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USB6008 cannot switch SSR.

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usb6008.pngaqz205D.png

I am trying to use the digital output of a USB6008 to switch a SSR (AQZ205D), and it failed to do so.

 

I checked the relay, it works fine. I can switch the relay from the 2.5v or 5v of USB6008. How come the digital output cannot do so? Do I miss something here? Thanks much for your thoughts/commens. 

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Message 1 of 11
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Hello,

I am uncertain about what you are asking. You mentioned that you can switch the relay using the USB-6008, but what do you mean by the digital output cannot do so?

Marina B.
Technical Support Engineer
National Instruments
www.ni.com/support
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Message 2 of 11
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First of all, thanks for reply. 

 

On the digital side of the USB6008, there are two constant outputs: 5v and 2.5v. If I manually connect and disconnect one of them to the relay, I can switch the relay ON/OFF. So this means that the relay is functioning. 

 

Next, if I set the 'P0.0' as 'Digital Output' and connect it to the relay, now I can turn ON or OFF of the digital output 'P0.0' from MAX. But this digital output from 'P0.0' cannot switch relay ON or OFF.

 

My question is why the digital output can not switch relays. Not enough Amps? 

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Would you be able to help?

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Hello,

 

What is your process in MAX? Are you using a task or the test panel to turn the line on and off? Have you used a multimeter to check the voltage and current coming from each state?

Marina B.
Technical Support Engineer
National Instruments
www.ni.com/support
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Message 5 of 11
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See the user manual of the USB-6008. For the value of the resistor, be aware that the max current is ±8.5mA. This is fortunately enough to switch the AQZ205.

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I used the test panel. Set digital channel to output and flip. 

 

Before connecting to the relay, flip it on and it gives 5v; connected to the relay, flip on and the voltage is dropped to 2.3v. So does this mean that USB-6008 does not have enough current to flip the relay?

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I checked the user manual of USB-6008. Like you said, the max current is 8.5mA. The relay AQZ205 requies a max of 3.0mA to operate. So in theory it should work. What is missing? Sorry, but I am not following your comment. Are you saying it should work, or won't work?

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@MengHuiHanTang  a écrit :

I checked the user manual of USB-6008. Like you said, the max current is 8.5mA. The relay AQZ205 requies a max of 3.0mA to operate. So in theory it should work. What is missing? Sorry, but I am not following your comment. Are you saying it should work, or won't work?


I'm saying that it should work.

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Solution
Accepted by topic author MengHuiHanTang

In this screenshot from the user manual, LED can be replaced by the pins 2 and 1 of the AQZ205.

How have you connected the AQZ205 to the USB-6008 ? What is the value of the resistor ? You must select it to have about 5 mA.

 

USB-6008.jpg

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