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My PCB must be able to connect to and power another PCB. I found that when I first connect them the other PCB draws enough power (peak of about 60mA) to dip the voltage on my PCB for about 80us and causes my microcontroller to reset.

I want to add a capacitor (or a few) to my PCB to keep this from happening.

Any ideas on how to calculate the right capacitor value?

See picture for an oscilloscope trace of the voltage drop:

enter image description here

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  • \$\begingroup\$ It appears your 3.3V regulator has poor load regulation. Why? what is it? What Cap is used on output and it's ESR? \$\endgroup\$
    – D.A.S.
    Commented May 8, 2021 at 0:33
  • \$\begingroup\$ I'm using the TLV75533PDBVR regulator. I'm using a 0.1uF and 1uf capacitor on the output. Both are XR5, +-10%, 10V ceramic capacitors \$\endgroup\$
    – Ryan W
    Commented May 8, 2021 at 0:47
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    \$\begingroup\$ It's 60mA surge when plugging in...how much current does it draw normally? If it's relatively low you could use a series resistor. That would limit the current surge and thus the voltage droop. If nominal is 10mA and you use a 10 ohm resistor, you'd get 10×.01=0.1V drop across it. That wouldn't affect most loads. \$\endgroup\$
    – Kyle B
    Commented May 8, 2021 at 1:04
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    \$\begingroup\$ Another option is decouple the CPU power from the load. To do this you'd isolate CPU VCC with a diode and have a decent cap between it and the CPU. When the regulator droops, the diode would prevent that new local cap from discharging backwards and maintain VCC at the CPU \$\endgroup\$
    – Kyle B
    Commented May 8, 2021 at 1:06
  • \$\begingroup\$ I really like the add a series resistor idea since it is something I can solder on to my existing PCB. The load only draws about 5mA during normal operation(the 60mA on start up is it filling up all the capacitors), so the voltage drop from a 10ohm resistor would only be 50mV \$\endgroup\$
    – Ryan W
    Commented May 10, 2021 at 4:17

1 Answer 1

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The load regulator error is simply a resistance ratio.

This also applies to the transient brownout.

The TLV75533PDBVR in the DBV package has Load regulation spec = 0.060 V/A which means it has an output impedance of 60 mohms The 1 uF ceramic cap could be 10 mohm more or less.

Analysis

The output time constant of the source caps would be 60 mohm*1uf = 60 us which isn’t 63% of the voltage swing (58%) but close enough. This means your load has a similar or slightly better ESR and a slightly smaller capacitance.

So your load if it were the same ESR, the voltage dips 50%. if your source is 10x lower ESR, or 6 mohms , you might have a 5A supply or 10 caps in parallel or add a series resistance 10x bigger to the switch if you don’t need such load regulation as the 10 mOhm 1 uF cap.

recommendations

  • It depends on your voltage error tolerance, but to reduce 50% to 10% you need 5x 1uF caps in parallel. (?) Will a bigger 5uF cap work or a 10uF cap.? ...
  • only if the datasheet says it has a proportionally lower ESR. But unlikely with the same size, so then it has to be 5x the physical size at least and spec the lower ESR
  • a better solution is to use an LDO in the auxiliary PCB from 5V and send an enable signal.
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