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Full Member level 3
Mar 20, 2021
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Hi all,
I have the circuit connected below, I am using the following AQY222R2T MOS Relay. Basically I am using the MOS relay to switch an output of 10V @ 400mA. The device is AQY222R2TY

I have worked out the input resistor as

R = Vin - VF

From the datasheet this works out as

10V - 1.05V
------------ = 1.79K or 1.8K

I have two questions:

1. Is my calculation for R correct? I have used the following (drop out voltage waveform diagram 7 in the datasheet i.e. at 5mA the VF is around 1.05V). Or should I be using 3mA (led operate current) instead of 5mA?
2. The Output power is rated at 250mW, I believe that my output of 10V @ 400mA = 400mW - hence the device is not suitable.
3. The FET used to switch the FET led has a RDS max of 3R, which I can add to the calculation, but is that really going to matter.
4. I can't see a max input voltage, so will the 10V be ok?

Finally will a G3VM-63G be a better choice? This has an output capability of 500mA @ 60V. I can't however see the output power dissipation of this device.
Any help would be appreciated.


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1) correct

2) 10V @ 400mA = 4000mW = 4W.
Not the output power is rated for 250mW, but the power dissipation of the relay.
The ouput power = load power does not matter for the relay.
The power dissipation is Relay_out_voltage x load_current.
So when the relay is OPEN, then the voltage across the relay is 10V, but the current is almost zero. thus the power dissipation is almost zero.
When the relay is ON, then the voltage across the relay is small, but the current is 400mA. Again the power dissipation is small.
The coninous relay current is 400mA, so you are at the edge.

3 Ohms at 1800 Ohms is about 0.17%, I guess the resistir´s tolerance is much bigger. Thus it does not matter.

4) max relay output voltage is 60V,
relay voltage input side is just 1.05V, the input never knows about the 10V, because the resistor limits the voltage.


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