5
\$\begingroup\$

Basically I have two questions related to decoupling capacitors:

enter image description here

  • In the circuit of the image you can see a simplified representation of a decoupling capacitor in parallel with a power supply, but how can I calculate the initial current to charge the capacitor?

  • Theoretically I_ch should have a very large value, since adding the trace resistance from the power supply to the capacitor with the ESR would be less than 1 Ω, but I will use a battery as a power source and I would like to know some way to limit the current charge of the decoupling capacitor so as not to consume the battery so quickly.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ How would charge current "consume" the battery? \$\endgroup\$ Mar 1, 2023 at 17:46
  • 1
    \$\begingroup\$ What one might be concerned in regard to the capacitors is not the initial charge current, but the leak current (esp. in electrolytic capacitors). While the leak current is small, it flows continuously and can leech the battery even when the device is not used. \$\endgroup\$
    – fraxinus
    Mar 2, 2023 at 7:35
  • 2
    \$\begingroup\$ @WalterPH: Are you concerned about the high current draw during initial charging of the capacitor? I think with most power sources and capacitor values it’s too short (and the resistances in the system too large, relatively speaking) to worry about. \$\endgroup\$
    – Michael
    Mar 2, 2023 at 10:05

4 Answers 4

4
\$\begingroup\$

I will use a battery as a power source and I would like to know some way to limit the current charge of the decoupling capacitor so as not to consume the battery so quickly

It makes no difference how quickly or slowly you charge the capacitor via a resistor. And it doesn't matter whether it's a trace resistance, a real resistor component or the internal resistance of the battery because, the energy put in from the battery/source is \$C\cdot V^2\$ and, the energy acquired by the capacitor is only \$\frac{C\cdot V^2}{2}\$.

In other words, 50% of the energy is burnt each time you charge the capacitor up to the battery voltage from a discharged state.

There are more sophisticated ways (using a buck-boost converter) to save that wasted energy but, it's possible that the small inefficiencies of the converter (over time) will outweigh the energy savings you might make.

how can I calculate the initial current to charge the capacitor?

Use the battery's internal resistance (effective series resistance or ESR) and the ESR of the capacitor to estimate the peak current.

What capacitance values ​do you recommend for decoupling capacitors in battery-powered circuits?

It's purely down to how much "instant" energy is required for each chip or sub-circuit. Decouplers are used to provide localized short-term energy dispensers thus avoiding long power lines carrying these currents. This reduces EMI (due to long trace inductance) and localized short-term voltage droop that might disrupt circuit functionality.

\$\endgroup\$
6
\$\begingroup\$

To the question in topic, you will use a capacitance value high enough to keep your circuit working properly under all normal conditions you need it to work, such as DC, AC and pulsed load. And you need to take some margin to allow for full and almost empty batteries, temperature etc. So there can be no recommendations unless you know what you need.

For the other quesrions:

  1. With that circuit, in ideal conditions the current will be infinite and the capacitor will charge in zero time to battery voltage. In real world, batteries, wires and capacitors have resistance, so it will be limited by the resistance. Which is unknown, as there is no description of what batteries, wiring, or capacitor you will use.

  2. There is actualy no question, but it really does not make any difference if the capacitor is charged via 0.1 ohm or 0.01 ohm or what the initial current is, the current will be zero after capacitor is charged. Or would be, as there will always be some non-zero leakage current through the capacitor. Any added resistance would just waste energy by dissipating power due to the current drawn by the actual load passing through any added resistance.

\$\endgroup\$
5
\$\begingroup\$

Decoupling caps are used because of wire (or trace) inductance and resistance. Currents traveling through the parasitic impedance of the wire lowers the voltage at the load which can become problematic.

It really depends on the wire length and the switching current. If the wire length is short, and the current isn't a large proportion of the maximum battery current, then you could just use the battery as there wouldn't be a lot of ripple.

If you are switching a heavy load and have a long wire, you'll need a higher valued capacitor.

So lets say you have a switching load of 100mA and a wire resistance of 1Ohm, you'll get a 100mV drop on the wire (plus whatever the inductance is).
A switching load 100mA and a wire resistance of 0.05Ohm, you'll get a 50mV drop on the wire (plus whatever the inductance is). You could use an RLC equation to estimate what the drop would be with frequency.

\$\endgroup\$
3
\$\begingroup\$

Batteries and capacitors both have values of ESR which combined with wire or load inductance and the risetime of desired current determine your noise voltage spectrum.

To design this for a radio demands less noise than a motor.

Basically if you need a net ESR lower than the battery at some range of f then C tends to decrease in value from uF to pF. Thus your options might be 1, 2 or 3 spaced caps values over thus range. Each design must carefully consider all sources of ripple and what is required. So that there is no one fix-for-all. Low ESR e-caps tend to be 1% of general cheap ones.

By running the return wire twisted with the supply wire, the common radiated pulse noise induced in sensors are reduced by partial cancellation. Sometimes ferrite beads of a small size might perform as well as 100 pF. These too may be considered in this high Q LC filter to dampen the shape by adding AC resistance.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.