Switching & Powering Relays

[R.G.]

Maybe I can help with a little bit of "how it works".

Power transformers may be accurately modeled to any reasonably needed accuracy as an ideal transformer plus some other parts added on to model the imperfections.

In the absence of imperfections, an ideal transformer converts Vp (that is, "V"olts on the "p"rimary) into Vs (ditto, secondary) in the ratio of the number of primary turns (Np) to secondary turns (Ns). So an ideal transformer with 100 primary turns and ten secondary turns would convert 115.00Vac into 11.500Vac on the secondary. If it had instead 11 secondary turns, not 10, it would make 115.00* (11/100) = 12.650V. If it had 101 on the primary and 10 on the secondary, it would make 115.00 * (10/101) = 11.386Vac on the secondary. I belabor this to make the point that (a) turns are quantized - you can't easily get half a turn and have things work out right - and (b) the basic transformer ratio *is* the turns ratio.

But all real transformers have those pesky imperfections. The really big one for power transformers is copper resistance. We think of copper as a short circuit, but in reality, all metals are just low value resistors. Copper is no exception. A good set of wire tables will list the resistance per length of copper wire. The resistance of the copper wire in a transformer winding is real, and it acts just like we had an ideal transformer and put a resistor equal in value in series with the primary (for the primary winding) and the secondary (ditto, secondary). Where the ideal transformer makes perfect voltage ratios happen, real transformers can only have the perfect Np/Ns ratio happen when there is no current flowing to drop voltage across the copper resistance. As soon as you start to draw current through the windings, you get a voltage sag according to ohm's law. This is the fundamental reason that transformer voltages sag. I have simplified out pesky details like leakage inductance, because it's rarely an issue with power line transformers because of the low frequency. Not so for transformers for higher audio frequencies.

Mostly. If you're feeding a resistive load with AC. Heaters are a good example, and secondary voltages are pretty well predictable if you specify a heater current. If you have a good low-ohms meter, you can measure the actual resistance of the windings, measure the no-load voltage transformation ratio and make very accurate predictions about output voltage at various currents. What goes into that calculation is the voltage (same as turns) ratio, the actual value of the input voltage, the resistances, and the current pulled out of the secondary. Notice that transformer makers routinely lie to you. Well, OK, tell you info that is not all you really need. They will say things like "6.3Vac at 7A", which is probably OK as far as it goes.

But what they really mean is "for the *exact* 120Vac input we specify on the primary, we have taken into account the resistive sag of the primary and secondary copper windings, and added extra turns to the secondary so that when you load it up to a full 7A, it will sag down to 6.3V; or close enough to 6.3V that you won't sue us or send back a big batch of transformers you bought. We know you won't have exactly 120.0000Vac, and that you won't have exactly 7.000A loading. So watch it. If you pull less current, the secondary voltage will be higher. Oh, yeah... about that voltage ratio - we could only use full turns, not partial turns, so there's the voltage of about half a turn uncertainty in how accurately we could set the secondary voltage anyway. But it's close enough we think you'll like it as long as you don't do anything we haven't guessed you will."

Rectifying a heater winding to DC is most assuredly not what they guessed you'd do. They may have set that heater winding voltage up to, say, 7.2Vac so that at the rated current it sags back down to 6.3V if and only if the incoming AC voltage is nominal. It will, of course, go up and down proportionately to the AC line voltage, which may be anything between 110 and 130 or so. Let's guess that it's really 120, and calculate the DC we'd get if we full wave rectified this hypothetical heater winding.

It's a sine wave if other stuff hasn't distorted the incoming AC power line waveform too much, and the "6.3V" is RMS for a sine wave. So if we really have 7.2Vac at no load, the peak of that sine wave is 1.414*7.2 = 10.368V. A full wave bridge rectifies it to that, minus the diode losses which we have to guess. Call it 0.7V per diode, and there's always two diodes conducting in a bridge, so the output of the diode bridge peaks at 8.968V. Notice that the uncertainty in the diode losses make the three decimal places in that number absolutely bogus. The error in our guessing diode drops eats up at least a couple of tenths of a volt, but we need to be able to calculate *something*, even as an estimate of the center of the expected result.

So unloaded, we might get 8.5 to 9Vdc on a filter cap. When we load that filter cap, it starts sagging. It does that for two reasons; first, the capacitor runs down between charging pulses at the peak of the AC wave, and second, the current through the windings and diodes causes them to subtract their respective voltages from the available secondary peak voltage. If we use a HUGE filter cap to keep ripple down, it has the confounding effect of causing both lower ripple and lower DC voltage. This is because the bigger the filter cap, the shorter the time it gets charged when the diodes conduct. That means that the currents through the diodes when they do conduct gets HUGE too. It's pretty easy to get 10-20 times the DC average current out of a filter cap as a diode current in. In practice this is something you measure, not predict, because of the highly uncertain nature of the diode drops with increasing current, capacitor ESR, wiring resistances, connection resistance, and phase of the moon. Notice that the 10-20 times the peak current causes 10-20 times the voltage losses in the copper wire resistances too. So the voltages sag even more from that.

I say all that to say: the DC voltage you get out will sag more than you think it will or should, and changing the filter cap to make it better can make it better or worse. Changing diodes to Shottky is a better choice, but it makes the peak current thing worse too. And the peaky nature of the current heats the winding, too. The secondary winding is rated not to melt down at its rated current, which is 7A in my made-up example. However, you cannot pull 7A of DC out of rectifying and filtering this winding without endangering it from overheating. This is because the spiky nature of the current pulled into the filter cap has a higher RMS rating than a sine wave; The RMS /heating value of the current in the secondary winding feeding a full wave bridge with a capacitor input is usually 1.6 to 1.8 times the DC current out of the filter cap. So your 7A winding can now only support 3.9A of rectified and filtered DC out of the bridge.

All this is easy to compute, given that you can or have accurately measured the transformer turns ratio, copper winding resistances, diode drop with currents (especially high ones), capacitor value and ESR, and load currents. It's a common sophomore EE kind of homework problem.

With that as background:

Perhaps you were missing the "other half" of the equation... For example, for a transformer rated at 6.3 volts, the rating could be something like 6.3 volts at 4 amperes. So, with a proper input voltage of 115 VAC, and you put a 4 ampere load on it, then the device is rated to provide 6.3 VAC ; under those conditions...

And we have a winner. Yes, Gary, you are right!! Transformers are rated in their output voltage at full specified load current. At light loads, the voltage goes up, for the reasons above.

Having said that, input voltages vary, and actual loads placed on the transformer depends on the part installed.
Unloaded, the output voltage tends to measure a little higher.. Also, if "unloaded", and DC rectified with filter caps charged, then the measured voltage is higher still.

And we have a winning STREAK!! Yes, Gary, it DOES! Again, it's calculable, as above.

It all depends upon the load. It's a sliding scale.
No load ; 8.8 volts dc
At rated load 6.3 volts dc.

Well, actually, the "scale" is the internal losses in the path from the AC line voltage, through the primary resistance; what's left over as voltage is transformed by the turns ratio, losses in the secondary resistance are taken, diode losses taken, and after that capacitor ripple.

It is entirely possible to calculate what you'll get by making measurements of the transformer. It's possible in fact to make good estimates if the transformer maker tells you the no load (and hence no-current!) output voltage as well as the rated output voltage and current. This tells you the resistances by Ohm's Law. Smart guy, that Ohm. No real need to hand wave if you can measure.

Most common approaches are dc regulators.

Yep. Those let you *calculate* what you'll get and know ahead of time. Highly recommended.

Me personally, I was looking at an A/C regulator. {less problematic I'm thinking}

That's interesting. What kind of AC regulator? Active AC regulators are quite complicated (inverter style), heavy and clumsy (tap changers) or both expensive and heavy (ferroresonant). Which were you personally looking at?

[loudthud]

Thanks for the lesson RG.

I've never met anyone who could accurately calculate the RMS current drawn by a rectifier circuit, they were only guessing. With the right equipment you can measure it though. Like inserting a 1 ohm resistor between the center tap and ground of a high voltage power supply and measuring it with a true RMS meter. For low voltage circuits the resistor method is a little messy.

I have two of the clip-on DC coupled scope current probes. One I got at a swap meet for $50. It only gives a flat line. Proprietary Tektronix IC is dead. The other one I paid several hundred bucks for on ebay. It gives an output but I don't trust it.

So the most accurate tool available to me is PSUD2. In this thread: http://music-electronics-forum.com/t20730/ I tell what you have to do to get accurate numbers from it. You'll need to measure the actual transformer being used as RG outlines above.

[R.G.]

I've never met anyone who could accurately calculate the RMS current drawn by a rectifier circuit, they were only guessing.

Neither have I. The ones who knew what they were doing would tell you that the inaccuracies in knowing the part values and parasitics would prevent any real accuracy.

What you can do is bound it, estimate it, and allow for the variations, while understanding what might happen if your luck is off that day.

With the right equipment you can measure it though. Like inserting a 1 ohm resistor between the center tap and ground of a high voltage power supply and measuring it with a true RMS meter. For low voltage circuits the resistor method is a little messy.

I have two of the clip-on DC coupled scope current probes. One I got at a swap meet for $50. It only gives a flat line. Proprietary Tektronix IC is dead. The other one I paid several hundred bucks for on ebay. It gives an output but I don't trust it.

Yep. I have one that works after I messed with it a bit. I use it for an indicator, not a measurment device. One really, really accurate measuring device is in every circuit - the transformer supplying the current. It's just slow. You measure the cold resistance of the wire, let it stabilize with some load - which may take a few hours - then measure the copper resistance and from that compute the temperature rise. Then you remove the primary power, remove the secondary load, and apply DC currents through the primary and secondary in the inverse of the turns ratio (this being to keep the heat in the right places). The DC current which keeps the temp steady at the value measured for the AC-powered and loaded conditions is by definition equal to the RMS current in the AC-powered case. The heavy dependence of the internal temperature rise/fall on thermal balance makes it pretty accurate. And it does not use a series resistor. But this method is a true PITA to actually do.

So the most accurate tool available to me is PSUD2. In this thread: http://music-electronics-forum.com/t20730/ I tell what you have to do to get accurate numbers from it. You'll need to measure the actual transformer being used as RG outlines above.

I use PSpice or a similar simulator. Even then, you need to vary the part parasitics to get a reasonable feel for what real-world parts might do.

[mooreamps]

I've never met anyone who could accurately calculate the RMS current drawn by a rectifier circuit, they were only guessing.

The RMS current, if measured before the rectifier, and the DC current, if measured after the rectifier ; "is supposed to be the same value" LT...

-g

[loudthud]

The RMS current, if measured before the rectifier, and the DC current, if measured after the rectifier ; "is supposed to be the same value" LT...

-g

Here we go again...

Where did you get that? If you look at the example posted in the link I provided, you would see that the RMS current in the transformer is (in that example) over twice the load current.

Just to show how little you know Gary, calculate the RMS value of an ideal triangle wave with a peak to peak voltage of 2.000V (from plus one volt to minus one volt). I don't remember enough calculus to do it, but I can come pretty close with a spreadsheet.

[mooreamps]

You take a 60 watt light bulb. and feed it 120 volts rms, and observe it to illuminate to a certain brightness.
Then, you feed the same light bulb with 120 volts "dc", and observe it to illuminate to the same level of brightness.

or do it with 12 volts and a front panel lamp..... But, this is what we mean by "the rms value."

-g

[loudthud]

The RMS current, if measured before the rectifier, and the DC current, if measured after the rectifier ; "is supposed to be the same value" LT...

-g

You take a 60 watt light bulb. and feed it 120 volts rms, and observe it to illuminate to a certain brightness.
Then, you feed the same light bulb with 120 volts "dc", and observe it to illuminate to the same level of brightness.

-g

So you would agree that the light bulb functions like a 240 ohm load. 120V/240ohms=0.5amps. 120V*0.5amps=60W. Lets run a simulation and see what we find. You'll notice that the load voltage is 120.08 volts and the load current is 500.33mA (underlined in red). But look at the RMS transformer current (underlined in blue), it's 1.0474 amps. The peak transformer current is 2.74 amps! And you say that 0.5A load current "is supposed to" equal 1.0474A transformer current?

Do you know how to calculate the RMS value of an arbatrary waveform? Turns out the RMS value of a triangle wave can be found in about 10 seconds by a google search. It's one over the squareroot of 3 times the peak value.

An engineer I worked with who had an MS degree in oscillator design said to me, "Rocket science isn't all that difficult. All you have to know is what goes up must come down. The rest is all done with computers."

[R.G.]

The RMS current, if measured before the rectifier, and the DC current, if measured after the rectifier ; "is supposed to be the same value" LT...

It may "be supposed to be", but it isn't.

Use your math, Gary, use your math. You remember - it's all that squiggly writing stuff you do as a rocket scientist to calculate rocket science stuff. Concentrate - it'll come to you. I believe in you Gary.

You take a 60 watt light bulb. and feed it 120 volts rms, and observe it to illuminate to a certain brightness.
Then, you feed the same light bulb with 120 volts "dc", and observe it to illuminate to the same level of brightness.

or do it with 12 volts and a front panel lamp..... But, this is what we mean by "the rms value."

Yep, equal heating value. That's the basis of the "get it the same temperature with DC" trick I mentioned earlier.

The value of DC out of the rectifier/filter setup is not the same as the RMS/heating current INTO the rectifiers, and can't be. It would violate a number of things, including the conservation of energy. The math says so, and in fact measuring the actual values by experiment says the same thing. I've actually done that experiment myself, to prove it to an EE that should have known better.

The reason "rocket science" got a name for being complicated in the first place is that the math describing rocket functions describes things that are often counter-intuitive. That's why rocket scientists get the big bucks - they know and understand the math.

Right, Gary?

[Gingertube]

Guys,
I'm not exactly a rocket scientist - just the Senior Electronic Design Eng. for a Laser Airborne Depth Sounder System.
I'm with Gary, the AC RMS Current into the rectifier not only SHOULD be the same as the DC current output from the rectifier but in fact it MUST be the same, unless you guys are in fact inventing a whole new physics.
Cheers,
Ian

[mooreamps]

I was in Adelaide a few years back... I was working on the Collins Submarines. Just a lovely city... Excellent rack of lamb. I even took the ferry to Kangaroo Island.. Very nice.......

Oh, ya, there is a pokey machine at the hotel I stayed at ; still owes me 20 bucks. :}

-g

[R.G.]

Guys,
I'm not exactly a rocket scientist - just the Senior Electronic Design Eng. for a Laser Airborne Depth Sounder System.
I'm with Gary, the AC RMS Current into the rectifier not only SHOULD be the same as the DC current output from the rectifier but in fact it MUST be the same, unless you guys are in fact inventing a whole new physics.

Ian, I think you're confusing equal energy transfer/power with equal current. It's the same mistake that the EE I demonstrated it to made, and I won a modest bet on the issue; he was hung up on the same chain of reasoning that you are. It's an easy thing to latch on to.

An inductor-input filter forces the output voltage to be lower than the peak of the incoming sine by stretching out the conduction period of the diodes. But inductor-input filters are almost never used.

A capacitor-input filter is a peak detector, so the output DC voltage is quite close to the peak of the AC sine wave. Because the output power is Vdc x Idc and the combination of the incoming (sine) waveform, diodes and peak-holding nature of the caps force the charging current into the cap to flow at pulses much shorter than the period of the incoming sine. The average current into the caps must equal the average current out in steady state, therefore, the peak of the charging pulses must be approximately equal to the average DC current out of the cap times the ratio of the relative time the DC flows out (i.e. 100% in steady state) to the time the pulses flow in, which is quite short. So the peak of the charging pulses is quite high; in fact, it's limited only by the resistances and leakage inductances in series with and associated with the transformer, the wiring to and from the diodes, the internal resistances and voltage drops in the diodes, and the ESR/ESL of the cap. The biggest driver of the peak of the pulses is the capacitance value; this forces the ripple voltage down, which means the diodes turn on at nearer to the peak of the next AC wave, shortening the charging pulse time.

The math which describes this says that the average of a low duty-cycle pulse is less than the root of the mean square; you're averaging in one case, but squaring, averaging, then square-rooting in the other. The RMS value of the current in the transformer secondary, which is what heats the wire inside, is on the order of 1.6 to 1.8 times the DC current out of the filter caps for a full-wave -rectified capacitor-input filter for a sine wave input. If the input is not a sine wave, the ratio of RMS current out of the transformer to DC out changes, in the same way that it changes for half-wave rectification, full-wave-center-tap rectification, and in fact anything that changes the duty cycle of the charging pulses to the average DC out.

This is one of those things a power supply designer has to know if he/she is designing AC-power-line input rectifiers; otherwise, he burns up transformers and diodes. It gets really important for the input rectifier/filter for an off-line switching power supply. It's not something that one necessarily knows for systems design, logic design, other specialties in EE. In fact, the higher in the systems chain an EE gets, the less likely they are to know it.

Gary is in a unique position. I'm sure he's a great guy to raise a few beers with, but his technical pronouncements here are frequently just flat wrong. Worse yet, he is resistant to looking at the technical details of what he may be wrong about and correcting himself, at least in public on technical forums. That's a good thing in playing poker, but it doesn't work all that well in technical matters. We had a couple of sayings about making technical pronouncements that were not supported by reality in reports and meetings. One of them was that you can't B-S electrons (this is, after all, Texas ). The other is that Mother Nature waits for you at the end of the assembly line.

For some reason, Gary will make a mistake, then defend the mistake to the end, sometimes in spite of easily-available information to the contrary. I make a lot of mistakes, but I try to both admit them when I make them, and to remember that little lesson from Mother Nature the next time. Defending an error by denying the error is nearly always a losing proposition. I'm trying to convince Gary to think about the techie stuff he says here. I believe he's a sharp guy, and that he could do even better if he'd pick up on this. Not knowing everything is merely the common human position we all share. Not admitting (and worse, not *believing*) that you don't know everything can lead you into really, really bad situations.

As in all technical matters, the math which describes the situation (which I personify as "Mother Nature") does not care in the slightest what we humans think ought to happen. In the end, Mother Nature always wins.

[mooreamps]

Actually, I have a preference to Victoria Bitter, and Sam Adams.
Howsome ever, I did set up the experiment just before writing this post..12 volts to a spare front panel lamp. The results were as expected.
I suggest others do the same, and to draw their own concusions.

-g

[R.G.]

Howsome ever, I did set up the experiment just before writing this post..12 volts to a spare front panel lamp. The results were as expected. I suggest others do the same, and to draw their own concusions.

So do I. There is no substitute for doing your own work.

However, you have to do the right experiment. I think you may have confused that issue.

You don't describe your experimental setup. If you'll read back through what I said, you'll find that the actual circuit makes a difference. A full wave rectifier bridge causes a transformer RMS current higher than the DC current out of the filter cap for a capacitive input filter. I suggest you look at this reference from Hammond. See particularly "Transformer Current Ratings" and "FULL WAVE BRIDGE Capacitor Input Load".

Here's another reference from Power Volt. Go down to the place where it says

To accurately determine the secondary RMS cuffent rating, one must go through complex calculations. However, for all practical purposes the chart below can be safely used.

and read from the table of transformer secondary RMS currents for various rectifiers and filters.

These are the first couple of references I ran into on google that addressed the issue succinctly; notice that they're both transformer manufacturers, so they have a vested interest in getting this stuff right. So if you still think this is incorrect when you do the right experiment and measurements, don't take my word for it. Pop back in with the details of how both of these transformer makers got it wrong and what the right answer really is.

By now you should have guessed that there are a large number of references on both the net and in textbooks on the topic. I think you may be missing the point that you are defending your opinion against the tested and verified, theory and experimental results of about 3/4 of a century of people who made their livings using this technical material. You're not defending your opinion against me - you're refuting that chain of experts.

So tell me how that chain of folks got it wrong. I'm willing to learn.

[Enzo]

At the risk of generating a tangent, this would be why they state things in terms of VA (volt-amps) instead of watts? And then there is power factor correction or some such. I learnt what that was about once, and it doesn;t come up - directly anyway - in the repair game. But I suspect that is involved.

[R.G.]

That is pretty much correct, Enzo. The power ability of a transformer is almost entirely defined by its internal heating. That comes from core losses, proportional to frequency and flux excursion (which implies volts in most cases); and heating/RMS current in the windings.

Originally, the idea of using VA instead of watts came up when transformers powered mostly resistive (light bulb) and motor loads. The inductive loads had a lagging current phase angle, so the product of volts and amps did not correspond to how much real power was delivered. In spite of that, the inductive current heated the transformer wires, and limited how much current could flow. A transformer powering a big inductance might have very small real power transferred out of it, but be limited as to how much current could flow because of heating.

When rectification to DC became a larger part of actual power use, the idea was found to still apply. Whatever current flows in the transformer windings heats the copper and contributes toward the overheat limit. So the product of secondary voltage times actual current was still a better way to characterize transformer power than watts. Watts and VA are the same for a purely resistive load. Not so for others.