1
\$\begingroup\$

I have a circuit in place for the Linear Technology LT3502A switching regulator (datasheet here). To prevent blowing up the whole thing during hot-plugging, I have chosen to add a small parallel ceramic cap and series resistance instead of an unnecessarily large electrolytic. As so:

Schematic

The LT3502A is the 3V3 output version.

I shall be using 100V rated ceramics (1206 size) (or perhaps 50V is also okay, to meet the maximum input voltage of 40V for the chip).

I was just wondering what size resistor I should slap on there. I suppose I should be fine with an 0603 (0.125W)?

My logic being that the maximum output power of my chip at full load (which it will never reach) is 3.3V*0.5A = 1.65W. Multiply that by say, 2, to account for power losses and an additional safety multiplier (~ 3W). Presuming VBUS >= 12V (not going to put a 1W resistor there), the worst-case current through R7 is 250 mA, implying 0.0625W dissipation. An 0603 rated for 0.125W should be okay, I think? Just making sure not to release any magic smoke :)

\$\endgroup\$
  • \$\begingroup\$ I'd look at direct connection for VIN, and have an RC network on the ~SHDN pin so that startup is delayed, but perhaps I misunderstand the problem you are trying to solve. \$\endgroup\$ – Jasen Mar 25 '18 at 0:58
  • \$\begingroup\$ @Jasen the problem is a resonant circuit when hot-plugged, which can cause the input voltage across the ceramic cap to even double - destroying both the cap and the regulator. The solution is either an electrolytic, or this. \$\endgroup\$ – Shreyas Mar 25 '18 at 9:46
0
\$\begingroup\$

First of all it should be a flame-proof resistor. Based on a maximum 250 mA load 1 ohm would be 250 mV * 250 mA = .0625 watts, so 2 ohms is 125mW.

Usually one would try for 4.7 ohms or 10 ohms, but that would be a through-hole resistor, so it looks like 2 ohms is your limit for a 0603 size resistor.

You can stack resistors, such as two 4.7 ohms stacked, but that is your finite limit.

\$\endgroup\$
0
\$\begingroup\$

The resistance has to be enough to critically damp the system. so the optimum answer depends on the length, impedance, and quality factor of the transmission line between the voltage source(s) and the capacitor.

A larger ceramic capacitor may be able to swallow all the excess energy available from the supply without an excessive voltage excursion.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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