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Power Wattage Consumption

03/18/2010 1:27 PM

I'm trying to calculate the wattage a radar ultra-sonic device takes. The radar instrument is loop powered (4-20ma).

Data sheet only shows 24VDC with max 550 ohms.

P = V*I = 24VDC * 20mA = .48 Watts

Can someone please confirm instrument power consumption?

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Anonymous Poster
#1

Re: Power Wattage Consumption

03/18/2010 2:36 PM

I would consider P-P, ERP, SWR loss. The mathematics becomes a little complicated. Consult a basic electronics or communications text.

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Anonymous Poster
#2

Re: Power Wattage Consumption

03/18/2010 11:30 PM

Hai

What your said that is correct.But the loop power consuption varied based upon the loop current(4-20ma) .

But the max power is .48watts only.

Regards

GIRITHARAN.R

giritnsf@gmail.com

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Anonymous Poster
#3
In reply to #2

Re: Power Wattage Consumption

03/18/2010 11:32 PM

hai

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#4

Re: Power Wattage Consumption

03/19/2010 12:34 AM

Not sure I understand your reasoning. For me the spec says power the device with 24Vdc across the 2 leads and the maximum load it can drive is 550ohm.

Inside the sensor there is a device that controls the current flow as well as some circuitry that controls that device. The circuit probably uses little current, so most of the power loss is across the pass element.

At 24V with max current (20mA) there will be 11V across the load and 13V across the sensors internal electronics. Power in sensor P = VI = 260mW , power in the load is P= i^2R = 220mW

So, my guess is that the power 480mW

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#5
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Re: Power Wattage Consumption

03/19/2010 12:41 AM

When using the power formula: P= U*I - than put P in Watts, U in Volts, and I in Amperes. 20 milliamperes = 20mA = 0,020 A (Amperes)* times your voltage = 24 V makes 0,480 Watts or 480 mW (milliWatts)

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#6
In reply to #5

Re: Power Wattage Consumption

03/19/2010 2:15 AM

True, that's the total.

But some power is dissipated in the sensor and some in the PLC's load. It depends on what Rabat is trying to work out.

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