I could not sleep over the question:
How do we get 24Volts on relais, when you have 40Volts after IRF9610 on power supply.
Also others reported their relais are not working – and it is needed to lower the series resistors.
Well, I figured it out:
With 2x30V AC transformer we get 30V * sqrt(2) = 42 Volt DC.
We loose some 2Volts at the Diodes – so we end up with 40 Volts DC.
A Nais DS2Y-S-DC24V relais has a coil resistance of 2280 Ohm.
With a series resistor of 2200 Ohm we have a total resistance of: 4480 Ohm.
With a voltage of 40Volts, we have a current of:
I = U/R = 40/4480 = 8,9 mA
This gives a voltage on the relais coil of:
U = R * I = 8,9mA * 2280 Ohm = 20,4 Volt
Both current and voltage values are within the allowed ranges for DS2Y-S-DC24V.
Also the pick-up voltage of this relais type is 70%*24Volt = 16,8 Volt. As our 20,4 Volt is well above 16,8 Volt – this will work fine.
A bit strange is that a 1.5K series resistor is used for K7 instead of 2.2K. I think that in the original X0.2 a DS2E-M-DC24V relais is used for K7. This relais has a coil resistance of 1.44K.
So:
DS2Y-S-DC24V with 2.28K coil resistance and 2.2K series resistor
DS2E-M-DC24V with 1.44K coil resistance and 1.5K series resistor
If you have different types of relais, you can check the relais datasheet to see what the minimum voltage and coil resistance of the relais is.
[ 本帖最後由 dsdjoy2 於 2011-12-7 Wed 05:28 編輯 ]
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