DC current transducer

Discussion in 'General Electronics Chat' started by ectmlk, Feb 12, 2008.

  1. ectmlk

    Thread Starter Member

    Feb 11, 2008
    21
    0
    Hey Forum
    I've got a number of dc current transducers for PCB mounting, i want to
    use them to replace resistor shunt circuits. These transducers are open-loop and based on the hall-effect sensor. From the block diagram it appears that an in-amp is used to condition the signal from the hall sensor. Depending on how the transducer is configured you can get 30mV/amp or 60mV/amp.

    First question is if any one here has experience of such devices whats your opinion of them and would you have any recommendations?

    2. The sensor has a Vref input (probably to the in-amp), what is the purpose of this, what advantage does it offer to me and how should i make effective use of it? I plan to use a 2.5volt reference IC anyhow. The Vref does put an offset on the output i just dont get why this would be required? I think it may be that i can input an AC wave thus this Vref will offset it so that i get the wave swinging around 2.5V rather then 0 (the transducer can only be supplied with +5volt.

    3. Some of the specs include RL = Load Resistance >= to 2kohm
    Rout = output internal resistance < 10ohms
    How should i interpret this data? Should i take RL been the minimum load i should place on the output? Rout not relevant?

    4. With 60mV/Amp and a typical range of 0-10Amps i get approx 0-600mV signal, i need to input this to an ADC, should i simply gain this, buffer it and feed straight in to the ADC?

    Any comments appreciated!!

    mlk
     
  2. kender

    Senior Member

    Jan 17, 2007
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  3. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Wow, lot's of questions :D

    Be careful with hall sensors when replacing shunts, they do not offer superior performance. Obviously, they do a nice job of isolating and stop resistive losses. You need to respect their bandwidth, which may or may not be suitable for replacing a shunt.

    You've got it right with the instr. amp, the ref pin just offsets. If you take a look at what makes an instr. amp, you'll see that quite clearly. I'm not sure what you mean by the RL and stuff, I'd really have to read the datasheet myself.

    You should try to match your sensor to your current range, it gives the best SNR and sensitivity. You should amplify the signal close to the sensor with the necessary gain, this will give some noise immunity along the path to the ADC and will make better use of your ADC resolution. So, roughly a 6.8 V/V gain for use with a 4.096V reference.

    Consider adding a pole to the amplifier to your ADC, as it will minimize noise a bit. This pole would be chosen to match your maximum bandwidth of your sensor or the maximum response you want to have.

    Steve
     
  4. ectmlk

    Thread Starter Member

    Feb 11, 2008
    21
    0
    Cheers for the replies

    At the moment my current range can vary from 0-10A or 0-40A, this is why i'd like to go with these devices because i can pretty much use the same device for 0-10A or 0-50A. Hopefully on the PCB i will be able to configure it for either the 0-10A range when needed or configure it to be 0-50A when needed.

    If iam only interested in the DC current would i really need to be concerned about bandwitdth? I get the impression that a resistive shunt is a much better way to measure DC Current from the app notes and scubasteve however i think with my range of current a DC Transducer is more favourable?

    I think its information i need to match the in-amp to say another op-amp or ADC? I have attached the datasheet!

    i just look at tieing it to ground?

    A pole? Meaning a lPfilter?

    Thanks
    Mlk
     
  5. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Mlk,

    The bandwidth determines your response time, which may or may not be a concern for you. If it isn't critical and you are fine with a ~10Hz response, then you can configure a low-pass filter for operation at 10Hz. You can see from the noise specs that as frequency goes up, so does noise. If you use a filter, you cut off this noise and will have better resolution. Even if you wanted to run this at full bandwidth (50KHz), then you can use a lp at that frequency to improve operation.

    Do you mind saying what your application is for? That way one could recommend this measurement technique or not.

    Steve
     
  6. ectmlk

    Thread Starter Member

    Feb 11, 2008
    21
    0
    Hey
    The transducer monitors the current on a device that has similar chracteristics to a resistor of about 3ohms. It should be purely a DC current!!

    I will implement a LPF thanks for the suggestion!! ;)

    Regarding the Output internal resistance Rout, do i need to match the gain circuit to this value or will the op-amp simply have a high enough input impedance?

    Thanks
    Mlk
     
  7. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Instrumentation amplifiers, if used, have remarkable input impedance. The only time you really need to worry about input impedance is for an inverting amplifier.

    I'm confused about the Rout and RL specifications though. If you use an inverting stage, ensure that you do not use less than 2k resistors for Rin.

    Steve
     
  8. ectmlk

    Thread Starter Member

    Feb 11, 2008
    21
    0
    Why is it that i would only need to worry bout that while in the inverting mode? (If you can explain something like that in a short sentence? :) )

    I will try to implement the LPF and gain, if i have any trouble i will post a circuit!!

    Thanks again
    Mlk
     
  9. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    For non-inverting mode, you directly connect the input signal to the opamp's + terminal, which is extremely high impedance. An instrumentation amplifier is composed of two non-inverting amps on the inputs and then a differential amplifier fed by these two amplifiers.

    For inverting, you are connecting the input through a resistor and the node past the resistor is the feedback point. This means that the input resistor connects between your signal and virtual ground, thus the input impedance seen by your amplifier.

    Steve
     
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