I am new to use Solid state relays (SSR). For my application i am using Panasonic 'AQV252G3' relay. i want to measure ON Resistance of this relay, according to the datasheet it ranges between 0.033 Ohm - 0.06 Ohm for A-type circuit connection.


i attached circuit diagram showing how my connections are with relay. i applying Load Voltage 5V/6V with Load Current 3.3 A (As suggested in Datasheet also LED Forward Current 5mA). When Relay is closed i measure voltage drop using DMM (its in the range of mV). Here I use Ohms Law V=IR to Calculate ON Resistance.

I use 5 relays, when i measure voltage drop for each relay it differs..

Is this measurement procedure correct?? if not can suggest improvements

  • 1
    \$\begingroup\$ You can use the schematic entry tool rather than a hand drawn picture, might make things a touch clearer. How have you got 3.3A flowing when two different voltages are used but it's the same resistance? \$\endgroup\$ – Puffafish Jun 5 at 13:34
  • \$\begingroup\$ @Puffafish its external DC supply. i mean i have two DC sources 5V/3.3A and other source 6V/3.3A \$\endgroup\$ – ki.j Jun 5 at 13:38
  • \$\begingroup\$ but these supplies don't push 3.3A into your circuit, as they would mean the votlages would always be the same (about 792V). Instead the current will be limited by the voltage and resistor, so that would mean currents of around 20mA and 25mA. \$\endgroup\$ – Puffafish Jun 5 at 13:59
  • \$\begingroup\$ @Puffafish i am apply same principle as they mentioned in this website , except using SSR instead of Electroechanical Relay panasonicelec.wordpress.com/2012/02/21/… \$\endgroup\$ – ki.j Jun 5 at 14:07
  • \$\begingroup\$ That link is using a power supply which will provide the voltage up to the 1A current limit, and then they reduce the voltage until only 1A is drawn. They do not have a limiting resistor between the supply and the replay. This means that they have a steady 1A current, they can then measure the voltage across the relay, apply Ohm's law and they have the resistance. 1A is a large current, which means the volt drop across the relay will be substantial even with a very low resistance. Your circuit is not the same, as you would need to remove the 240 resistor and have a current limited supply \$\endgroup\$ – Puffafish Jun 5 at 14:15

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