HIGH VOLTAGE IN-LINE BIDIRECTIONAL CURRENT SENSE

Hello,

Thank you very much for reaching out and for your participation in this community.

I will be transferring this inquiry to other group - Amplifiers. Hopefully, your questions could be answered more accurately. 

Regards,

Nino

Hi BestUser ,

Is this the voltage divider configuration you have in mind?

(Sample circuit only, the voltage divider circuit of R5, R3 and R6, R4)

Yes, this is what i have in mind

Hi,

With this configuration, a slight change in the discrete resistor pair values (With 1% tolerance) will have an unexpected output voltage value if your next system or device needed a precise voltage value especially if this is connected to a precision ADC. The external input resistors should be evenly matched to achieve high CMRR value.

You can check our Difference Amplifier portfolio for suitable devices on your application (120V common mode range and -55°C temp. range) and we suggest using AD8479.

The AD8479 has ±600 V common-mode voltage range with -55°C operating temperature range but the Gain is only 1.

Hope that helps.

Regards, 

JE

Thanks for the reply. Unfortunately AD8479 have low bandwidth.

By choosing the solution of the voltage divider at the input of LT1999, which formula define the output voltage?

In the following graph an LT1999-10 is used with 15A load and 10mR resistor with 28V (green curve) and 100V (blu curve)

LT1999 is supplied at 5V

The green curve is 4V as expected without voltage divider at the input (=(15A * 10mR * 10V/V) +2.5V).

The blu curve is 2.83V with 10K-10K voltage divider at both +IN and -IN (like your circuit example) but i can't find the formula that give as result 2.83V

What is the formula that define the output voltage of the LT1999 with a voltage divider at its inputs?

Thanks

HI,

Based on the formula (page 15 of datasheet), it is VOUT = VSENSE * (RG/RIN) +VREF

VSENSE is the voltage difference of IN+ and IN-. The value of VSENSE of the blue curve on simulation is about 33.346mV. RG/RIN is 10 since RG = 40k ohm and RIN = 4k ohm (p.15). VREF is 2.5V

VSENSE = 49.276105V - 49.242758V

VOUT = 33.346mV * 10 + 2.5V

VOUT = 2.833V

The input bias current also has effect on the voltage going into the input.

Regarding the bandwidth, what is your expected value?

Thanks.

Yes, the bias current is very high and is not predictable so the voltage divider solution is not the best solution.

Our application is a motor control system. The LT1999 is used to sense the inline motor phase current. The problem arise since the regenerated voltage can reach approx 100V.

For sense the motor phase current we do not need so much bandwidth (300KHz is sufficient) but we use the LT1999 also as high side current measure and we need a fast response in case of high current event out of our control and 300KHz is not enough.

P.S. just noticed that the EP version of AD8479 has lower bandwidth (130KHz) than the non EP version (310KHz).

Probably 130KHz is too small for our application.

Thanks

Hi BestUser 

I communicated with my colleague regarding this issue, we can study this article and it may help.

https://www.analog.com/en/analog-dialogue/raqs/raq-issue-168.html

The article is about bootstrapping, and it may help you improvise your circuit.

Regards,

JE