Derating transformers for rectifier use.

Discussion in 'General Electronics Chat' started by The Electrician, Apr 15, 2012.

1. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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A favorite project on the forum is the construction of a power supply. If you're going to be building electronic projects, it soon becomes tiresome to always be buying and using up batteries, so the desirability of a grid powered supply quickly becomes apparent!

The kind of power supply project to be built is going to be transformer based for safety reasons. It may be regulated or unregulated, but the basic circuit is a small, relatively low voltage transformer, followed by a rectifier and an electrolytic capacitor for filtering.

I'm going to be dealing with only one very specific topic, but the beginner should read this description of power supply design before continuing here:

http://www.zen22142.zen.co.uk/Design/dcpsu.htm

Beginning power supply builders soon discover that a given transformer can't supply as many DC amps when used in a power supply as the transformer is rated for by the manufacturer. This is because the transformer rating is for an AC resistive load which draws a current whose waveshape is a sine wave, like the waveform of the grid voltage. When the transformer is used in a power supply with a capacitor filter, the current drawn from the secondary is no longer a sine wave; it's a narrow, peaked waveform, and such a waveform causes extra heating in the transformer. This is why the transformer must be derated when used in a rectifier/capacitor power supply, and that's what this post is all about.

I've attached an image showing a scope capture of the voltage and currents in a rectifier/capacitor power supply project. The blue trace shows the grid voltage, slightly flattened on the peaks. The orange trace shows the primary current and the purple trace shows the secondary current. You can see that the currents are very non-sinusoidal. To correctly measure their heating effect you need a true RMS responding meter. You can see some extra waviness in the primary current compared to the secondary current; this is the exciting current that supplies the magnetic core.

Derating factors for various rectifier circuits (bridge, full wave, etc.) can be found on the web, but it's almost never explained just how these factors are determined. I'm going to show how the derating factor can be determined for a real power supply.

I will assume the project builder has a DMM capable of making "True RMS" measurements of current and voltage. You shouldn't be attempting to build a power supply without having a meter! You should also have some power resistors to use as loads. You might be able to use light bulbs, such as automobile tail light bulbs for this.

Transformers have ratings; they are designed to supply a certain maximum current at a certain secondary voltage. What determines the ratings? The answer to that question is simple; it's just a matter of how hot the transformer gets. The transformer is constructed by winding insulated copper wire around a core made of iron. The iron is in thin sheets called laminations. The copper wire is different from the wire you use to hook up parts of your circuit; it is insulated with very thin plastic-like insulation, and is known as "magnet wire". There are at least two separate windings of copper wire on the core. There is no electrical connection (also known as a "galvanic" connection) between the two coils (primary and secondary) in the type of transformers used to make our power supply; this is an important safety feature of a transformer and provides "isolation" from the grid voltage.

One particular type of transformer, a transformer which is adjustable and known as a "variac" is an auto-transformer, and does have a galvanic connection between primary and secondary. NEVER use a variac alone as the transformer in a project power supply. You may use a variac between the grid and another isolated transformer to adjust the voltage applied to the primary of the isolated transformer, but this should seldom be necessary. There are also auto-transformers that are not adjustable; they also have a galvanic connection between primary and secondary. NEVER use an auto-transformer alone as the transformer in a project power supply.

The copper wire used to make the windings has some resistance. Copper is the second best conductor of electricity (other than superconductors); that's why it's used. But its resistance is not zero, so whenever a current passes through a copper wire, the wire is heated; this heating is called "copper loss" because some of the electrical energy is lost in the process. The amount of heat generated is given by the formula P = I^2*R, where P is the power heating the wire, I is the current in the wire and R is the resistance of the wire. In texts you will find this loss referred to as "I squared R loss".

When a transformer is supplying current to a load, the primary and secondary windings get hot due to the copper loss. The primary and secondary are separated by insulation which is a special paper. If the transformer gets too hot, the plastic insulation on the magnet wire and the paper insulation will be damaged, and parts of the windings may become galvanically connected where they shouldn't be. This will result in a short circuit in the winding and could cause a fire. The manufacturer has designed the transformer so that if the rated secondary current is not exceeded, the highest temperature in the middle of the windings will not exceed a certain value. There are various classes of transformer temperature ratings. As you might expect, transformers intended for special uses, such as military use for example, are constructed to withstand higher temperatures. The typical transformer the hobbyist will use will be a class A rated transformer, with a maximum hot spot temperature of 105 degrees centigrade.

There is another source of heat in transformers. The iron core is magnetized when a voltage is applied to the primary. Since the applied voltage is AC the magnetization of the core is constantly reversing direction, 120 times a second in the U.S. Every time the magnetization is reversed, a little energy is lost and converted to heat. This is called "core loss" and its amount is dependent only on the magnitude of the voltage applied to the primary; it doesn't change noticeably with the current drawn from the secondary. Since we have no control over it, we'll ignore it; it will be a constant in our transformer.

The sort of transformers used in the project may have a center tapped secondary and the designations printed on the transformer will indicate the secondary voltage and current rating. The designation may be something like 12-0-12, or 24v CT. These two designations represent the same voltage rating, which is 24 volts AC between the outer legs of the center tapped winding, and 12 volts AC from the center tap to each of the outer legs.

To start, let's make a few measurements. I'll show the measurements I made on a transformer rated to supply 12.6 VAC at 3 amps. First measure the resistance of the primary and secondary winding. Be aware that the secondary will probably be difficult to measure accurately because its resistance will be low. Most DMMs have test leads, and the leads themselves have a resistance of a few tenths of an ohm, comparable to the resistance of the secondary winding. What you should do is to set the DMM to measure ohms and connect the ends of the test leads together; this will give you a value for the resistance of the leads (wiggle the connection of the leads and press them together with force to get a stable reading). Then connect the leads to the secondary wires of the transformer. Subtract the resistance of the leads from the reading you get on the secondary; this will be the true resistance of the secondary.

CAUTION: We need to make a measurement on the primary of the transformer while it is energized. Be very careful while doing this. You will need a line cord to plug into the wall outlet and connect this cord to the primary of the transformer. If you use "wire nuts" to make the connection, the possibility of the ends of the line cord touching will be prevented, but you can still make contact with the wire inside the wire nut with your meter probes. See here in case you don't know what a wire nut is:

http://www.bing.com/images/search?q=wire+nut&qpvt=wire+nut&FORM=IGRE

In your final version of the project, you might make soldered connections with a fuse in the primary side, all in a nice box.

EXERCISE CAUTION: Make sure the line cord is unplugged from the wall socket and connect it to the primary with wire nuts. There should be nothing connected to the secondary. Now set the DMM to measure AC volts; if the meter isn't auto-ranging, set the range to a suitable range to measure the grid voltage. Plug the line cord in so that the transformer is energized. Carefully probe into the wire nuts with the meter probes, one probe to each wire nut. This will give a reading for the grid voltage; write it down. Next, measure the voltage across the full winding of the secondary of the transformer; write this down. These two readings will allow you to calculate the turns ratio of the transformer.

The transformer primary will actually draw some current even when there is no load on the secondary. This is the current that magnetizes the core and supplies the core losses; it's called the exciting current and adds to the primary current caused by the load when the secondary is loaded.

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2. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Here's what I measure on my little transformer:

Primary resistance = 12.24 ohms
Secondary resistance = .306 ohms
Primary voltage = 119.4 volts
Secondary voltage = 14.536 volts
Turns ratio = 119.4/14.536 = 8.214 = N

These resistance measurements should be measured when the transformer is hot, after operating at full load for 3 hours. Using measurements made at room temperature will lead to some error, but will be good enough for a first trial.

Note that the secondary voltage is higher than the rated 12.6 volts of the secondary. This is because there was no load on the secondary when the voltage was measured. The secondary voltage of a transformer will always be higher with no load than with a load. The rated output voltage is what you'll get with a load which draws the rated secondary current (3 amps for my transformer).

Now we can calculate the primary current when the transformer is under full load. If the secondary current is 3 amps AC, the primary current will be 3/8.214 = .365 amps. We should increase this value a little to account for the fact there is some exciting current; let's add 5%, so the primary current under full load will be about .383 amps.

Knowing the primary and secondary currents under full resistive load, we can calculate the copper losses in both windings:

Primary loss = (.383)^2 * 12.24 = 1.795 watts
Secondary loss = (3)^2 * .306 = 2.754 watts
Total copper loss = 4.55 watts

The total heating due to copper loss is 4.55 watts when the transformer is providing the rated current to a resistive load. This is what the manufacturer has rated the transformer for. Notice that the losses in the primary and secondary windings are not equal; this shows that the transformer was not optimally designed. It has been known since the early days of transformer design that the best design divides the copper loss evenly between the two windings. But when using a center tapped configuration, this rule must be modified; the transformer designer may have had other requirements also.

The key to derating the transformer for rectifier use is to simply see to it that the heating in the transformer is no more than the manufacturer specifies. What this means is that if we're not using the center tap, we must see to it that the current in the secondary is not more than 3.00 amps RMS. It's that RMS requirement that comes into play when a transformer is used with a rectifier. The current in the secondary is no longer a sine wave (see the attached image), but is a narrow, peaked wave shape. The heating effect of this current is determined by its RMS value, and this is why we need a meter that can measure true RMS.

I connected a bridge rectifier to the secondary followed by about 10,000 uF of filter capacitance. I connected a power rheostat to the DC output of the filter capacitor so that I could smoothly adjust the load current. I set my DMM to measure AC current and connected it in series with the secondary winding of the transformer. Then I adjusted the rheostat until I measured 3.0 amps RMS AC current in the secondary winding. Next, I set the DMM to measure DC amps and moved it (reconnecting the break in the secondary circuit where the DMM had been connected in series) to be in series with the rheostat without changing the rheostat. This gave me a measurement of the DC load current which would cause an RMS AC current of 3.0 amps in the secondary. The result was a DC load current of 1.84 amps.

So we see that the transformer has 3.0 amps RMS in the secondary winding, leading to the maximum allowable heating according to the manufacturer's rating, but we're only getting 1.84 amps of DC from the rectifier/capacitor. If we calculate the ratio 3.0/1.84 = 1.63, there's our derating factor. If you want 1.84 amps of DC, you need a transformer whose AC rating is 3.0 amps. If you want 5.0 amps of DC, you need a transformer whose AC rating is 1.63 * 5.0 = 8.15 amps.

A derating factor of 1.6 is often specified in various places, and that's a good start; now we see where it comes from. It depends on many factors, but as I've shown, it's easy to actually measure given a real power supply. Just vary the load on the DC output and measure the RMS current in the transformer secondary. When the DC load is such that the secondary current is just equal to the transformer's AC current rating, that's how much DC current you can safely get.

A slight cause of error is that the resistance of the DMM leads is in series with the transformer secondary when you measure the secondary current. If the project is more than just an amp or so, the lead resistance may be more than the resistance of the secondary and this will cause too much error. In the case of high currents, use a current shunt. A low cost alternative to professional current shunts is to use low value resistors as shunts. You can buy .01 ohm 5 watt 1% resistors from Digikey or Mouser; they are quite reasonably priced. You just put one in series with the transformer secondary and leave it there while making all your measurements. You can also put one in series with the DC output and leave it there also. You measure the voltage drop across the resistor and use ohm's law to calculate the current in the resistor; the current is just 100 times the voltage. Measure the voltage on the resistor leads close to the resistor body. When using shunts (or .01 ohm resistors for shunts) if you want to measure DC current, set the DMM to measure the DC volts across the shunt and multiply by 100. If you're wanting to measure AC current, set the DMM to measure AC volts. If you want to measure True AC+DC RMS current, set the DMM to measure True RMS AC+DC volts across the shunt.

The case of a two rectifier full wave configuration using the center tap is a little more complicated, and I'll deal with that in another post.

Last edited: Nov 25, 2012
3. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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When the rectifier configuration being used is the two diode full wave arrangement using the transformer center tap, Figure 3 on this web page:

http://www.zen22142.zen.co.uk/Design/dcpsu.htm

the procedure for determining the derating factor is a little different than for the bridge rectifier arrangement used in post #1.

The circuit used in post #1 has the property that the current in the secondary winding is related to the current in the primary by the turns ratio. If you examine the primary and secondary currents with a scope, they have the same waveshapes (ignoring the magnetizing current in the priamry) as can be seen in the image attached to post #1. The result of this is that all we have to do to determine the heating due to copper loss is to measure either the primary current or the secondary current; if we know one we can determine the other from the turns ratio. The ratio of the primary copper loss to the secondary copper loss remains the same whether a pure resistive AC load is applied to the transformer (and the current waveshape is sinusoidal), or a rectifier/capacitor load is applied (and the current waveshape becomes narrow peaked pulses). We need only measure the secondary current with a true RMS meter and see if it's less than or equal to the rated secondary current; we don't have to also measure the primary current.

But when the two diode full wave circuit is being used, the currents in the two halves of the secondary winding are not duplicate images of the primary currents. The currents in the secondary consist of alternate half cycles of current. The copper in the two halves of the secondary are not being fully utilized because they only carry current half the time, and this causes the heating in the secondary winding to be a larger proportion of the total copper loss heating than was the case in the bridge rectifier circuit. This will result in the temperature rise of the secondary being somewhat larger than the primary, but because the primary and secondary are fairly well coupled thermally, the difference won't be much.

The best thing to do in this case is to allow the TOTAL copper loss to be the same as it is in the bridge rectifier case (and the resistive AC load case). It can be shown that there is a factor k by which the primary current and secondary current can be determined to make this so.

Let:

Ip = primary current under full rated AC load
Is = secondary current under full rated AC load
Ip' = primary current with two diode full wave circuit which gives rated temp rise
Is' = secondary current with two diode full wave circuit which gives rated temp rise
Rp = primary DC resistance
Rs = secondary DC resistance
N = primary to secondary turns ratio

where

$k=\sqrt{\frac{Rp+N^2 Rs}{Rp+2 N^2 Rs}}$

and

$Is' = \sqrt{2} k Is$
$Ip' = k Ip$

Then if the DC load current on the two diode full wave rectifier/capacitor circuit is adjusted so that the true RMS secondary current (measured in either half of the secondary) is Is', the total heating loss in the transformer will be equal to the heating loss when the transformer is loaded with a resistive load giving the rated secondary current.

Using the small transformer I referred to in the first two posts, we have:

Rp = 12.24 ohms
Rs = .306 ohms
N = 8.214

Then k = .7838 and Is' = 3.325

If the secondary current is not greater than 3.325 amps true RMS, the transformer will not overheat. We could also check to see that the primary current is about Ip' = k Ip = .300 (times 1.05 to account for the magnetizing current) = .315 amps true RMS when the DC load is such that the secondary current is 3.325 amps true RMS.

The DC load current which gave this result was 2.95 amps, which corresponds to a derating factor of 3/2.95 = 1.02

Notice that the primary current should be .315 amps true RMS, whereas when we were using the bridge rectifier circuit the primary current at rated load was .383 amps true RMS. The ratio of the secondary current to the primary current (these currents give the same total copper loss as the rated AC load gives) is now 3.325/.315 = 10.54, which is not equal to N, the turns ratio. This is what happens when the two diode center tap circuit is used.

Last edited: Apr 19, 2012
4. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Stash point number two.

5. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Stash point number three.

6. #12 Expert

Nov 30, 2010
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Holy smoke! You're the second person to do this in a week. Is there a worldwide shortage of lithium? Was there a question in there? Have you considered posting this as one of your "blogs" for this site?

(That's what I do when I'm in the manic phase...I post a blog.)

7. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Can you give me a link to the other person's thread or post? I'd like to see it. I didn't notice it, I guess.

Nov 30, 2010
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9. bountyhunter Well-Known Member

Sep 7, 2009
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Most currently available meters can only measure true RMS waveforms with a crest factor of about 3 or less.

In a bridge rect design with a large input cap, the current is conducted in narrow pulses which used to be called haversine pulses. They are not actually sinusoidal.

The only meter you can trust to read their true RMS value is a thermal TRMS meter, and those are rare and expensive.

10. mik3 Senior Member

Feb 4, 2008
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Nice analysis The Electrician.

11. Wendy Moderator

Mar 24, 2008
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If you would like when you are finished I can soft delete the extra threads, then bring them back on demand. I would have liked to be able to do this on some of my articles before I was a mod. As is, always set aside more posts than you will ever need. I still make that mistake.

Would you like me to insert this on my blog, or do you have other plans for it?

12. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Why would a crest factor limitation of 3 be a problem when measuring rectifier waveforms?

Haversine pulses are quite sinusoidal to my eye and I wouldn't describe them as particularly narrow either. They are a negative cosine with an offset:

http://mathworld.wolfram.com/Haversine.html

See the attached image which shows a scope capture of rectifier current pulses in blue and a haversine waveform in purple. I don't ever recall having seen any references that described rectifier current pulses as haversines. Perhaps when inductor input filters were more common, but the rectifier current pulses typical in capacitor input filters don't look much like haversines.

A haversine shaped pulse would be easier to get a true value for than an actual rectifier current pulse.

The better true RMS DMMs claim that they have accuracies better than 1% at 50/60 Hz. The Yokogawa TY720 is one of the very few DMMs that I'm aware of that allows the user to choose true RMS or average responding. It's rated accuracy in average responding mode is 1%, but in true RMS mode, the rated accuracy is .4%--it's interesting that its accuracy is better in true RMS mode than in average responding mode.

The lower cost ones are typically around 2 or 3 percent accurate for 50/60 Hz waveforms. Are you telling me that the manufacturers are lying to us, that their meters can't be trusted to provide their rated accuracies?

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13. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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OK, thanks for the help.

I was thinking this sort of information would be good to have available somewhere. That submission you're working on might use it.

The long post that someonesdad started: