How to amplify a 10VDC/35mA signal to a 1.5A output?

[JW8]

I want to activate a 12VDC solenoid trigger from my current experimental set-up with labview. However, the trigger draws about 1.5A at 10VDC, and my current set-up with Labview can only generate a 10VDC 35mA (0.035A) signal from a BNC connector. What are some ways I can amplify my current signal to obtain the current I need (1.5A) to generate the force required to activate the solenoid trigger? I've heard suggestions about transistors or Op-amps, but I don't know what these do or how to apply them to my BNC connector signal because I am unfamiliar with them. Any ideas??

[billko]

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@JW8 wrote:

I want to activate a 12VDC solenoid trigger from my current experimental set-up with labview. However, the trigger draws about 1.5A at 10VDC, and my current set-up with Labview can only generate a 10VDC 35mA (0.035A) signal from a BNC connector. What are some ways I can amplify my current signal to obtain the current I need (1.5A) to generate the force required to activate the solenoid trigger? I've heard suggestions about transistors or Op-amps, but I don't know what these do or how to apply them to my BNC connector signal because I am unfamiliar with them. Any ideas??

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Sure.  Take a few courses in electronics.

[Ben]

Use an external supply capable of the voltage and curent and control it using s Solid State Relay that is spec'd to activate with drive you have and is capable of switching the exteranl supply.

Talk to an electronics type to select the proper relay.

Ben

[SnowMule · Accepted solution]

10V is kind of an oddball voltage; i'm assuming you mentioned this because it's the highest your DAQ can go.

You mentioned the solenoid is 12v... so use 12v to make it toggle.  A <$20 switching regulator like this can pump the current to the solenoid.  A 5V power relay like the VE5HS-K or G5SB can handle the 12V/1.5A power (up to 30V/5A), while the coil side can be driven from the DAQ's 5V output.  You'll also need a relay driver like the ULN2003A to sink the current from the relay's coil.

Some solid-state relays can be driven right off a digital output, some play nicer with the additional current from an analog output on the DAQ.  For the cost/flexibility(normally-open and normally-closed terminals)/isolation of a mechanical relay and driver circuit, that's the route I take.

Then it's just a matter of toggling the solenoid on/off by driving a digital line high or low in software.

[JW8]

Ah! I would have never thought of that. Thank you so much for the post, suggestions, and the links as well. This sounds like the pefect solution. Extremely helpful.  

[Sparkman2]

As one of the responses indicated, use the ULN2803 for the driver to a pilot relay which will control the final output relay. Each output is rated to sink 500mA,  and the supply voltage can be set at 10VDC. The input will be driven by one of the digital outputs of the NI device. I have used this arrangement to sink the current required for a 24VDC relay that I used to control a larger relay to operate a 120VAC power supply.

I used 24VDC control relays as we use many of those in our products. In my arrangement, I applied 24VDC to pin 10 of the ULN2803 and to one side of the DC relay. Pin 9 of the ULN2803 is tied to ground of the power supply and to the NI device. The digital control signal from the NI device ( I have used both the USB-6008 and USB-6501 ) is used to supply pins 1 - 8 of the ULN2803. The other side of the control relay has its current sunk through the corresponding pins 11 - 18 of the ULN2803.

In fact, I have one arrangement that uses 9 digital outputs from a USB-6501 to directly control any one of 64 relays using some chip-select circuitry and four 82C43s. Of course, the ULN2803 was used on the outputs of the 82C43 to drive the control relays.