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Ideal Diodes - WTF?

Discussion in 'General Electronics Discussion' started by a4596415, Sep 15, 2016.

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  1. a4596415

    a4596415

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    Jun 27, 2014
    I have been having a discussion on another forum on how to produce an ideal diode using a MOSFET. In my case I have wanted to use a P-channel, but N-channel FETs share the same problem of a body diode which allows current flow in the revers direction, thus only any use if the polarity remains the same?

    I appreciate it is possible to get FETS which don't have the body bonding and therefore can avoid the body diode effect but how does a ideal diode controller such as the LTC4352 use a common N-Channel FET and yet operate as a near ideal diode?

    I am sure that there is an obvious explanation, I just have not found anyone who can tell me what it is!
     
    Last edited: Sep 15, 2016
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    It's a mattter of orientation (connection) of the MOSFET. Source is connected to the input (positive) which requires a gate-source voltage higher than Vin. The LTC4352 generates this voltage using an internal charge pump.
    A common method, also used e.g. by the LM5050.
    This way the body diode is oriented correctly, anode to Vin, cathode to Vout and is bypassd by the MOSFET being "on" to reduce power dissipation.
     
  3. a4596415

    a4596415

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    Jun 27, 2014
    Got-ya, I knew it would be simple. Much appreciated!

    With that in mind would you say the attached circuit would work to eliminate the potential difference between two charging pins when a charger is removed. (I'm having issues with electrolysis). The principal of operation as I understand is:

    1. With the charger disconnected and the uC output to Q1 low, R3 (high Ohm ~MOhm) brings the gate of Q2 up to +Ve cell, and +Ve charge tab is brought to ground through R5.

    2. When the charger is connected through -Ve and +Ve charge the uC senses the increase in voltage at +Ve Charge and switches Q1 base high. Q1 brings the gate of Q2 low with respect to Q2's source and so allows current to flow from drain to source, bypassing the body diode?
     

    Attached Files:

  4. Harald Kapp

    Harald Kapp Moderator Moderator

    11,642
    2,690
    Nov 17, 2011
    Your circuit looks plausible.
    An issue may exist with R4, R5: As it is, The sense input of the µC will see the full "+Ve charge" with the charger connected. Depending on the voltage this may damage the µC.
    I suggest you use R4, R5 as a voltage divider with the "hot" end connected to "+Ve charge", the tap connected to the µC's sense pin. Adjust the divider such that with max. "+Ve charge" the input voltage to the µC does not exceeed Vcc (2.5 V). This setup will at the same time limit any residual current into the sense pin. Meanwhile, the lower leg of the divider will still hold sense at 0 V with "+Ve charge" disconnected.
     
    a4596415 likes this.
  5. a4596415

    a4596415

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    Jun 27, 2014
    Thank you very much for your input!
     
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