high side current sensing

[--BawA--]

plz tell me any high side current sensing opamp , which is easily available in market?
task is to sense a current ranging from 0 to 10 ampere and voltage is 12v.

 

[D.A.(Tony)Stewart]

Ground side current sensing is easier, but many solutions exist with Rail-Rail input Op Amp used as a differential amplifier.

Commercial IC's also exist.

Or you can use common quad OA' that only sense near ground and use 0-5V output voltage scale for 10A with this circuit that has a gain of 50 overall, I believe.

**broken link removed**

 

[--BawA--]

Can u tell me any easily available ic no. For high side current sensing ?

 

[Audioguru]

A TL081 or TL071 opamp (made by Texas Instruments and copied by many semiconductor manufacturers) has inputs that work up to its positive supply voltage.

 

[FvM]

Or you can use common quad OA' that only sense near ground and use 0-5V output voltage scale for 10A with this circuit that has a gain of 50 overall, I believe.

I guess you have used the 9:1 voltage divider to extend the high side voltage range relative to OP supply respectively common mode range. Unfortunately the offset error will be also multiplied by a factor of 10.

It's better to keep the divider ratio as low as possible. 1:1 (four equal resistors) would already work for the said 12 V input voltage. The shown circuit has negative Vout, by the way. You probably want to flip the Rsense connection for positive output.

Final remark, the differential amp/ level shift and the final gain stage can be combined to a single OP circuit.

 

[--BawA--]

thanks FvM ,
one more confusion , i have to measure the input current for push pull stage , which can be at max 50 ampere, so which is the best method to measure this current?

 

[D.A.(Tony)Stewart]

I agree FVM, that schematic wasn't my choice of R's



A better choice is the use of MAXIM's high side sensor chips. They key is to use 100mV full-scale for current sensing, which for 10A translates to 0.1/10 = 0.01 Ohm. To means you can choose a length of copper that will not heat up with the resulting 1Watt of heat dissipation 10*10A*0.01Ohm=1W and use twisted pair for the 2 signal wires coming back to the current sensor. Filtering is advised with series resistors in each leg and a big cap across the sense +/- input.

Note the MFG DOCUMENTATION listed on this page for more details and choice of chips. https://www.digikey.com/product-detail/en/MAX4374FEUB+/MAX4374FEUB+-ND/1702192
Since copper has a thermal resistance, that may change your resistance, calibration can be done with a 10A load and tune the length for 100mV drop to achieve 0.01 Ohm. This example uses a thin copper track

Since heat loss is major factor. I would re-consider the 100mV drop requirement and go towards 100mW dissipation in the copper track, which means 0.001 Ohm drop either on board , or perhaps external, short length of thin copper relative to the heavy cable, close to the circuit board.

Sense Cable pair must be at right angles to the current to avoid inductive pickup.

The tradeoff is input offset voltage of the chip and offset voltage due to copper temperature coefficient, so more details on your layout and wiring method is required. The resistance temperaturecoefficient of copper is fairly high (approximately 0.4%/°C)

( I recall using this method to monitor up to 100kA (100,000A) on a diffusion bonder which welded zirc-steel welding two cylindrical tubes together with forced water cooling. I used two screws in the solid copper trunks of the welder for my current sense R and used shielded twisted pair at right angles to the flow of current leading to differential amp. ) It was a monster welder with 16" diameter copper electrode wheels 2" thick. and a smaller wheel for inside the tube with a huge step-down transformer from 480V to 4V. But it worked due to very balanced wires and high CMMR.