[SOLVED] Power supply voltage dip protection design

• Oct 21, 2016

• #1

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Hello again...we have a system running in the production. Sometimes we can't avoid a power dip due to some reasons. When power dip happens one contactor(K1) trips. The contact part number is LP4K1210BW3, it has a 24Vdc triggered coil. We would like to design a power dip protection that will hold the K1 for 3 sec when power dip occurs.
Does anyone can help us, really appreciate it...thanks a lot.

 

• Oct 21, 2016

• #2

KlausST

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Hi,

Some calculations:
1.8W @ 24V means 350 Ohms
Time is 3s, but for a voltage drop of about 50% you need about 0.5 tau (raw estimation).
So if you need 3s then tau needs to be about 6s.

Tau = R x C --> C = tau / R = 6s / 350 Ohms = 0.017 F = 17000 uF

I recommend to build a 24V capacitor buffer with this 17000uF and a decouple diode just to supply the relay.
Don't connect the capacitor in parallel to the coil.

In doubt post your actual circuit. (Power suppl to coil, with switches)

Klaus

 

• Oct 22, 2016

• #3

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Hello KlausST...thanks for the response...I had attached the schematic...but i don't have schematic for the power failure protection module...Actually two contactors that i need to hold on power dip, the KM1 and KM2. From the drawings, is possible to insert separate RC for node 54 and 59 (encircled with blue)?

 

• Oct 27, 2016

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Hello KlausST and other readers, i sketched the schematic for the power protection module and looks like in the image. Based from my understanding from the circuit, when 24Vac fails, the charge from capacitor take over and hold the coil for a short period of time...So is there a good calculation how long the capacitor charge can hold the contactor coil? Hoping for response..thanks a lot.

 

• Oct 27, 2016

• #5

KlausST

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Hi,

Calculations: I did it in post#2.
* capacitor charge calculation: wikipedia and 1 million of other sources
* RC time constant: wikipedia and a million of other sources
What else do you need?

****
Circuit:
Why that complicated? A simple diode and a capacitor should do the job.

Klaus

 

• Oct 27, 2016

• #6

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Hello KlausST the circuit is the original design...i wan't to modify it to increase discharging time.
So are you suggesting to replace the capacitor to 17000uF and the resistor to 350 ohms?
Thanks a lot.

 

• Oct 27, 2016

• #7

KlausST

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Hi,

no resistor. 350 Ohms is your load = relay coil

Klaus

 

• Oct 29, 2016

• #8

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Thanks mr KlausST....by the way, i used this formula --> Vc = Vs(e(-t/RC) to calculate the Vc after 3sec. It show like 33000uF got 18.5V which is what i am looking for. So do you think if i use 33000uF there would be any problem with the circuit? Thanks.

 

• Oct 30, 2016

• #9

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What to consider if I will increase the capacitance from the circuit

Hello Team...would like to ask you opinions about my plan to increase the capacitance. We have the original design in our power sag protection. But the design says it can hold only for 1 sec under power dip situation. I want to increase it to 3 sec. So I use the discharge formula of capacitor Vc=Vs * e (-t/RC). It shows that 33000uF fits the 3 sec requirement. So my question is, based on the circuit from the image, if i increase the capacitance from 10000uF to 33000uF can i have any problem from the circuit? Thanks team.

 

• Oct 30, 2016

• #10

KlausST

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Hi,

The capacitor circuit itself will work.
But if it works in your circuit is impossible to answer, because we don't know what's inside the box "Option / power failure protection".

Klaus

 

• Oct 31, 2016

• #11

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Hello mr KlausST...i dismantled the power protection module box and inside is what i draw in post #9 and #4.

 

• Oct 31, 2016

• #12

KlausST

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Hi,

You want a guarantee?
--> I can´t give.

Klaus

 

• Nov 1, 2016

• #13

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Thanks Mr KlausST i think this most we can do for this topic...Really helping all your inputs.Thanks again and more power edaboard...