125VDC TRIP CIRCUIT 1

[tulum]

I have two 125vdc bridge rectifiers that are bieng used to supply a toshiba VCB (125vdc control). One is on the close circuit and the other is on the trip circuit.

Right now we are just doing a point - point on the system. It was found that the rectifier supplying the close circuit (with circuit connected, breaker racked out) measures 166vdc. When the breaker is racked in, the closing coil operates momentarily, and secondary voltage drops to 107vdc momentarily (returns to 166vdc after).

These seems normal to me (besides the 166vdc being slightly high). Problem being, the only things in the circuit are the breaker close circuit, and a GEC instantaneous relay (GEC MCGG instantaneous overcurrent relay)AND WHEN I DISCONNECT THE GEC RELAY THE (REST OF THE CIRCUIT STILL CONNECTED) THE RECTIFIERS SECONDARY VOLTAGE DROPS TO 130VDC...WHICH IS RATED VOLTAGE. This to me doesn't make sense...disconnect a load...and the voltage goes down??? Any suggestions??

 

[ScottyUK]

Do your rectifiers have any smoothing, or is it just how you describe - two rectfiers. Also, are the rectifers single or three-phase?




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Never look down any at anybody, unless you are helping them up.

 

[stevenal]

If you're using a full wave bridge rectifier on single phase, the output waveform will look like a sine wave with the bottom half inverted. I'm not sure how most DC meters treat this, they may average it or filter out the AC components. (jump in here if you know) The result will be something less than the peak of the wave form. My guess is your relay is somewhat capacitive. Adding a capacitor smooths the waveform close to the peak.

 

[dpc]

I agree with stevenal that you may be seeing metering errors from your voltmeter due to the ripple in the rectfier output.

Breakers typically have a spring-charging motor that runs whenever the breaker is racked in and after each closing operation. You need to make sure your rectifiers are up to the job of supplying this current.

How will you trip your breakers if you lose incoming power to the rectifiers or the rectifiers fail?

 

[jghrist]

These questions probably don't relate to your main concern, but:

Why is the instantaneous overcurrent relay in the close circuit?

What causes the closing coil to energize when you rack the breaker in.

 

[tulum]

Thanks for the tips and advice...

ScottyUK:

The recitfiers are 240V AC, however in the current system the primary is wired to 120vac.

The input & output both have Diac's across their terminals. I believe the diac's breakover voltage is set to turn off at the end of each positive and each negative alternation of the ac wave in turn smoothing the waveform. However, just not my area.

Stevanal,

I should have mentioned on the one rectifier we read 130vdc which is probably close to peak with 120vac on the input (which makes sense for an unloaded rectifier). I also buy the fact that the relay is capacitive, it is rated I believe for 115vdc to 150vdc..so something must be going on inside of there.

dpc,

Rectifiers seem to be supplying the spring charging motors fine. There is an under-voltage trip circuit on the breaker to trip it in case of power failure.

Thanks again for the help. After taking all the comments in, the rectifier may be behaving fine.

1. Gec relay is probably capacitive: 166Vdc... reasonable
2. Unloaded - close to peak: 130vdc... reasonable
3. Loaded - slightly less than rectified ...107vdc...ok

I will take... for now...Thanks

 

[peebee]

I agree with stevenal, I would not necessarily trust your VDC readings. Your voltage source is probably not a true flat DC, it probably has some ripple, which could seriously affect the reading you're getting. Any chance you could put a scope on there and see what the "DC" waveform really looks like?

Regarding the voltage increase, yes, it sounds like you're adding some capacitance.

 

[buzzp]

Generally capacitors are part of a rectifier circuit (especially half wave, which you dont have but just as common or needed with full wave rectifier). This could explain the increase in voltage.

 

[ScottyUK]

It sounds like your rectifiers are unsmoothed, and that your relay has a capacitor across its input terminals. The capacitor is probably quite small. When there is a light load on the system the capacitor stores enough charge to smooth the ripple, giving you the high measured voltage. When the load increases, e.g. when a tripping or closing coil operates, or the spring charging motor runs, the capacitor is too small to provide effective smoothing at the higher current, and the measured voltage will drop.

The measurements don't exactly tally based solely on the presence or absence of smoothing, so I guess your power supply does not have very close regulation or is entering current limit when the tripping coils operate. The whole system sounds wrong for the application and doesn't follow accepted practice for provision of critical tripping & control supplies. Normally a tripping & control supply has a battery supplied by a trickle charger. This gives high immunity to loss of AC supply and ensures that the switchgear can be operated under such conditions. Is it possible that you actually have a battery charger without the battery being present?



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Never look down any at anybody, unless you are helping them up.

 

[busbar]

ScottyUK — Sometimes the device you are describing is termed a “charger/eliminator.” Typically, stored energy in such a device during an AC collapse is close to zero.

 

[jbartos]

Question: What is an input impedance of the voltmeter being used?
Please, notice that 166VDC/107VDC=1.55 is close to sqrt2=1.41 considering that 1.41 = Vmax/Vrms. Then, when 1.55 x .9 = 1.395 ~ 1.41.
Vav(=Vdc)/Vrms=0.9
Vav=2xsqrt2xVrms/pi=0.9xVrms; for sinusoidal wave

 

[tulum]

ScottyUK,

I am not as familiar with the codes and regulations surrounding battery supplies as you may be. However, the equipment I am working on is not my design. Furthermore, in the mining industry (where I am) a separate DC source is rarely, if ever used as you describe. Almost all tripping comes from recitfiers supplied by the main feed. As the exception, I do believe the surface transformers have pure DC control from a battery. Whether this is right or wrong...this is how it is.

Jbartos,

Thanks again for the enlighting look into the situation. You always seem to think a little bit out of the box. As for the input impedance, I am not sure...I can check..

 

[tulum]

Jbartos,

If I echoe back your thoughts...Is my understanding reasonable?

Meters read average voltage (Vavg). If the recitfier has a capacitor on the output (as with the GEC relay as a load) the average voltage at the output of the recitifier is smoothed close to the peak voltage (Vp) - in our case 166V. If we now disconnect the Capacitive load we will still read the averge voltage, only this time not smoothed. Vavg = 2/pi*Vpeak which in our case is 107V. What we measured.

I realize the numbers are correct...but is the logic?

Thanks again Jbartos and everyone else.

 

[ScottyUK]

Hi Tulum,

Yes, the principle of what you are saying is what jB, myself, and others have suggested. The numbers don't quite work out if you use the mathematical proof, but there are real-world explanations for the deviation from theory. Is there any chance you can try one of the following tests:

Connect an oscilloscope to the DC bus and see how the waveform changes when the relay is absent or present?

Connect a multimeter measuring AC across the DC bus, and see how the measurement changes with and without the relay present?

The results will give a fair indication of what is going on. The first test is the more informative, but the second is easier if you don't have a lot of test gear available. Post your findings if you can do either of the above.




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"Never look down any at anybody, unless you are helping them up."

Jesse Jackson.

 

[tulum]

ScottyUK,

I don't have a oscilloscope and I tried the meter but it did not produce the results I thought it would...

With or without the GEC the DC terminals (measured with an AC voltmeter) cycled from 210-38 volts in increments of around 30v? What this means, I don't know.

ScottyUK, what do you meant the numbers don't work out?

Peak voltage (Vp) = 166V. What we measured.

Vavg = 2/pi*Vpeak = 107V. What we measured.

Is this just coincidence?

 

[jbartos]

Suggestion to tulum (Industrial) May 17, 2004
Jbartos,
If I echoe back your thoughts...Is my understanding reasonable?
Meters read average voltage (Vavg). If the recitfier has a capacitor on the output (as with the GEC relay as a load) the average voltage at the output of the recitifier is smoothed close to the peak voltage (Vp) - in our case 166V. If we now disconnect the Capacitive load we will still read the averge voltage, only this time not smoothed. Vavg = 2/pi*Vpeak which in our case is 107V. What we measured.
I realize the numbers are correct...but is the logic?
///The logic is correct since the Vav,no-load=Vpeak=166VDC, meaning the ripple of the full-wave rectifier is negligible. This can be accomplished by thorough filtering, i.e. large capacitors.
The full-load conditions approach to Vav,full-load=(2/pi)xVpeak=(2xsqrt2xVrms/pi)=107VDC. It is necessary to notice that 166VDC is an envelope to Vmax peaks. This is accomplished by the thorough filtering at no load. However, at full load, the Watt power is Iav x Vav in Watts no matter how thorough passive filtering is since it the passive filtering cannot deliver Watt power.\\

 

[ScottyUK]

Hi Tulum,

Yes, you are correct the proof does work out. My brain had fallen out of gear and made a simple arithmetic error in my earlier post which I've carried forward into the later one. Apologies.

I can't explain the strange variation you are seeing on the AC range. I wonder if the presence of a large DC component is confusing the digital instrument? What happens if you connect a blocking capacitor of a few uF in series with the meter while measuring AC? I'm guessing that you will see very little AC component with the relay in, and 50V or so with the relay out.



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"Never look down any at anybody, unless you are helping them up."

Jesse Jackson.