Capacitor in parallel with resistor

Thread Starter

Dritech

Joined Sep 21, 2011
907
Hi,

I was looking for a shunt sense resistor to use with my H-bridge. While doing my research, I noticed that some use a capacitor in parallel with the resistor. What is the reason for this please?
 

bertus

Joined Apr 5, 2008
23,027
Hello,

I can imagine that the capacitor will make the current sensing less sensitive for spike currents.

Bertus
 

Thread Starter

Dritech

Joined Sep 21, 2011
907
Thanks for the reply. As an H-bridge I am using the BTS7960. It has the a status flag diagnosis feature. This is used to output the full current when when a fault occurs. Does anyone know how to get 5V from the IS pin? There is an equation in the datasheet, but I cannot full understand it.

Any help will be appreciated.
 

Gibson486

Joined Jul 20, 2012
360
I have used that chip before. Just to let you know, that chip is being phased out. They are already 2 iterations away from that chip.

Your output for that pin is dependent on your source voltage and your load current. I remember I spent a good two hours figuring out how to derive the resistor values like you are doing. I have it in my notebook at work on how to derive the equation. I'll update tomorrow.
 

Thread Starter

Dritech

Joined Sep 21, 2011
907
I have it in my notebook at work on how to derive the equation. I'll update tomorrow.
I will appreciate that a lot :) I was planning to get 5V when the flag is active.

Another quick question. The datasheet states that the ON resistance is approx. 16m ohms. Assuming that the load consumes 20A, does that mean that the voltage drop is of 0.32V?

Just to let you know, that chip is being phased out. They are already 2 iterations away from that chip.
Is the BTN7970 a better alternative?
 

Gibson486

Joined Jul 20, 2012
360
The BTN8962TA is the replacememnt. I am not sure they released it to distributors yet, but it is what the manufacturer told me. In the interim, I have used BTN8960TA.
 

Thread Starter

Dritech

Joined Sep 21, 2011
907
So, the equation becomes V(current) = current(out) / 85000 * R
Thanks for the reply. Is V(current) is output voltage when the flag is active? and I'm assuming current(out) is the current output from the pin not the load current, right?
 

Gibson486

Joined Jul 20, 2012
360
Yeah, the whole point is that you adjust it so your max current leads to you flag voltage. At least that is how I understood it. It worked when I tested it, but we have not really implemented it since...
 
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