Electrical – Capacitor in parallel with relay coil backfires on contacts

You can place the RC either at the B side or the A side. When components are placed in series the order of them doesn't matter for the working.

About the diodes. When you switch off the relay it will cause a (possibly large) negative voltage on the FET's drain, and a flyback diode is used to limit that voltage to a 0.7 V diode drop. So the diode(s) don't serve to protect the coil, but the FET. Using the zeners will allow this voltage to go to -5.7 V or -15.7 V if you'd use the 15 V zeners. There's no reason for taking risks here, even if the FET can handle -30 V. So I would just use a rectifier or signal diode, or even better a Schottky diode.

edit re your comment
You can indeed use a zener (combined with a common diode, D1 doesn't have to be a zener) to decrease switch-off time, and Tyco also mentions it in this application note, but I don't read it as if they insist on it. The scope images in the first link show a dramatic decrease in switch-off time, but that measures the time between deactivating the relay and the first opening of the contact, not the time between first opening and the return to the rest position, which will change much less.

edit re the 6 V relay and the RC circuit
Like I says in this answer you can operate a relay below its rated voltage, and since its operate voltage is 4.2 V the 6 V version of your relay can also be used at 5 V. If you use a series resistor not higher than 9 Ω you'll have that 4.2 V, and then you don't need the capacitor (keep an eye on the tolerance for the 5 V!). If you want to go lower you're on your own; the datasheet doesn't give a must hold voltage. But let's say this would be 3 V. Then you can use a series resistor of 32 Ω and you'll need the capacitor to get the relay activated.

Operate time is maximum 15 ms (which is long), so as the capacitor charges the relay voltage shouldn't go below 4.2 V until 15 ms after switching on.

Now we have to calculate the RC time for that. R is the parallel of the relay's coil resistance and the series resistance (that's Thévenin's fault), so that's 19.3 Ω. Then

\$ 3 V + 2 V \cdot e^{\dfrac{- 0.015 ms}{19.3 \Omega \text{ C}}} = 4.2 V \$

Solving for \$\text{C}\$ gives us 1500 µF minimum.

Re switching off:
You can't violate Q = CV, it's the Law. Your clamping voltage is 3.3 V + 0.7 V = 4 V. That means that when you switch the FET off the low side of the capacitor momentarily will be pulled to -4 V, and quickly rise again to 0 V. The high side is 2 V higher, and will simply follow that 4 V drop while the capacitor discharges through the parallel resistor. The capacitor won't even notice the drop. The discharge time constant is 1500 µF \$\times\$ 32 Ω = 48 ms, then the capacitor will discharge to 20 mV (1% of its initial value) in 220 ms.

The 62 mA won't charge nor discharge the capacitor. We often apply Kirchhoff's Current Law (KCL) to nodes, but it also applies to regions:

Draw a boundary around C1 and R1, and you'll see there's only one path to the outer world since the way to the FET is cut off. Since the total current has to be zero there can't be any current through that unique connection. The coil has to take care of the 62 mA on its own, and it does so by using the loop formed by the zeners.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Redrawn circuit with negative rail at bottom.

I'd say your circuit analysis is pretty good and your explanation of circuit use is a great help. Convention on circuit schematics is that current generally flows from top to bottom and since most of us reads left to right we generally put inputs (logic, analog, power) on the left and outputs on the right. What that in mind, I've redrawn your schematic.

What you have described is a "shunt regulator". If the voltage across the regulator rises above the preset voltage the regulator turns on and "shunts" current away from the power source to hold the voltage in control. They are not all that common because they waste power in the shunt circuit - hence the large heatsink and ceramic resistor in your circuit. Also, the shunt circuit relies on the source power-supply having some internal resistance otherwise a huge current would flow through the shunt as the power supplies try to keep the voltage up and the shunt tries to keep it down.

This circuit is a rather unusual, however, because there is no complete power circuit through the shunt circuit. Based on your Hall thruster link (page 338) your circuit schematic regarding the non-closed loop is correct.

So, if all the above is correct, what's happening is as follows:

• The circuit is designed to maintain the V1, 2 and 3 negative terminals \$V_{CLAMP}\$ somewhere between 0 V (actually + 0.7 allowing for the diode forward-voltage drop) and -50 V DC.
• C1 is a local bypass capacitor and will shunt any high-frequency noise away from the regulator. It won't have any effect on the DC.
• If \$V_{CLAMP}\$ rises above GND then D1 will be forward biased and prevent it from rising any futher.
• If \$V_{CLAMP}\$ falls below -50 V then D2 Zener diode will break down feeding current into the base of Q1. This will allow current to flow through R3 and Q1 to "pull \$V_{CLAMP}\$ back up" to -50 V.
• If the voltage is between 0 and -50 V the shunt circuit will take no action.

Please let me know if you think this is inconsistent with any other information you have. I know nothing about thrusters.