555 monostable configuration: why shorting a charged capacitor is not a concern?

Question

In the typical monostable configuration, 555 timers are shown with the discharge pin (7) connected directly to the capacitor. The idea is that, when the output pin (3) goes low, both sides of the capacitor are grounded, so the capacitor is basically shorted (through the internal transistor in the 555 chip). The capacitor discharges instantly, and this is taken to be a good thing, as it means that the whole circuit is instantly ready for the next rising edge on the trigger pin (2).

However, the idea of shorting a charged capacitor strikes me as highly undesirable. It should produce a nasty current spike which could potentially damage the transistor inside the 555 chip. I know that capacitors typically used with 555 chips are very small and current spikes from them should be feeble. However, given the popularity of the 555 chip I can only assume that there is a crowd out there playing with (and abusing) 555 timers, and that certainly includes a fair share of large capacitors in monostable configuration. I'd expect to see reports of people blowing out transistors, and warnings about some maximum safe capacitor value. What I find is not a single report or warning, and everybody assuming that shorting a capacitor in this way is always safe regardless of the capacitor value.

So, my question is: why is it not a concern to short a charged capacitor?

Edit: I didn't know there are different types of 555 timers. While my question is probably applicable to all types, the 555s I'm used to are NE555 with bipolar NPN transistors.

Answer 1

There are limits to everything. Capacitors also have a resistance internal to them (ESR) preventing instantaneous discharge. The transistor in the 555 was designed for this task and appropriate die area was allocated to it. Read note 6 on TI LM555 datasheet Feb 2000-Revised Jan 2015. It states: "No protection against excessive pin 7 current is necessary providing the package dissipation rating will not be exceeded".

Answer 2

Simply put, it is not a direct short, and the chip is just designed to handle it.

The current spikes may be large, but on the other hand, the discharge transistor is not driven very hard.

In steady state, the currents and voltages of the discharge pin indicate that the output impedance is in the order of 10-20 ohms.

And since a BJT just amplifies the base current, if the base current is limited accordingly, also the collector current to discharge the capacitor is limited, so it is not destructive.

The capacitor cannot also be arbitrarily large, as many datasheets have charts with electrolytic capacitors that end up at 100 uF. Larger electrolytic capacitors than that could exhibit too much leakage to be used for timing. So since that already limits the practical capacitance that may be on discharge pin, it can be assumed that it can discharge this practical capacitance.