# When an open-drain output asserts low, where does the line capacitance discharge?

I was looking at the auto-retry functionality of the TI part TPS2553 (datasheet http://www.ti.com/lit/ds/symlink/tps2553.pdf) and came across something I didn't understand and thought it a good opportunity to ask a general question.

I am going to refer to the picture below to ask this question using it as an example but really I am unsure about this in any situation where you have a line pulled up with a resistor and connected to an open drain output with some similar capacitance.

Referring to the picture below, when the part senses an overcurrent condition, the FAULT pin pulls low. My question is, when this happens where does Cretry discharge to? The only place I could see it discharging is directly into the FAULT pin. However there is no resistance between Cretry and that pin and the FAULT pin specifes a 10mA max current from the datasheet. So wont there be a big current spike when it discharges since there is nothing to limit the current, and this would damage the pin as it's outside the recommended sink current for the pin (datasheet actually specifies absolute max of 25mA sink current for FAULT)? Am I wrong in my understanding about how/where the cap discharges, the fact that the current spike exists assuming I am right about the previous point, or am I misunderstanding the significance of said discharging current spike if I am right about previous 2 points?

Again I am not just asking about this specific case but using this as an example to ask this question. I wonder too about I2C lines. When the I2C driver pulls the line low, where does the charge that was on that line discharge to?

• currents travel in loops from an internal switch (Fault_) to the return path of ground, dissipating in the switch. – Tony Stewart EE75 Sep 17 '17 at 1:51
• The switch ( drain when closed to source =gnd) has low enough resistance to discharge the cap quickly. ( usually around 50 ohms for 74HCxx) – Tony Stewart EE75 Sep 17 '17 at 2:07
• But spec actual says VOL Output low voltage, FAULT I/ FAULT = 1 mA 180 mV thus 180 ohms (Max) – Tony Stewart EE75 Sep 17 '17 at 2:10
• You are right. The cap discharges directly into the FAULT pin. In this case, since the vendor is telling you to do it, it is probably OK. In other cases, you normally would not put 0.1uF on an open collector output. If you did, you could also insert a small resistor (10 to 100 Ohms) between cap and output to limit the current. To be honest, there is not much energy in a 0.1uF cap charged to 3 or 5V or whatever. All the energy will be dissipated as heat by the open collector output. But it may not be enough energy to destroy the internal structure. – mkeith Sep 17 '17 at 2:12
• Yes but not ONLY limiting the current. The other point is controlling where the dissipation occurs. The cap internal series R, for a ceramic cap, is negligible. So the energy stored in the cap will be dissipated between the external series R and the internal structure of the open collector output. The bigger the external resistor, the larger the percentage of energy will be dissipated in the external resistor. Hope that makes sense. – mkeith Sep 17 '17 at 2:22

The charge on $C_{RETRY}$ goes to ground through the $\overline{FAULT}$ pin.
The absolute max current specification on that line specifies a continuous current. This puts a lower limit on the size of $R_{FAULT}$.
The energy on $C_{RETRY}$ ends up being dissipated as heat in the pin driver transistor. This is a small and limited amount of energy which has to be distributed in the thermal mass of the transistor, unlike the energy due to a continuous current which is unlimited, so has to be dissipated.