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Last Post 12/03/2012 10:46 AM by  Tony Hodges
Model 19
 1 Replies
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Author Messages
Tony Heweston
New Member
New Member

12/02/2012 10:31 PM
    Basic 240v single phase wattmeter

    Hi am looking at using an AD633 four quadrant multiplier
    and driving the input reference voltage from a resistor
    network across the supply and the current reference via
    a Model 19 current transformer with the possible addition
    of a LM324 to provide some gain for the Model 19 output.
    Max current should be about 20amps. Would this seem to be a workable application for the Model 19 product?
    Tony Hodges
    CRM Staff
    Basic Member
    Basic Member

    12/03/2012 10:46 AM
    Dear Sir,

    Please refer to the attached Model 19 specs. The maximum driving voltage available is 2 volts, the DCR is 3 ohms, and it has 230 turns of wire on it. For a 20 amp input, the device will output 20/230 or 87 mA of current. The maximum available to the user would then be 2 – (0.87 X 3) = 1.74 VAC. To still stay in a reasonably accurate range, it is recommended to use 80% or less of the available output voltage, so for this app and CT, a burden resistor of 16 ohms would generate an ac voltage of 1.4 volts. This would give you about a 1% accurate part. You could improve on that slightly by running the CT at say 100 mV and then amplifying as you discuss. However, the amplifier chosen will add inaccuracies as well. We recommend high accuracy differential amplifiers like the OP07 products to produce the best results. Additionally, the model 19 was never designed to be a measurement style CT. A more appropriate choice is the CR8448-2500-N. Use of this part with an OP07 pre-amp would create a very accurate power transducer. I have attached the catalog page of our CR8400 Series for your review.

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