I recently opened up a linear power supply and found out it has a resistor of 8.2M ohm between the primary winding (specifically in the neutral) and the secondary winding of the step-down transformer.

What is that resistor doing there?

I couldn't fine anything related on the internet.

  • 1
    \$\begingroup\$ There is a tiny parasitic capacitance between the primary and secondary coils. And because it's not much capacitance, it doesn't take a lot of built-up charge to yield quite a voltage. Usually, there is something on both primary and secondary sides to discharge (or otherwise limit the voltage build up on) their own coil's parasitic capacitance. But nothing in particular deals with the trace capacitance between primary and secondary. A bleeder resistor cheaply removes the problem if the circuit doesn't have another means. Whether it is the right thing to do is a designer's question, though. \$\endgroup\$
    – jonk
    Sep 2 '19 at 3:55

This resistor provides a leakage path for any charge that builds up on the secondary side.

If this leakage path was not provided then the secondary could potentially float (with respect to primary) to a large potential

  • 4
    \$\begingroup\$ Can you recommend me a book that talks about this?. How can a charge build up in the secondary respect to the primary? and why is it uncommon to see such resistors in transformers? \$\endgroup\$
    – MLuna
    Sep 2 '19 at 0:17

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