How to Measure Circuit Output with SRS830 Lock-in Amplifier Using Labview?

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To measure circuit output in volts using a SRS830 lock-in amplifier with LabVIEW, the user seeks guidance on setting up the system and creating a .vi file. While the necessary LabVIEW drivers are available, the user expresses confusion about the setup process. Participants suggest verifying the ability to run example programs included with the drivers, highlighting that the SRS830 is generally easy to program. There is a discussion about whether the user is facing communication issues with the amplifier or difficulties using LabVIEW itself. Clarifying these points could help resolve the user's challenges effectively.
meanyack
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Hi to all,
For a project I need to measure output (as Volts) of a circuit. I want to do this with a SRS830 lock-in by using Labview application. Google search gave me nothing. If anyone knows how to set this up, please help. Thanks
 
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I am not sure what you are asking. Have you downloaded the Labview drivers from SRS?
Btw, the address is -for some unknown reason- www.thinksrs.com
The drivers include some examples.

The 830 is easy to program so you shouldn't run into any major problems.
 
Problem isn't drivers. I have the necessary drivers. The problem is how I can make a setup and make a .vi file.
 
I still don't understand what the problem is.
Are you able to run the examples that came with the driver? The 830 is -I wrote above- a "well behaved" instrument so controlling it is fairly easy.

Do you know how to control other instruments from Labview (I am assuming you are using GPIB to connect the computer to the 830)?

I am basically trying to understand if you are having problems communicating with/controlling the 830, or if you don't know how to use Labview.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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