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I am trying to figure out the correct size capacitor (and resistor, but that's easier) to make an economizer for a Kilovac EV200A1ANA relay. The relay coil specs are as follows:

• Coil voltage: 9-36 V
• Hold voltage (min.): 7.5 V
• Inrush current (max.): 3.8 A
• Holding current (avg.): 0.13 A@12 V
• Inrush time (max.): 130 ms

I will be using this on a 12 V (4S LiFePO4 battery) circuit.

To calculate the capacitor's value, I used 3.8 A x 13 V = 49.4 W
To get Joules, I took 50 W x 0.13 s = 10 Joules
To get farads, I took 10 J / (13 V² x 0.5) = 0.118 F
In µF, that's 118000 µF. The largest µF capacitor I could find on McMaster Carr was 2000 µF.

Do I really need to solder nearly 60 of these in parallel, or is there something off in my math? I know this is elementary stuff, but that seems like a really large result, and need a sanity check.

I am trying to figure out the correct size capacitor (and resistor, but that's easier) to make an economizer for a Kilovac EV200A1ANA relay.
The coil will burn up in minutes without any current limiting, so a resistor is added to limit the current to near the "holding current" amount. A larger current spike is needed to pull the relay closed, so a capacitor is often used in parallel with the resistor to dump the needed current (3.8A for at most 130ms, in this case).

The relay coil specs are as follows (copied from the relay spec sheet):

• Coil voltage: 9-36 V
• Hold voltage (min.): 7.5 V
• Inrush current (max.): 3.8 A
• Holding current (avg.): 0.13 A@12 V
• Inrush time (max.): 130 ms

I will be using this on a 12 V (4S LiFePO4 battery) circuit.

To calculate the capacitor's value, I used 3.8 A x 13 V = 49.4 W
To get Joules, I took 50 W x 0.13 s = 10 Joules
To get farads, I took 10 J / (13 V² x 0.5) = 0.118 F
In µF, that's 118000 µF. The largest µF capacitor I could find on McMaster Carr was 2000 µF.

Do I really need to solder nearly 60 of these in parallel, or is there something off in my math? I know this is elementary stuff, but that seems like a really large result, and need a sanity check.

I am trying to figure out the correct size capacitor (and resistor, but that's easier) to make an economizer for a Kilovac EV200A1ANA relay. The relay coil specs are as follows:

• Coil voltage: 9-36 V
• Hold voltage (min.): 7.5 V
• Inrush current (max): 3.8 A
• Holding current (avg.): 0.13A@12 V
• Inrush time (max): 130 ms

I will be using this on a 12 V (4S LiFePO4 battery) circuit.

To calculate the capacitor uF value, I used the 3.8 A x 13 V = 49.4 W
To get Joules, I took 50 W x .13 sec = 10 Joules
To get Farads, I took 10 J / (13 V^2 x .5) = .118 Farads
In µF, that's 118,000 µF. The largest µF capacitor I could find on McMaster Carr was 2,000 µF.

Do I really need to solder nearly 60 of these in parallel, or is there something off in my math? I know this is elementary stuff, but that seems like a really large result, and need a sanity check.

I am trying to figure out the correct size capacitor (and resistor, but that's easier) to make an economizer for a Kilovac EV200A1ANA relay. The relay coil specs are as follows:

• Coil voltage: 9-36 V
• Hold voltage (min.): 7.5 V
• Inrush current (max.): 3.8 A
• Holding current (avg.): 0.13 A@12 V
• Inrush time (max.): 130 ms

I will be using this on a 12 V (4S LiFePO4 battery) circuit.

To calculate the capacitor's value, I used 3.8 A x 13 V = 49.4 W
To get Joules, I took 50 W x 0.13 s = 10 Joules
To get farads, I took 10 J / (13 V² x 0.5) = 0.118 F
In µF, that's 118000 µF. The largest µF capacitor I could find on McMaster Carr was 2000 µF.

Do I really need to solder nearly 60 of these in parallel, or is there something off in my math? I know this is elementary stuff, but that seems like a really large result, and need a sanity check.

I am trying to figure out the correct size capacitor (and resistor, but that's easier) to make an economizer for a Kilovac EV200A1ANA relay. The relay coil specs are as follows:

• Coil voltage: 9-36v
• Hold voltage (min.): 7.5v
• Inrush current (max): 3.8A
• Holding current (avg.): 0.13A@12v
• Inrush time (max): 130ms

I will be using this on a 12v (4S LiFePO4 battery) circuit.
To calculate the capacitor uF value, I used the 3.8A x 13v = 49.4w
To get Joules, I took 50w x .13sec = 10Joules
To get Farads, I took 10J / (13v^2 x .5) = .118Farads
In uF, that's 118,000uF. The largest uF capacitor I could find on McMaster Carr was 2,000uF.
Do I really need to solder nearly 60 of these in parallel, or is there something off in my math? I know this is elementary stuff, but that seems like a really large result, and need a sanity check. Tia.

I am trying to figure out the correct size capacitor (and resistor, but that's easier) to make an economizer for a Kilovac EV200A1ANA relay. The relay coil specs are as follows:

• Coil voltage: 9-36 V
• Hold voltage (min.): 7.5 V
• Inrush current (max): 3.8 A
• Holding current (avg.): 0.13A@12 V
• Inrush time (max): 130 ms

I will be using this on a 12 V (4S LiFePO4 battery) circuit.

To calculate the capacitor uF value, I used the 3.8 A x 13 V = 49.4 W
To get Joules, I took 50 W x .13 sec = 10 Joules
To get Farads, I took 10 J / (13 V^2 x .5) = .118 Farads
In µF, that's 118,000 µF. The largest µF capacitor I could find on McMaster Carr was 2,000 µF.

Do I really need to solder nearly 60 of these in parallel, or is there something off in my math? I know this is elementary stuff, but that seems like a really large result, and need a sanity check.