I didn't try to understand your circuit, but your PMOS sources and drains are swapped.Well, I figured out a way to wire add PMOS's into the circuit for each relay coil and it's NMOS driver to chain relays together so that each relay in a chain can be activated individually, but activating a relay will also activate all relays that came before it in a chain. All without using the primary terminals of the relays which would distort the current readings. This would allow a SPxT switch to be used.
Still a lot of parts though which could be replaced by a single strange switch.
I figured as much. I just worry about some NOOB copying your design and ... Well, you know the rest.Ah, yeah the PMOS drains are swapped. Yeah i know you need diodes. I just quickly drew this up to show the method than the actual implementation.
Have a look at the bottom of page 12 of the datasheet.
The charge current is determined by:
The charge current sense resistor, and, the voltage on terminal Viset.
The Viset can be between 0 and 2 volts, thus allowing a wide range of adjustment of the charge current.
JimB
Hi there dk,
So this is what you are going to use after the transformer for your chargers?
Did you consider yet using a simple voltage divider for the current sense? That would give you easier control over changing the feedback so you can switch currents a little easier. Taking it to the extreme, if it works, it could make it a LOT easier.
The idea is to use (say) a 0.1 ohm resistor as the current sense. A voltage divider built around this resistor cuts the voltage down to 0.1 volt when 10 amps flows to the output. If you want to match the resistance of the 0.1 ohm, you would use a 0.08 ohm in series with a 0.02 ohm and take the left hand side current sense lead from the junction of these two. This way when 10 amps flows to the output 5 amps goes through the 0.1 ohm and 5 amps through the 0.08 in series with the 0.02, which puts again 0.1 volts across the 0.02 resistor so the current feedback sense gets the same voltage as before.
Taking this to the extreme, we might get away with using (say) a 0.1 ohm resistor and even possibly 90 ohms and 10 ohms, taking the left current sense lead from the junction again and the right current sense lead from the right side of the 10 ohm resistor. This way when 10 amps flows to the output almost the entire 10 amps flows through the 0.1 ohm resistor yet the differential current sense voltage is still around 0.1 volts.
Whether this works or not partly depends on how the internal current sense it really built up. From the data sheet it looks like a true differential current sense and if it really is then this would definitely work.
After you get it working (perhaps with the 0.08 and 0.02 first) you can proceed to figure out how to adjust the output current, which should come out much easier than having to switch in or out a bunch of resistors.
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