power supply you are building will be good for about 0.86 KVA at the DC output using a bridge rectifier and a capacitor input filter. If your new secondary winding produces a DC voltage of 70 V, then your maximum load current would be about 860/70 = 12 A.

However, was the original 2 KVA rating a continuous duty rating? This factor and whether or not your load is continuous may determine whether you can load the transformer to 12 A. It could be less if your need is a continuous 12 A and the original duty cycle was not 100%. On the other hand if you only need short bursts of high current, like during acceleration, and these are maybe a couple seconds long and do not occur too often, then maybe you can greatly overload relative to 12 A.

.

[gar]

070529-1943 EST USA

Tiger:

What is the steady state DC current that you require with no movement with a 70 V source?

Have you found any indication that the transformer is rated 2 KVA for continuous duty?

.

[tiger762]

I'm not sure what the quiescent current draw is, but the servos draw 7 amps maximum.

I don't know yet what the service factor of the transformer is.

070529-1943 EST USA

Tiger:

What is the steady state DC current that you require with no movement with a 70 V source?

Have you found any indication that the transformer is rated 2 KVA for continuous duty?

.

[gar]

070529-2126 EST USA

Tiger:

When you get your transformer there are two useful experiments to run.

First, measure the internal resistance of the DC output. This not linear, but not too far in deviation.

At this time take measurements of voltage vs load current. Use digital meters, 4 digit accuracy for the voltage at least. Current is less critical so a 3 digit or analog meter.

Pick load currents of about 0, 3, 6, 9, 15, 20 A. Exact values are not necessary you just need to know the voltage at a measured current, and we want points distributed over the desired range.

It is assumed that your diode bridge and filter capacitor are adequate for the maximum load. The slope of the curve at any point is the internal resistance near that point.

Second, measure the average internal temperature rise of the primary coil at 12 A. If this is not excessve then try 15 A, and maybe 20 A if it is unlikely to burn out the transformer. All this assumes your secondary won't be the maximum rise.

Allow the transformer to stablize at room temperature for maybe an hour or more. My wild guess for the 277 V 2 KVA primary resistance is 1 to 2 ohms.

Use a 4 terminal method to measure the resistance. Here accuracy of both current measurement and voltage is very important. I would use about 1/2 A for the test current. The temperature coefficient of copper is 0.00393 per deg C. Suppose the maximum average rise you will considerable acceptable is 50 deg C, then the % change in resistance is 0.00393 * 50 / 100 = 19.65%. If your initial room temperature resistance was 1.75 ohms, then 1.75 * 1.197 = 2.095 ohms. You may need to allow an hour or more under load for the resistance reading to stablize.

You apply AC input with your test load foir a while, possibly 15 miutes (maybe make this shorter at first to get an estimate of how fast temperature rises), remove the AC and connect your 1/2 A source to the primary and measure the voltage. Keep this resistance measurement time as short as possible. Reapply AC input. Continue until virtually no change between measurements. Your experience will tell you how to adjust timing intervals.

.

[tiger762]

070529-2126 EST USA

Tiger:

When you get your transformer there are two useful experiments to run.

First, measure the internal resistance of the DC output. This not linear, but not too far in deviation.

Do you mean just measure the resistance of the secondary windings?

At this time take measurements of voltage vs load current. Use digital meters, 4 digit accuracy for the voltage at least. Current is less critical so a 3 digit or analog meter.

Pick load currents of about 0, 3, 6, 9, 15, 20 A. Exact values are not necessary you just need to know the voltage at a measured current, and we want points distributed over the desired range.

Do you have a recommendation for a DC load? I had thought about using a length of rubber tubing (like surgical tubing), filling it with tap water, and threading bolts into each end. Then measuring the resistance. Adjusting the length and/or salinity of the water as necessary. I figure this should be able to dissipate vast amounts of heat without changing temperature (and thus resistance) appreciably.

It is assumed that your diode bridge and filter capacitor are adequate for the maximum load. The slope of the curve at any point is the internal resistance near that point.

Yes, a 26,000 uF electrolytic and a 200PIV/50amp full-wave bridge.

Second, measure the average internal temperature rise of the primary coil at 12 A. If this is not excessve then try 15 A, and maybe 20 A if it is unlikely to burn out the transformer. All this assumes your secondary won't be the maximum rise.

Allow the transformer to stablize at room temperature for maybe an hour or more. My wild guess for the 277 V 2 KVA primary resistance is 1 to 2 ohms.

The biggest problem I'll have is coming up with a DC load of constant and kn own impedance.

[gar]

070530-1220 EST USA

Tiger:

The equivalent internal resistance of the power supply when viewed from the two output terminals is a composite of a number of items. These being the primary resistance referred to the secondary, turns ratio squared, leakage inductance, secondary resistance, diode resistance, quite non-linear at low current, and maybe other minor things. Part of the minor things is ripple on the output.

In a DC electrical circuit a linear network of resistors and energy sources can be replaced by a single constant voltage source and a single series resistance. This is called Thévenin equivalent circuit. Allternately this same network can be represented by a current source and a shunt resistance.This is called Norton equivalent circuit.

This one equivalent resistance is the one that I am referring to.

This is measured as I described by varying the load on your supply and measuring the change in output voltage for various measured current values.

70 V and 20 A is 3.5 ohms. A 1500 W electric heater with no fan motor at 120 V is 12.5 ohms. You could use four of these heaters to get 25, 12.5, 6.25, 4.17, and 3.13 ohms for an appropriate number of test points.

Salt water tanks have been used for a high power variable resistance by having a pair of large copper plates that can be variably inserted into the tank.

A single large resistor, 2000 W 0.5 ohm, with a series pulse width modulated Hexfet can provide on average a variable load.

You do not need to know the resistance. You can get this from V and I measurements.

.

[tiger762]

This one equivalent resistance is the one that I am referring to.

Yes, it is the impedance that if connected to a load of the same amount, would deliver the maximum amount of power possible.

70 V and 20 A is 3.5 ohms. A 1500 W electric heater with no fan motor at 120 V is 12.5 ohms. You could use four of these heaters to get 25, 12.5, 6.25, 4.17, and 3.13 ohms for an appropriate number of test points.

Once it is at its elevated temperature it will be 12.5 ohms. I think that at room temperature it is a lot less. But as you say, the resistance is not important, the instantaneous voltages/currents are what's important.

What I am trying to do is find some home-expedient method for the DIY-er to characterize his/her homegrown power supply.

One thing I have noticed is that the voltages change slowly, as load changes, and as such will require holding a constant load for several time constants (whatever that might be), to get accurate measurements.

[gar]

070530-1930 EST USA

Tiger:

Yes when the internal resistance equals the load resistance you have maximum power transfer to the load.

It also needs to be pointed out that in a power distribution system you do not want to operate at this point. Operating at maximum power transfer means your efficiency is 50% and half of the input power goes into the source. This is done in radio transmitters because you want the maximum power in the antenna, but in a power distribution system it is different.

In a power distribution system you want the minimum practical internal impedance. This improves output voltage stability with respect to load, and improves system efficiency. This Edison understood and his contemporaries did not.

I have measured the primary DC resistance of a Dongan 3 KVA continuous duty 480 V transformer model # 85-1050SH at 0.7 ohms. If we scale this to 277 V and 2 KVA I would estimate a primary resistance of 0.75 ohms. This Dongan weighs 73 #. Thus an equivalent 2 KVA might be 50 #. I suspect the design is high on iron and low on copper to improve efficiency by reducing copper losses.

My discussion on measurement of primary resistance change as a means of measuring average internal temperature rise is a very valid technique when you can not place a thermocouple at your expected maximum hot spot point.

If primary DC resistance of your 277 V transformer is the 1.75 ohms as I previously guesstimated, then I will guestimate the DC output internal resistance at 2 * 1.75 / Turns ratio squared = 0.7 ohms. Assuming this at a 20 A DC load the DC ouput voltage would drop about 14 V from no load.

Internal power dissipation would be about 280 W at this 20 A DC load.

To get better values we need actual measurements from your transformer and assembled power supply.

Your time constant for the electrical circuit is about 0.026,000 * 3.5 = 0.1 second. This is about 12 half cycles at 60 Hz.

The thermal time constant is very much longer.

.

[tiger762]

Hopefully the transformer will be here soon and then I can do the experiments you mention.

[bbutcher85710]

Do you mean just measure the resistance of the secondary windings?



Do you have a recommendation for a DC load? I had thought about using a length of rubber tubing (like surgical tubing), filling it with tap water, and threading bolts into each end. Then measuring the resistance. Adjusting the length and/or salinity of the water as necessary. I figure this should be able to dissipate vast amounts of heat without changing temperature (and thus resistance) appreciably.

I would avoid the water load, the resistance will change wildly and heating can cause it to boil and maybe explode. Instead try using a bank of standard household light bulbs. Just get a bunch of 100 watt bulbs or whatever and some cheap light bulb sockets from the hardware store, and connect them all in parallel. Then you can screw in as many bulbs as you want to get the current you desire in the load. Don't worry if you are running 120 volt bulbs at 70 volts, they just won't be very bright, and they work on AC or DC. Light bulb resistance will vary with filament temperature, so the load current may not be what you expect, but all you really need is to measure the voltage and current to get the answer you want. You can mix in some 60 watt or 25 watt bulbs to fine tune if you want. The best part is that when you are done you can use the leftover bulbs around the house or shop. It is hard to beat a 100 watt load for under a buck.[/QUOTE]


Yes, a 26,000 uF electrolytic and a 200PIV/50amp full-wave bridge.



The biggest problem I'll have is coming up with a DC load of constant and kn own impedance.[/QUOTE]