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Diode selection

Discussion in 'General Electronics Discussion' started by anroop, Dec 12, 2013.

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  1. anroop

    anroop

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    Dec 11, 2013
    Need to sort out diode (forward biased) for use in a 12v dc circuit. can a 1n4001 diode sustain this. it has 50v reverse breakdown voltage, 1v forward voltage. but no spec given on the max supply that could be given to its input ie., +12v to anode....
    :confused:
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    There is no maximum voltage at the output. As long as the forward current and the reverse voltage limits are honoured, you can do anything!

    Typically though, if you have a bridge rectifier, the maximum DC output voltage is going to be about 45V.

    If you use a single diode half wave rectifier to convert AC to DC, the limit is more like 20 volts.

    Note that the 1A average forward current is a very important rating! There are many other specifications for the device (look up a datasheet) but typically only a very few are important in any given application.
     
  3. anroop

    anroop

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    Dec 11, 2013
    I am sorry that I did not post my question completely!;
    I need to run a bipolar stepper motor using a driver ic. To prevent the flywheel current due to motor, I need to implement diode bridge, as I am unaware of diode's withstanding voltage I am asking you what diode(eg. 1n4007) can be used in 12v circuit.
     
  4. Raven Luni

    Raven Luni

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    Oct 15, 2011
    I'm fairly sure the 1N4001 forward voltage is closer to 0.5V. I tested a few recently with the only bad one reading 0V.
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,269
    Nov 28, 2011
    If you're referring to a diode connected "backwards" across each winding of a stepper motor to protect the driver IC from back EMF from the coil, the important characteristics are:

    1. Maximum reverse voltage. When the winding is energised by the driver IC, the diode is reverse biased. The full supply voltage appears across it in the reverse direction. The diode's maximum reverse voltage rating needs to be at least equal to the supply voltage, and preferably 50~100% higher than it.

    2. Continuous forward current. When the driver switches off and the magnetic field in the stepper motor coil starts to collapse, the "flywheel" effect causes current to flow through the diode. This current is initially roughly equal to the current that was flowing from the driver into the motor winding. The maximum value of this current can be calculated using Ohm's Law, from the supply voltage and the DC resistance of the winding. For example if the supply voltage is 30V and the DC resistance of each winding is 10 ohms, the maximum current will be 3A. The forward current rating of the diode needs to be at least this high, and preferably 50~100% higher.

    3. Switching speed is important, especially if the motor is running fast and/or efficiency is important. This is why I would not use a 1N400x diode. A fast recovery rectifier or a Schottky diode would be a better choice.

    This diode is typical for this kind of application. It's a Schottky diode, rated for 5A and 50V. http://www.digikey.com/product-detail/en/SB550/SB550FSCT-ND/3907898
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Beware that if you're using a bipolar stepper motor you may need to put the diodes across the transistors in your H bridge rather than across the coil(s).

    You can also use a snubber network.
     
  7. anroop

    anroop

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    Dec 11, 2013
    thanks krisblueNZ, that explained me exactly what I was asking. and steve, what is a snubber network? and in case of a unipolar stepper motor is it must to add freewheel diode s?
     
  8. mursal

    mursal

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    Dec 13, 2013
    If you use a freewheeling diode across the bipolar winding, the diode will conduct when the motor is reversed. This will short out the circuit.
    A typical snubber network has a resistor in series with a capacitor and connects across the coil (sometimes across the switch) allowing the spike voltage to short out, protecting the switch. Because of the capacitor is looks like an open circuit to DC voltage regardless of polarity.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    If you're using an H bridge, you can also place diodes across the transistors (BJT or mosfet).

    Rather than "shorting" the inductive spike, this redirects it to the power source (which is typically low impedance and can handle it without the voltage rising to a huge extent. Beware that this causes more noise on the power supply than a diode across the windings (which you may not be able to use).

    Yet another option is to place back to back zener diodes across the winding. These need to be of a voltage significantly higher than the supply voltage. There are several caveats for this (for one the dissipation in them is much larger).
     
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