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Ground loop question

  1. Jun 7, 2010 #1
    When the grounds are different at source and destination, how do you connect the signal source to your circuit?
    I know there is a difference in the ground potential of source and destination. But is this treated as noise or just an additional voltage that adds with the source signal?
    I thought we could use an isolation transformer or an opto-coupler. But how does the isolation transformer help?
     
  2. jcsd
  3. Jun 7, 2010 #2

    vk6kro

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    Take an extreme example.

    Suppose you have one car battery being used for two devices but the case of one device is connected to the negative terminal of the battery and on the other device the case is connected to the positive terminal of the battery.

    You can see that if you join the cases together, you would be shorting out the battery by joining the positive and negative terminals together. This may cause smoke and flames.

    This is the type of problem you can get, although it may be more subtle. It can result in no output from one device because the output is connected to the ground of the other device.

    If each circuit has an independent power supply, isolation transformers and optocouplers are common ways of linking them together.
     
  4. Jun 7, 2010 #3
    But I want to know how the isolation transformer helps.
    Say, there is a potential between ground 1 and ground 2 as their grounds are different.

    Assuming the potential is DC, then the isolation transformer will help. In case my signal is AC, I just connect it to the isolation transformer. This appears across the secondary of the isolation transformer. The DC ground potential won't.
    What if the ground potential is AC? Can it be AC?
     
  5. Jun 7, 2010 #4

    berkeman

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    AC noise will typically be AC Mains frequency, 50/60Hz. Hopefully your signal AC frequency is far away from this low-frequency noise.
     
  6. Jun 7, 2010 #5
    Ground loops often are the result of multi-path grounds between various sources and destinations. Stray ac (50 Hz and 60 Hz) magnetic fields can induce voltages in these low resistance loops, which in turn produces ac currents and IR drops in the various cables. One successful method of getting rid of these loop currents is to get a high-permeability (over 10,000) thin-lamination (~1 mil) toroidal core with about a 5-cm dia aperture and winding ~20 turns of signal cables (e.g., RG-58) or power cables through the aperture. This greatly increases the 60 Hz impedance of ground loops, without affecting the differential signal amplitude.

    Bob S
     
  7. Jun 7, 2010 #6

    vk6kro

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    If you had an isolation transformer connected to some circuit that had an AC voltage on it, the transformer would not pass this AC voltage unless the voltage was different at each end of the transformer winding.

    If both ends of the transformer winding were at the same AC potential, nothing would be passed to the other side of the transformer.
     
  8. Jun 9, 2010 #7
    yup! exactly.
    While we are discussing this, can we talk about common mode chokes.
    From what I understand, the current is common to both the lines and the B field from the common mode current at the choke is 2x. I don't understand how this will cancel/reduce the common mode current?
    The differential mode current flows only in one direction. so it's not affected by the choke?
     
  9. Jun 9, 2010 #8

    sophiecentaur

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    A good, though expensive and possibly 'old fashioned', way to avoid the problems caused by Earth loops is to send 'balanced' signals from place to place. You can do this using a signal (bal-un) transformer at each end (also referred to as a 'rep coil'), eliminating any 'common mode' differences between the ends of the transmission line. The signal arriving at the other end just consists of the wanted voltage and can, according to how well the transformers have been wound, be immune to almost any level of Hum which could be caused by variations of equipment ground voltages at each end.
     
  10. Jun 9, 2010 #9

    berkeman

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    The common mode current sees the inductance of the CM choke; the DM signal sees no inductance. You use the impedance of the CM choke as an impedance divider, to attenuate the CM interfering signal. You have to have some impedance to divide against, which is usually the CM input impedance of the receiving device. If the input Zcm of the receiving device is high, like around the CM impedance of the CM choke, then you will not get much attenuation of the CM signal.
     
  11. Jun 10, 2010 #10
    I thought it was just the CM choke. Never considered the input impedance of the receiver.
    But why do application notes have a CM choke at the input stage of an op amp, particularly instrumentation amp or ADC driver.
    Op amps have huge input impedance. It would make the choke useless.
     
  12. Jun 10, 2010 #11

    berkeman

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    Correct. If the input impedance is high, adding series impedance does very little to attenuate the signal.
     
  13. Jun 10, 2010 #12
    Last edited by a moderator: Apr 25, 2017
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