Switching & Powering Relays

[mooreamps]

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.

You're "over-reacting"....

I never said anyone was wrong. Where as most think of RMS as it applies to the sine function of y = a sin (2 pi f t) ; where a = a / square root 2. I was thinking more of the RMS as it applies to the DC function of just simply y = a.

In the Hammond link as referred to : they point out losses due to "generic silicon diodes" and they say Idc is greater than Iac in a capacitive input filter circuit. Well, maybe perhaps... But that would imply more algebraic variables inserted into the equations, above and beyond the primary mathematical definitions ; as I have described above.

-g

[R.G.]

I never said anyone was wrong. Where as most think of RMS as it applies to the sine function of y = a sin (2 pi f t) ; where a = a / square root 2. I was thinking more of the RMS as it applies to the DC function of just simply y = a.

OK, I'm good with that. One always has to describe the experiment. The RMS value of a sine wave and DC are different.

You said:

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

That was the statement that I was referring to. In the case of a full wave bridge rectifier feeding a capacitor input filter, the RMS current before the rectifier is in fact larger than the DC current out of the filter cap, and significantly larger. This is true for a sine wave, like the typical tube amp power supply;I used that example because the subject of the thread was making DC for relays from a heater winding.

If you meant that one could feed a full wave bridge rectifier DC and use that to feed a filter cap, then yes, that would work, and the RMS into the bridge would be the same as the DC out. However, it's not very interesting to feed DC to a full wave diode bridge. It's more interesting to feed square waves or trapezoidal waves to a bridge. That case gives an RMS that's very nearly the DC value too, and is actually something a power transformer could do - as long as you fed it 170V peak square waves on the primary.

Not impossible I guess. But I think sine waves are so much more common that it didn't occur to me that you meant feeding a bridge rectifier either pre-made DC or a square wave output.

In the Hammond link as referred to : they point out losses due to "generic silicon diodes" and they say Idc is greater than Iac in a capacitive input filter circuit.

Actually, that's backwards. They say that Iac is larger than Idc for a full wave bridge with cap input filter; that's the point. Whether the diodes are generic silicon diodes, Shottky, germanium, or silicon carbide, that doesn't affect the issue that the Iac is larger than the Idc. Diode losses amount to a pure subtraction of the diode drops from the peak of the AC waveform feeding the rectifiers, so they have no effect on the issue of Iac being bigger than Idc.

Well, maybe perhaps... But that would imply more algebraic variables inserted into the equations, above and beyond the primary mathematical definitions ; as I have described above.

No maybe about it. It very much does require more algebraic variables. Gotta use the right equations.

Understanding this requires understanding the way a rectifier charges a filter cap when fed with a non-rectangular waveform. The cap charges until the peak of the incoming rectified signal. At the point the incoming signal starts decreasing, the diode shuts off and current stops flowing into the cap through the diode, which becomes reverse biased. The load DC flows out of the filter cap, discharging it (relatively) slowly until the diode gets forward biased again by the power signal rising more than a diode drop above the voltage the capacitor has run down to. Only when the diode gets forward biased again can diode current flow into the capacitor.

So the incoming signal seen by the cap from the rectifier(s) is not a sine wave or a full wave rectified sine wave; it is the several degrees of the sine before the peak value, and maybe a very little afterwards, depending on the perfection of the diodes, the size of the cap as compared to the load current. At 60Hz sine wave fed to a full wave bridge, with an 8.66mS period for the half-sine, the diodes generally conduct for much less than a millisecond with capacitors sized to keep ripple down to low percentages of the DC output voltage. The bigger the cap, the less it runs down over a half-cycle, and the shorter the pulses are that charge it each voltage peak.

The waveforms of current into the cap are very, very non-sine waves, with more algebraic variables inserted into the equations. That's the point - it's not a sine or a DC level, it's a short, sharp pulse of current, which, when you do the integration to figure out its value for the RMS calculation, produces a value that's larger than the DC output current.

The only reason I keep explaining this is that it's a common misconception that the RMS current into a bridge/cap filter is the same as the DC current out. That's quite wrong, and counter-intuitive.

[Raybob]

...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...

I hope you meant to say square root of 2, not 3.

[loudthud]

I hope you meant to say square root of 2, not 3.

The square root of 2 rule only applies to sine waves. Actually it's when you take the Root of the Mean Square, you need the square root of 1/2 which is 1/square root of 2. For a perfect triangle wave, it's the 1/square root of 3. Google it!

If you look up the definition of Volt in the dictionary it says something like "the voltage that will cause 1 Amp to flow in a 1 Ohm resistance." That will cause 1 Watt (1 Joule per second) to be dissipated by the resistor. Now if you connect a sine wave of 1 Volt peak to the 1 Ohm resistor you would find that it only dissipates 1/2 Watt (average over several cycles). So what DC voltage would make that resistor dissipate 1/2 Watt? 0.707 Volt would cause 0.707 Amps to flow. Multiply Volts times Amps and you get 1/2 Watt. So we call that 1 Volt peak sine wave "0.707 VRMS" because it does the same amount of work (create heat) as 0.707 VDC. Actually, math is used (just short of calculus) to find the amount of heat a sine wave would create in a resistor.

A 1 Volt peak triangle wave only causes 1/3 of a Watt to be dissipated by the 1 Ohm resistor. You have to know the average value of a parabola from zero to one.

[diagrammatiks]

sooo...has anyone managed to power a relay yet.

[R.G.]

I thought I'd let the topic cool for a bit.

Yes. I did it long ago. It's not a problem. You measure what you have in terms of available power, make DC out of it, then use relays that match, or else make the DC into what you want for the relays you got. Have I done it with that amp? No. But there really isn't any mystery there.

I liked Enzo's simile -

Is it better to scratch my back with a regular screwdriver or a phillips screwdriver?

[wizard333]

Loudthud:

You won't get very far in the electronics industry or in amp building until you get the ability to figure things out for yourself. You can't post every question on a forum.

I figure out about 99% for myself thanks, don't post here much. You on the other hand wont get far in life with that attitude.

As far as explaining, you did not, you simply got your panties in a wad because you didn't like the question or thought it was too basic? Fine, don't grace me with an answer then. Unless I missed something, this forum exists for questions and for people to learn from each other, not for jackasses to pontificate to each other and try to impress each other as to how much they already know. As for "exercises for the student", I've tought for almost 30 years. I'd never respond to a student I was trying to teach with the attitude you displayed. It's clear you have never taught, or god help anyone you tried to, because frankly your skills in that area flat-out suck balls. I had a professor years ago that said "if you can't explain it to an intelligent non-expert, you dont know it". I'm left with wondering how much you really know vs. how much you plug things into a computer simulator w/out knowing what goes on behind the simulation, similar to plugging things into a calculator without having any idea how to do them by hand. Running it through the simulator gives an answer, it doesn't explain the answer.

Contrast that with RG who provides detailed, interesting, and informative answers. I hardly knew squat about transformer losses, current draw on the transformer vs. rectified current, etc. before reading those posts, but I sure have a solid idea about it now. My guess is RG has been a teacher at some point, or at least shows good aptitude in the area. It's also obvious he knows WTF he's talking about, and doesn't have to try to cop a 'tude in order to try to make anyone think he does.

[Jazz P Bass]

"Panties in a wad".
I like that phrase.

[loudthud]

Perhaps some of my remarks were taken a little too personally however, most of those were made after, in post 7, you demanded a detailed explanation. Did anybody else answer your question? I delayed responding to your question in post 6 hoping someone else would post an answer. I'm sorry, I don't have the capability to do animation or produce videos, and I have no idea what your background is because you left that information off of your profile. In post #3 I wrote "Duncan's PSUD-II has a model for a voltage doubler, run a simulation and see what you get for output voltage. Or just build it." You ignored that advise. These days simulation plays a big role in electronics education. In post #9 you seemed to understand all the concepts necessary to put the puzzle together.

PSUD2 is the best tool I can come up with. As a first step, it provides a schematic. Next, you can plot the voltage or current of any component and get the peak, average and RMS values. I wish this was available when I was introduced to power supplies in 1967 because I didn't get it the first time even though I got straight A's in that class. For many years I thought much like Gary does as posted in post #19. Then one day I tried to make a 5 Volt 2 Amp power supply. I wound a new secondary on a transformer. It got too hot and eventually burned up. So I had to figure out how things really worked. It started with a program I wrote to predict ripple voltage based on current draw and filter capacitance.

This thread got a little off topic, but I don't like to see people post incorrect information. Quite a few people don't understand the idea behind RMS numbers. Sure they will tell you "Root Mean Square", but they can't tell you how to compute the RMS value of an arbitrary waveform. Reading back on Gary's comment in post 19, "The RMS current, if measured before the rectifier, and the DC current, if measured after the rectifier is supposed to be the same value." is actually not wrong if you measure the current between the rectifier and the filter capacitor. The trouble is I was referring to the current after the filter capacitor, the load current as compared to the current in the transformer winding.

Back to your question. If you take the time to use PSUD2, you can see how on the first half cycle of the incoming line one capacitor gets charged to the peak and on the next half cycle the other capacitor gets charged to the negative peak. The capacitors are in series so the voltages add. You can view current or voltage of any component by checking the boxes below the schematic. To see the voltage across both capacitors, you must look at the voltage of the load. If the voltage drops too low at a specified load, you can increase the value of the two capacitors to decrease the voltage drop. PSUD2 allows you to predict the voltage drop with different loads. But to get an accurate result, you need to know the DC resistance of the transformer winding. That is the true value of PSUD2.

Do you think Ceriatone would post a layout that didn't work? That company is pretty good at producing products that work. Do you think PSUD2 would produce a decent result if the voltage doubler didn't work. You brought both of these things into question. Sometimes you just have to accept the fact that a circuit works even if you can't figure it out right away. You seem to be the one who's underwear is in disarray.

[Enzo]

I for one find Loudthud's comments useful and informative, and I very rarely have any reason to disagree with him.

[mooreamps]

Perhaps I should said the output power was the same...
In a dc power supply, unloaded, yes the caps charge up, and the measured voltage is higher. But if the secondary winding is spec'ed for say, 2 amps, and you put a 2 amp load on it ; then the output voltage is supposed to be abouts that value...

-g

[R.G.]

Perhaps I should said the output power was the same...

Yes, this is correct. For all non-nuclear processes, power out equals power in minus losses; Mother Nature says "The sum of matter and energy is always conserved".

In a dc power supply, unloaded, yes the caps charge up, and the measured voltage is higher.

Good. For a diode-rectified capacitor-input filtered DC power supply, the caps charge to the peaks of the incoming waveforms, minus any diode forward drop and resisive losses.

However, for an inductor-input filtered DC power supply, the inductor forces the current to match the DC average current out. It does this by the inductor's field forcing the inductor voltage to take up the differences between the incoming rectified signal and the voltage on the following caps, if any.

But if the secondary winding is spec'ed for say, 2 amps, and you put a 2 amp load on it ; then the output voltage is supposed to be abouts that value...

I don't understand that statement.

If a secondary winding is rated for 2A, the most common practice in the transformer industry is to rate it at a specific voltage when it's supplying that 2A. I *think* that's what you're saying.

However, there are implicit assumptions in that rating. Transformer makers know that their products have internal resistances. They commonly specify them a couple of ways, the most common being that the output voltage is some amount higher than the specified output voltage when there is no load. Then, under load, the voltage sags down to the specified minimum at the maximum load. This sag is (confusingly to non-transformer-literate people) called "regulation". Unless otherwise specified in the datasheets, this rating assumes that the load is resistive and the load current is sinusoidal (well, as long as the primary voltage is sinusoidal to start with). Also unless otherwise specified, the voltages and currents are RMS, not peak or average, values.

A transformer rated at, say, 12V at 2A output will have an output that's 12VRMS --ONLY-- when (a) the nominal input primary voltage is applied to the primary, (b) the loading is resistive or a reasonable facsimile thereof, and ( c) the output current is 2A RMS. If the load is reduced to 50ma RMS, the secondary voltage will rise. At no load, it may be 13.2Vac RMS, the no-load voltage having been designed in to make up for resistive losses in the windings. Such a transformer would be rated as having a "regulation" of 10%. If the no-load voltage was only 12.6Vac RMS, the "regulation" would be rated at 5%.

However, things get funny when you put a diode bridge on the output and a filter after that. If you use a capacitor input filter (like the vast majority of power supplies do), and the capacitor is charged up to some DC voltage, then the diodes will not let the transformer secondary conduct at all until the instantaneous secondary voltage forward biases them by at least a diode drop. Then they let current flow into the capacitor, limited only by the resistances of the transformer wires, connecting wires, diode resistance, and capacitor ESR.

It's been done many times, but I wrote up my own version of this little dance back in 2001. You can see it here.

Notice that the diode current (and hence, current from the transformer secondary) only flows for a short time near the peak of the incoming secondary voltage. The capacitor charges from that short pulse, and supplies the DC load until the next pulse, half a cycle later. The quick charge up and long run-down of the capacitor voltage give the classical sawtooth ripple voltage waveform.

With that as a background, here's the main point: Mother Nature requires that both the power into the cap and the charge on the cap be conserved in steady state. If power into the cap is consistently greater than power out, the cap heats up and explodes. If charge on the cap is consistently increasing, the capacitor voltage is increasing by V = Q/C, or decreasing the same way. So for a stable average output voltage on the filter cap, the charge into the cap during diode current charging must equal the charge out of the cap for the long stretch where the cap is supplying the load by itself. Charge in equals charge out. OK so far?

Charge in through the diodes flows for a tiny fraction of the time. Charge out to the load flows all the time. The only way these can balance is if the charge in from the diodes is much higher than the average current out, because it can only flow for a shorter time.

Here's the kicker: the RMS value of the peaks of current coming through the diodes (and hence the transformer secondary) is 1.6 to 1.8 times the DC average current out of the capacitor. This is because the heating is proportional to the instantaneous current squared, not the average current. So for a short period of time each cycle, the transformer windings are heated proportional to current squared, and that current is many times the DC average out of the filter cap. The longer time the current is not flowing in the transformer secondary lets the transformer mass average the current-squared heating down. The transformer's average heating is about the same as it would be if the secondary was supplying a sine-wave RMS current numerically equal to 1.6 to 1.8 times the DC average out of the filter caps.

But if the secondary winding is spec'ed for say, 2 amps, and you put a 2 amp load on it ; then the output voltage is supposed to be abouts that value...

The waveform of the "2A load" matters - a lot. And whether the "output voltage is supposed to be abouts that value" matters with it. Beginner engineers know to look at datasheets. Experienced engineers know to think of them as the advertising that they are and not take them as being literally true. The math involved tells you what Mother Nature will actually let happen. I'm "supposed to get" a big birthday party with a fancy cake and lots of presents every year. The reality is that sometimes that doesn't happen.

"Supposed to" is not equal to "really happens".

[Dave H]

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.

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.

What's the problem? You are both right

Dave H.

[diagrammatiks]

I can't take RG seriously because he hasn't factored the phase angle dependencies of rc filter networks.

HARUMPH.

[mooreamps]

No, I woundn't blame him. But if it were up to me, I'd be selling a lot more ad space on high hitting threads like this one...

-g