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I use this circuit as my device power supply:

non isolated SMPS

Here I have R2=5.6K as my SMPS load but at this load I have about 10KHz switching frequency which leads to an annoying audible noise, probably from L1 (the inductor.)

I tried to increase the output current by reducing R2 to 1.2K and I got more than 20KHz switching frequency which eliminated the noise from the human hearing range, but in this particular application the current consumption has to be as low as possible.

How can I reduce the noise without increasing the output current consumption?

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  • \$\begingroup\$ What does the switching frequency have to do with the load ? Long story short: just don't use audible range switching frequencies. \$\endgroup\$
    – tobalt
    May 11 at 5:54
  • \$\begingroup\$ Do you have that thing in an insulated box? It is not isolated, so despite having 12V and 5V outputs, those outputs could be at mans voltage. That circuit could kill you or destroy any grounded devices you connect to it. \$\endgroup\$
    – JRE
    May 11 at 5:54
  • \$\begingroup\$ Try to increase the C4, the capacitor between FB and VSS, to something like 1u or higher. \$\endgroup\$ May 11 at 8:20
  • \$\begingroup\$ thank you guys for your replies, the device is in a insulated box with a glass surface (it's a wall touch switch) \$\endgroup\$ May 12 at 3:34
  • \$\begingroup\$ I will try increasing the C4... would you please explain how will this work? \$\endgroup\$ May 12 at 3:36

1 Answer 1

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The HT7A6322 has a fixed 60KHz oscillator - it shouldn't change depending on load.

I think the problem is that your load is very light. It is drawing almost no current, causing the regulator to go into burst mode to keep the output regulated.

It runs at 60KHz for a bit, then switches off until the voltage drops, then runs at 60KHz again. Repeat constantly.

The bursts occur at a rate that depends on the load. The inductor likely vibrates in response to the bursts rather than the switching frequency.

To make it quiet, you could try a different inductor. You could also just live with the higher current consumption needed to keep the regulator out of burst mode. Your 1.2k load resistor only draws 10 mA amperes of current, wasting 0.12 watts of power.

Alternatively, find a regulator made to work with extremely low loads.


Please do remember to keep that circuit in an insulated housing with no user access to metal parts from the outside.

Your circuit is not isolated. There may be only 5V and 12V between the outputs and the circuit ground, but the output ground can be at line voltage depending on how the outlet and your power plug are wired.

That circuit can kill anyone who touches any conducting part of it.


There is a design for an isolated power supply in the datasheet. Consider using it instead of your unisolated circuit.

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  • \$\begingroup\$ thank you JRE for your reply.. this device is actually a wall touch switch and should consume as less energy as possible in standby state...cause it is always plugged. the ac current consumption is equal to about 4mA which is a little big for a wall switch.. I will try a different inductor for the noise. \$\endgroup\$ May 12 at 3:42

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