Switching & Powering Relays

[loudthud]

Can either of you guys run a SPICE simulation? In my example from post #22 the power factor is 60.86%. It takes 98VA from the line to run a 60W light bulb on DC.

[Steve Conner]

What's the problem? You are both right

Dave H.

Well, sort of. The average current in a wire doesn't have a fixed relationship to the RMS current in it. The RMS:average ratio depends on the waveform. For DC it is 1, and it goes downhill from there.

If you're an Internet pedant you'll instantly point out that the average value of an AC current is zero. So, think of "rectified average" current- the mean of the absolute value of the current. This is useful because it quantifies the amount of charge that will be transferred into filter caps (and hence to the load) through a rectifier.

Indeed, the RMS current on the AC side of a bridge rectifier is the same as the RMS current on the DC side, and the rectified average current on the AC side is the same as the average current on the DC side. This is just a roundabout way of saying that it does nothing magical to the waveform beyond flipping alternate cycles of it. (Other topologies like the voltage doubler can introduce a factor of 2:1 or whatever.)

The average current goes to the load, while the RMS heats the filter caps up from losses in their ESR.

Spice simulations of these kinds of circuits are usually pointless, because electrolytic caps have a +/-20% tolerance at best, and transformer makers never specify the leakage inductance either. If you're building a really big power supply, even the impedance of your AC line affects the RMS-to-average ratio in the same way as the transformer's own leakage inductance does.

[R.G.]

... This is useful because it quantifies the amount of charge that will be transferred into filter caps (and hence to the load) through a rectifier.

Indeed, the RMS current on the AC side of a bridge rectifier is the same as the RMS current on the DC side, and the rectified average current on the AC side is the same as the average current on the DC side. This is just a roundabout way of saying that it does nothing magical to the waveform beyond flipping alternate cycles of it. (Other topologies like the voltage doubler can introduce a factor of 2:1 or whatever.)

The average current goes to the load, while the RMS heats the filter caps up from losses in their ESR

Absolutely correct. If you insert a bridge rectifier into an AC line feeding a resistive load, the RMS current changes not at all.

What people often fail to appreciate is that putting a BFC after the rectifiers alters the current waveforms in the rectifiers from the resistive case. The cap holds the voltage up and turns off the diodes when the incoming voltage declines below the voltage that the caps holds up. It turns the diodes off until the AC voltage magnitude gets larger than the cap voltage again. Instead of flowing for full half-cycles, current flows for only the portion just before the peak voltage. Current waveform changes, RMS changes.

[diagrammatiks]

Absolutely correct. If you insert a bridge rectifier into an AC line feeding a resistive load, the RMS current changes not at all.

What people often fail to appreciate is that putting a BFC after the rectifiers alters the current waveforms in the rectifiers from the resistive case. The cap holds the voltage up and turns off the diodes when the incoming voltage declines below the voltage that the caps holds up. It turns the diodes off until the AC voltage magnitude gets larger than the cap voltage again. Instead of flowing for full half-cycles, current flows for only the portion just before the peak voltage. Current waveform changes, RMS changes.

RG would it be better to switchmode relay power?

[R.G.]

RG would it be better to switchmode relay power?

Probably not. Putting a noisy switchmode power supply in a tube amp is a chancy thing to do.

I'd probably go along with powering them from the heater supply, given that there is enough power and current available to do it right.

This stuff is not mysterious. It's just math. Knowing how things work let you figure out whether you can use what you have or not.

[Alex R]

I do recall a Mesa that came in with all sorts of shorts in the relay power supply tapped off the heaters, and burned traces in the DC supply to the relays also. A preamp tube had shorted plate to heater. So I guess that's a downside to rectifying the heater supply for your relays.

[Dave H]

Indeed, the RMS current on the AC side of a bridge rectifier is the same as the RMS current on the DC side, and the rectified average current on the AC side is the same as the average current on the DC side. This is just a roundabout way of saying that it does nothing magical to the waveform beyond flipping alternate cycles of it.

Yeah, that’s what I should have said. I was just trying to point out that Ian and R.G. don’t look to me as if they are on the same page. Ian talks of “the DC current output from the rectifier” but R.G. refers to “The value of DC out of the rectifier/filter setup” i.e. The current flowing in the wire from capacitor to load, Ian’s current would be the current flowing in the wire from rectifier to filter cap (if he had a cap). That’s how I read it but I could be wrong.

Dave H.

[Steve Conner]

Hi Dave

Yes, you're right. It's the BFC (Bigass Filter Cap? ) that does the magic. It shorts out the RMS component, but as mandated by conservation of charge, it doesn't touch the average.

Whenever you say "current" you have to be clear whether you mean average or RMS, as they do quite separate things. Generally, if a load needed DC in the first place, it's the average current through it that performs the useful work. So the average current quantifies work done. The RMS current determines the power dissipated in any circuit elements that obey Ohm's law. Hence, RMS-to-average ratio is a kind of tool for thinking about the efficiency and cost-effectiveness of power supplies.

The diodes in the rectifier don't obey Ohm's law too well, so the heating is less than you would expect from the RMS, but more than you'd expect if you just multiplied the average forward volt drop across them by the average current.

A power supply with infinite sized filter caps and a transformer of zero impedance would draw current in infinitely short pulses of infinite amplitude. It would also need infinitely beefy diodes. The power supplies in audiophile-grade hi-fi amps get pretty close to this "ideal".

[mooreamps]

Well, all I know is ; if it's a 12 volt dc relay, then I feeds it 12 volts. If it's tube filaments ;and the data sheet calls out for 6.3 volts : a/c or DC ; then I feeds it either 6.3 volt "rms" or 6.3 volts DC...

-g

[loudthud]

I want to make sure everyone is on the same page. Refer to the schematic below. I think everyone agrees that I1(RMS) = I2 (RMS). There is a small error caused by reverse leakage in the diodes. PSUD2 models this as 1 Meg resistor in parallel with each diode. The disagreement seems to be that some say that I1(RMS) = I3(RMS).

PSUD2 says I1 = 1.0474 A(RMS) and I3 = 0.50033 A(RMS). I maintain that the program is accurate to better than 5%. The power in the load is 60.44W and 97.3VA is being delivered by the transformer. The power factor is 63.1% (W/VA).

I am prepaired to conduct an experiment to verify the results of PSUD2 with a pair of Agilent 3458A 8-1/2 digit DVMs. Current price is $8673 USD each. These DVMs are frequently used as lab standards to calibrate other meters. Can anyone who disputes PSUD2s results offer any evidence other that "it must be"?

Refer to post #22 of this thread for a screen shot of PSUD2s results.

[Steve Conner]

Loudthud, you are dead right. I've got three 6.5 digit DMMs here, but I don't need to look at them.

It agrees with my casual assertion that the filter cap simply shorts out the RMS current, as you measured I3 = 0.50033A.

The average component is 120/240 = 0.5A, and the extra is the RMS component that the cap failed to absorb, as it was of finite size and so had to have a finite ripple voltage across it, which causes an AC current in the resistor by Ohm's law.

Note that RMS, by its actual mathematical definition, includes the DC component, which is the average value. The RMS function in the circuit simulator works like this. But the AC ranges on most "true RMS" multimeters don't include the DC component in the calculation. Only a few meters by Tektronix do it, and the mode is called "RMS AC+DC" or similar.

Some people object to the meters being called "true RMS" on these grounds, but I'm too busy designing circuits to complain. Someone wrote to Fluke about it, and their engineer said that to get the "real" true RMS from a "true RMS" meter, you take the AC reading, and the DC reading, and do the vector sum.

So, if you took a regular "true RMS" DMM and stuck it in series with your 240 ohm load, it would read 0.5000A when set to DC. When set to AC it would read 18mA RMS. The vector sum of 0.5 and 0.018 is 0.50033, which is what the fancy AC+DC RMS unit would read.

[R.G.]

Well, all I know is ; if it's a 12 volt dc relay, then I feeds it 12 volts. If it's tube filaments ;and the data sheet calls out for 6.3 volts : a/c or DC ; then I feeds it either 6.3 volt "rms" or 6.3 volts DC...

Yes, Gary. You are 100% correct with that.

That would have been a good place for you to start with this thread, but you did get to the right answer. Here's a beer for you.

[Dave H]

I can’t compete with you guys. I’ve only got 3.5 digits on my DVMs now I’m not at work.

PSUD2 is good enough for me. On Loudthud’s schematic I1 RMS = I2 RMS but I1 RMS != I3 RMS. If you've think I1 RMS = I3 RMS you invented a new physics without a conservation of energy law.

Dave H.