How to turn on relay.
What ?!? What kind of question is this? All you need to do is high or low side switch (a MOSFET or bi-polar transistor) and connect relay/contactor coil as a load. If you don't mind slowing down turn-off, you just connect a clamp diode and/or RC snubber to absorb inductive kick-back of a coil. Done. Right?
Not so fast.
This theory above does not apply to power relays or contactors. It certainly does not apply to the EV200 HAXXA01 type contactors used in Think's A306 battery. (it is basically more common EV200 AAAHA contactor with added auxiliary contacts that are used to remotely check status of the main contacts). These contactors have integrated economizer circuit and PWM driver, so do not represent a linear inductor at all. If you connect it to the power source (12...14 VDC), it will sink current in two distinct phases - pull-in phase and retention phase. Both are implemented by PWM, duty cycle depends on the voltage applied and decrements in discrete steps as input voltage ramps up. The economizer circuit is very noisy, a lot of ringing is going on.
The tests I've done were meant to determine the optimal type of driver I'd want to use in the I/O circuit for reliable operation. A discrete transistor would still be an option, but I chose specialized driver IC that has integrated suppression circuit, logic level input, favorable thermal de-rating characteristics and other desirable qualities. The test voltage was 14.4 VDC - about the maximum the contactor will ever see.
Below are few oscillograms revealing what's going on. Before getting to them, I wanted to mention that I ended up using noise suppressing capacitor across the contactor input. If you do this, make sure the cap has sufficient ripple current handling capability - a generic $0.05 multilayer ceramic cap of 0406 size won't do it.
Here is what happens when you turn contactor
on and off. Note current scale. First ~110 ms is pull-in phase followed by
retention phase until contactor is off.
Zoomed in section of the oscillogram above. The current ramp does not start for about 2 ms and then within 25 ms settles
Zooming in further. White section on green and yellow traces that captured transition from the pull-in phase to retention phase is expanded below (labeled Z1 and Z4). The current during pull-in is peaking at 3A and the average is about 2A (voltage dependent).After plunger pulled in and contact established, the coil current can be reduced. Kilovac does it by increasing frequency from about 4 kHz to ~18 kHz but decreasing duty to 20% and then also gradually decreasing current peaks to ~0.6 ADC. This averages ~ 0.12 ADC from the power supply - the value you will see if you just measure current consumption with Fluke meter.
High mag of above - pulses details.
Single current pulse details (retention phase).
This capture shows what happens if you try to suppress voltage noise by connecting a cap in parallel to the input of contactor. 40VDC peak voltage excursions get reduced to about 28V max, and -2V negative excursions totally disappear. All good, right? Again, not so fast.
From basic electric circuits theory you should remember that capacitor conducts AC current - the reactive resistance (reactance) is inversely proportional to the frequency, namely R=1/(2Π*f*C). And you might recall the Fourier theory stipulating that the sharper (faster rise/fall times) pulses are, the higher frequency contents they have. Basically meaning, capacitors have very low resistance for these pulses and conduct them right through. So what happens if you try to suddenly connect a voltage source to the cap (which is in parallel to the input)? You will see peak charging current at the moment the voltage is applied. On the left side there is about 0.7 ADC peak current that is due to internal capacitance of the contactor's circuit. As the contactor enters retention phase the current is peaking to about 3 ADC with small negative excursions. Guess what happens if you connect a cap in parallel to the input. This cap stores energy and works as a tank supplying it to the circuit when contactor consumes current and recharging from external power source when it does not. Such pulsing ripple adds to what contactor itself consumes - now the current peaking at above 4 ADC and discharging this cap results in negative 3 ADC pulses as well. This kind of ripple may make the cap hot and destroy it in short order!
So, you just traded reduced voltage spikes for increased current ripple. Which is better to allow ? Very much depends on what other circuits (if any) you feed from the same power source as contactors are fed, and the length/inductance of the wire from the main tank (12V battery or large capacitors on the DC/DC converter's output) to the contactors input. If your other circuits are sensitive, you better suppress voltage spikes - these circuits will not see current ripples and won't care about contactors. But if there is nothing else, you may save your caps and allow elevated ripple current along 12V net. This has to be carefully calculated compromise and usually determined by trial and error during commissioning testing when all the wiring is in place.
Not so simple after all, huh?
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