relay coil capacitor needed

i want to turn On a relay using a capacitor, the coil of the relay need 0.4W in 10 ms, i don't know how to calculate the capacitor in µF needed to turn on the relay

What is the coil relay voltage?
Could you measure the DC resistance of the relay coil?

Are you trying to delay the turn-on by 10mS or have the relay turned-on for only 10 mS? All this depends on the relay's pull-in/drop-out voltages, the supply voltage, and the coil's resistance and inductance. Please post a schematic of what you have in mind.

Ken

the voltage of the relay is 6V, to turn on the relay i have to maintain 0.4W in 10ms on the coil of the relay

Is this a latching relay? Can you post a part# or link?

Ken

here is a link **broken link removed**

You are using a 5V relay on a 6V supply? How are you switching in the capacitor? Are you putting the capacitor in series or parallel with the coil to switch it?
Coil resistance is 62Ω, supply is 6VDC, pull-in voltage is 3.5VDC (5v relay). Assuming the coil voltage must remain above 3.5V for 10mSec:
RC Time Calculator
Looks like a 0.22uF capacitor will give you ~12mS, so should work.

Ken

R_coil = V_coil^2/P_coil = 90 Ohm (relay type A06)
V_coil = 6V
P_coil = 0.4W

Let us assume:
L_coil (coil inductance) is relatively small.
V_cap = 6V (a bit higher is better)
T_dis = 10ms
dV_cap = voltage drop after T_dis = 1V
I_coil = V_coil/R_coil = 67 mA
Capacitor discharge is linear

C = I_coil * T_dis / dV_cap = 667 uF

Note: if V_cap = V_coil + V_extra, dV_cap = 1 + V_extra. This reduces the value of C.

@KerimF : please can you explain to me i didn't understand what you mean.

If it is for practical purpose, Experiment it practically observing results on scope for timing with different values of capacitors. It may take less than five minutes to evaluate the right values of capacitor. You may like to experiment with different realys of same part number for variation.

yassin.kraouch said:

@KerimF : please can you explain to me i didn't understand what you mean.

Click to expand...

I used two simple formulas:
P = V^2 / R
and
dV = I * t / C

And the calculations are made to give a value for C to start with since they are approximated. For example, 1000 uF capacitor will likely work. But the last word is what the test gives.

I use this technique even with normal relays to save power.
I turn on the relay by a capacitor. And with a series resistor between Vcc and C, the relay coil voltage drops to a level that keeps the relay on (I find the low holding voltage by experiment). The drawback of this method is that there is charging time (after the relay is turned off) during which the relay shouldn't be turned on (the capacitor voltage would be not high enough).

Anyway, I add my voice to ALERTLINKS in determining the practical value of C by experiment. The calculations just give a theoretical approximated value which is better than having nothing :wink:

Kerim