It might be worth looking at what the rail voltage is doing. The period you've labelled as '500nS Shoot-Through?' could be what the supply side is doing to sort itself out after the hit.
This is what my board looked like. The current sense resistor is a conventional part, so there's some inductance there with the partial loop round the leads, and it looks like it tries to keep the current going momentarily whilst burning off the energy with the resistor, taking the voltage negative in the effort (if I'm understanding it right). A better arrangement might have been an SMD part, straight down to the board from the pin, with maybe an improvised x10 coax probe.
From a design perspective, using a 555 would definitely benefit from good local decoupling, and be quite disruptive on a board without it.
I couldn't resist - I tried an experiment with a couple of the 555s I've got here. This is with a 50mR thick-film resistor between the part GND pin and the board ground. The yellow trace is the voltage across the resistor, the blue trace the output of the timer. I've done this on a piece of pcb material, so NOT a plug-in breadboard, but the part is in a socket and I'm probing with a 10M probe (might have been better with a piece of 50R co-ax).
This is a Fairchild bipolar part which might be from 1986. It's brutal. As the output rises, it gets to 30A for a few nanoseconds before limiting and falling. (The local decoupling is two 10uF tants and a 1uF ceramic - possibly not a very good combination.) It's a poor test, though, because the ground has lifted to 1.5V. I haven't shown the other end of the waveform, but, curiously, there's no shoot-through when the output drops.
This next is the TLC555 part. Here the shoot-through is worse at the end when the output comes down again, though it's nothing like as bad as the bipolar part. Something that's interesting is that there's a very short period before the output goes up when the chip is taking a little under an amp (perhaps that's a more limited shoot-through when the driver stage for the output MFETs starts transitioning) - by starting a fair current moving, it looks like it improves the situation when the output actually starts to move.
I think all that I'm showing here is that the shoot-through current will depend very much on what the decoupling is and how it's arranged. It's all very messy to interpret - the decoupling caps will be interacting, there's enough inductance to play a part, and my probing isn't going to truly reflect what's there. I think I've ended up with more questions than answers, but it's looking fairly certain that there IS a shoot-through current there.
Top Comments