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INA169A, shorting the Rs resistor

Dan Wolfe
Prodigy 80 points
Other Parts Discussed in Thread: INA169

I am working on an application to sense when some series connected switches operate.  We've wired the shunt resistor, Rs, in parallel with the contacts and when they open, a voltage is seen at the Vout.  When the contacts are closed, Rs is shorted.  This measurement circuit works for us but we are failing the INA169A current shunt monitor.  Sometimes Vout is stuck a zero volts, other times it seems that it is producing only half output.  Is there an issue with shorting the Rs on this component?

  • 0
    TI__Expert 4595 points
    Hi Dan,
    Can you provide a full schematic of the configuration you are describing. What gain resistor (RL) do you use for the INA169? What current do you expect through Rs? Also please try measuring the input voltage at the inputs of INA169 and note the output of the device. If this is already done, please let us know the results.
  • 0 in reply to Rabab Itarsiwala
    Prodigy 80 points

    Here are some of the circuit parameters:

    RL is 10k (stock on board)

    RS is 10  OHM

     ~120 ma   (.12 amps x10 = 1.2 v)

     .125 volts across RS (drop)

     Input to board 12v

     Output = 1.244 v

    We have seven INA169's wired in series (connecting Vin to Vout).  There is a set of contacts across each Rs. When the Contact opens, the INA169 sees the current and provides an output.  All seven of these contacts are driven by the same prime mover.  With the INA169, we are able to see when each contact opens (by monitoring Vout) and check the contact synchronicity (time from when the first one breaks to when the last one breaks).  This works well for us but it seems to be tough on the INA169.  Maybe it is somehow bad if Vin and Vout are shorted?

  • 0 in reply to Dan Wolfe
    TI__Expert 4595 points
    Hi Dan,
    You mentioned 1.2 V drop across RS when 120mA current flows. Is the expected drop 0.125V when the switch is closed? Also when you say seven INA169s in series, is the output of the first goes to the input of the second and so on? Is you supply and common mode both at 12 V? It would be helpful if you could provide a schematic, even a hand drawn one will be ok.
  • 0 in reply to Rabab Itarsiwala
    Prodigy 80 points
    The INA169s are actually connected in series by Vin+ and Vin- so the shunt resistors are in series with a switch in parallel with Rs; I mistakenly wrote Vout in my last posting instead of Vin-. There is a 0.125V drop across Rs, not a 1.2V drop. 1.2V is Vout after the amplification. Supply and common mode are both at 12V.
  • 0 in reply to Rabab Itarsiwala
    Prodigy 80 points

    I've attached a PPT file that shows how the INA169s are connected.  There are seven in total.  The first and the last are shown and one in the middle of the chain.

    INA169 circuit.pptx

  • 0 in reply to Dan Wolfe
    TI__Mastermind 26465 points
    Hi Dan,

    Rabab is out today and I will be helping as well. In the circuit diagram you have all the Rshunts in parallel and the contact in parallel. So if you close one contact it will create a short on all inputs. Is my statement correct? I just read the first statement and the see the diagram different and I want to make sure which one is correct.

    Javier.
  • 0 in reply to Javier Contreras4
    Prodigy 80 points
    Javier,
    It seems that my diagram is not so clear. There are seven INA169s, each having a switch in parallel with the Rs. If the switches are open, current flows through the shunt resistors and there is about 1.2V output seen on Vout of each INA169. If one of the switches that is in parallel with an Rs is closed, the current no longer flows through Rs but rather through the closed parallel contacts. This action causes the Vout of that particular INA169 to drop to nearly zero volts while the rest stay at 1.2V on the Vout pin. As other contacts close their outputs change. Each of the seven Vout lines are wired to AI channels on a DAQ card so the time between the contacts closing and opening can be monitored by looking at the output voltage of each of the INA169s. This circuit functions well for us but we are experiencing problems with the INA169 failing. Sometimes the failure causes Vout to stay at 0V regardless of the current flowing through Rs. Other times the Vout is not what is expected (about half voltage) for the same current in Rs. We can replace the chip and the circuit works again.
    We feel that the INA169 might not like having Vin+ and Vin- connected by the closed contacts instead of a small shunt resistance. Do you know the lowest resistance value that can be used for Rs?
  • 0 in reply to Javier Contreras4
    Prodigy 80 points

    I see what you mean about the diagram.  The Rshunts are actually connected in series, the diagram was incorrect.  I've attached another diagram that should be closer to what actually have in our circuit.  Sorry for the confusion.

    INA169 circuit-revA.pptx

  • 0 in reply to Javier Contreras4
    TI__Expert 4595 points
    Hi Dan,
    It shouldn't be an issue to short INA169 inputs. Depending on the current and on-resistance of the switches, the output will be at a certain level. Are the devices getting damaged? Do they intermittently produce the expected results?
  • 0 in reply to Rabab Itarsiwala
    Prodigy 80 points
    We usually get the expected results but we have had about eight INA169s fail where the Vout is stuck at zero volts. Just recently, we had four units fail where they would output about half the expected voltage. When we change the chip, everything works again. If shorting the Vin+ and Vin- is not bad for the INA169 we keep looking for the source of the trouble.
  • 0 in reply to Dan Wolfe
    TI__Expert 4595 points
    Hi Dan,
    If the units are getting damaged, you might want to check if there are transients at the inputs that are exceeding the maximum rating of the device during the switching. Also if the input current is 120 mA across 10 ohm shunt then the input voltage when the switch is open should be 1.2V and the output for a gain of 10 would be 12 V. Is it possible to capture a scope shot of the input voltage.
  • 0 in reply to Dan Wolfe
    Prodigy 165 points

    Hi,

    May I know how did you calculate the Rs Value and you got 10 ohm ? what is the maximum current did you require to sense ?

  • 0 in reply to user4353247
    Prodigy 80 points

    In this case, the value of Rs is a somewhat open.  All that is required is that the output voltage of the current monitor changes significantly when the series contacts open.  So the minimum criteria is that we can see a difference between when the contact that is in parallel with Rs opens and closes.  The maximum values should prevent the output voltage to exceed the maximum allowed by the DAQ card which in this case is 10V.  We chose the lowest reasonable Rs so that the overall circuit resistance does not change much as the contacts open and close.

  • 0 in reply to Dan Wolfe
    Prodigy 165 points
    I already understand the method to determine the Rs Value. I have used in few application current shunt monitors yet I have not much experienced at the industrial level in the analog design. Usually, I see the Rs Value must be very low ( milli ohm range) and I wondered when I read that you are using 10 ohm which is little far from milli ohm range. Therefore, I would like to know the method you used to chose the shunt resistor.

    Thank you.
  • 0 in reply to user4353247
    TI__Expert 4595 points
    Hi user4353247,
    Please watch this video to learn more about choosing the Shunt resistor.
    training.ti.com/getting-started-current-sense-amplifiers-session-4-how-choose-appropriate-shunt-resistor
    If you are interested to learn more about Current sensing please look at our video series here:
    training.ti.com/getting-started-current-sense-amplifiers