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I am designing a board for my project which is powered from 4.2V Li-Ion Battery. I am using a buck-boost converter ADP2504 for powering the board which outputs 4.0V constant voltage. I have buzzer on my board for alarm applications. Whenever I turn on my buzzer I get a huge drop of 1.4V in my regulator output (attached is my drop waveform) and comes back to 4.0V in 33 microseconds. enter image description here

When I turn on buzzer, current shoots up by 200mA (buzzer current = 200mA). Question is

How to calculate a capacitor to handle this transient with the above input? Currently I don't have any capacitor in my board. Above observations are without capacitor.

If the calculated value is very high in 100's of micro Farads, how to handle inrush current caused by that?

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  • \$\begingroup\$ I=C*du/dt. Solve for C. \$\endgroup\$
    – winny
    Commented Mar 31, 2018 at 19:50
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    \$\begingroup\$ Alternatively, you could power the buzzer directly from the battery and not from the regulator. \$\endgroup\$
    – Wesley Lee
    Commented Mar 31, 2018 at 19:55
  • \$\begingroup\$ @winny, I=C*dV/dt is for instantaneous current through capacitor. Is that correct? I think here we have to find capacitor based on energy. It has to store enough energy so that during transient I can provide the needed energy. Please correct if I am wrong. Can you please elaborate your explanation by relating energy or stored charge? \$\endgroup\$
    – Jahir hussain
    Commented Mar 31, 2018 at 20:04
  • \$\begingroup\$ Yes, you are correct. If you integrate the formula, that’s where you end up. This isn’t the axiom, but for all practical EE work, this one and U=L*di/dt is. \$\endgroup\$
    – winny
    Commented Mar 31, 2018 at 20:20

2 Answers 2

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A correctly designed regulator based on ADP2504 should be able to supply 200mA indefinitely.

In general, every buck regulator has to have some output capacitance, but in many applications the capacitor does not have to be particularly large.

However, if your load increases abruptly, the regulator needs some time to ramp up the current in the inductor and, for that time window, the capacitor should be able to hold the line without significant voltage droop.

Based on your diagram, the regulator needs 33uS to catch up. Say, you want to limit the voltage drop during this time to 0.1V. Then the minimum capacitance could be calculated as C=dQ/dV=200mA*33us/0.1V=66uF.

You can reduce this time window and therefore the requirement for the capacitance by reducing the size of the inductor, but that would require more detailed analysis of the circuit.

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According to the datasheet you should see less than 100mV drop provided your regulator is operating under similar conditions with the reference design:

enter image description here

Check that you are using the recommended 22uF ceramic output capacitor, that it is actually something similar to 22uF at 4V, the recommended input capacitor, and that you are using one of the exact recommended inductors as shown in Table 5.

enter image description here

Also check that the input voltage to the regulator is not dipping. If it is, it could be because the inductor is inadequate.

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