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At the moment when I insert the USB cable, at the output of the converter, 5 volts appears for a short time. Why is this happening?

Circuit: enter image description here Voltage Oscillogram:

enter image description here

BUT! When I connect the ground oscillograph to the board, this effect disappears. The ground of the oscillograph and the computer are connected in an electric outlet.

This effect is a measurement error or an error in the circuit?

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  • \$\begingroup\$ what is the time scale on that plot? \$\endgroup\$ – ratchet freak Oct 6 '17 at 14:19
  • \$\begingroup\$ Where is your scope ground connected when you see the issue? I.e. before you connect the ground to the board and the effect disappears? In general, these parts to not behave as you describe, so I suspect some kind of measurement error, but hard to say without more info. \$\endgroup\$ – John D Oct 6 '17 at 14:21
  • \$\begingroup\$ John D, the GND of the USB and the GND of the Oscillograph are a short circuit. I checked this with a digital multimeter \$\endgroup\$ – AndreyB Oct 6 '17 at 14:30
  • \$\begingroup\$ "what is the time scale on that plot?" 400 μS in cell \$\endgroup\$ – AndreyB Oct 6 '17 at 14:34
  • \$\begingroup\$ If the ground impedance was actually 0 between source and scope, there would be no difference, so in fact it is not 0 ohms including ground inductance and probe coax capacitance(est 100pF)//10M= 1us) \$\endgroup\$ – Sunnyskyguy EE75 Oct 6 '17 at 14:44
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Knowing the time scale would be useful.

  1. Using two channels connect scope ground to power supply ground NOT board ground) and monitor Board_+5V and board_Ground as the USB plug is inserted. Odds are they do not both make contact simultaneously. If the difference is about as long as the 5V pulse you see on the 3V3 output it explains your problem.

  2. Ground the scope ground near the IC ground.
    Ensure that the system ground is properly connected. If connecting scope ground to board_ground or a wire from PSU_ground to board_ground before the USB plug is inserted cures the problem then as in 1. you probably have a non simultaneous power supply connection.Note that Vout relative to board ground may be ~= 0V rather than 5V during the pulse period.

  3. Only if the above do not cure the problem then what you are seeing MAY be real.
    The IC regulates at 3V3 when the R15/R16 divider indicates that correct Vout is present. Depending on the time scale involved, the capacitance & time constant of the feeback circuit, and the energy stored in L1 compared to the load power, the output may exceed the correct value until the IC 'comes into regulation.

A faster transient response may be achieved by placing a small capacitor (start with say 1 nF) across R15.

Initial overshoot is more likely to occur with zero load.
Is there any load connected?
If not, try adding a modest load to see the effect (if any).

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  • \$\begingroup\$ "Knowing the time scale would be useful." 400 μS in cell \$\endgroup\$ – AndreyB Oct 6 '17 at 14:32
  • \$\begingroup\$ As a load microcontroller + several digital IC \$\endgroup\$ – AndreyB Oct 6 '17 at 14:40
  • \$\begingroup\$ Probably you are right, since this effect is unstable and the pulse width of 5 volts always varies. \$\endgroup\$ – AndreyB Oct 6 '17 at 14:46
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    \$\begingroup\$ Thank you. This effect was actually caused by the fact that the power wire is connected first. \$\endgroup\$ – AndreyB Oct 6 '17 at 14:53
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It's possible that the ground made contact slightly later so only the V5 line is connected to the board but and the ground if floating. Then it's not surprising that the voltage equalized across the entire board because there is only a single connection.

When you connect the ground of the oscilloscope to the ground of the board it closes the circuit.

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  • \$\begingroup\$ +1 I covered a range of maybes. You managed to target the correct one :-) \$\endgroup\$ – Russell McMahon Oct 7 '17 at 9:32

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