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Automated capacitor precharge circuit

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HR19

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I have an electric go kart. The controller has huge capacitors that have really high in-rush. My batteries can handle it but I'd still rather have a precharge circuit with a resistor to reduce the current. But I'd like to have it automated, so I turn the key, the precharge circuit comes on, when it's reached equilibrium it switches over to main power via some relays I guess. As an electrician I'm familiar with relays and a number of other controller elements. I also have an Arduino, but my battery voltage system is 54, much too high for an Arduino. What I would need is something that can essentially measure the voltage across the precharge resistor and once it levels off (the capacitors are charged) it triggers a relay to turn on the main power. I understand the physics, but I'm not sure if there's a simple/easy way to do it. Any ideas?

EDIT: I have 2 switches on it, perhaps the best option is just a resistor that activates on the first switch, and then that becomes my precharge sequence. Activate first switch, finish getting go kart ready, activate secondary switch, off I go. That would be, by far, the easiest, quickest, and cheapest option. But I'd still like to know if there's a reasonable way to do the precharge circuit though. Could be a fun little addition.
 
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A simple way would be to have a resistor in series with the battery with a relay contact across the resistor.
The relay is controlled by a simple voltage detector circuit connected to the capacitor that closes the relay contacts across the resistor when the desired capacitor voltage is reached.
How does that sound?
 
A simple way would be to have a resistor in series with the battery with a relay contact across the resistor.
The relay is controlled by a simple voltage detector circuit connected to the capacitor that closes the relay contacts across the resistor when the desired capacitor voltage is reached.
How does that sound?
Pretty much exactly what I was thinking. How do I make a voltage detector circuit?
 
Pretty much exactly what I was thinking. How do I make a voltage detector circuit?
One way would be a 12V regulator, a TL431 voltage reference used as a comparator, driving a transistor to control the relay coil.

How much current will the relay contacts need to carry?
 
I already have main power relays, so a small relay to switch them on would work. But I'm not sure I understand, I have a 54V nominal, 62V peak, battery system. Looks like a TL431 will only handle up to 36V. I could always make a voltage divider circuit I suppose. Am I just missing something?
 
Am I just missing something?
Yes, I also mention a 12V regulator from the battery supply.
This could be something like a 30-35V Zener in series with the input to a 7812 IC regulator.

What's the voltage and current (or resistance) of the main power relay coil?
 
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About 100, maybe 150 mA combined for the relays at peak voltage.

Also thank you for all your help.
 
About 100, maybe 150 mA combined for the relays at peak voltage.
Why relays (plural).
My scheme only uses one relay.

And I need to know the relay coil voltage as well as its current.
 
Why relays (plural).
My scheme only uses one relay.

And I need to know the relay coil voltage as well as its current.
It was much cheaper to get two 40A relays than one 60A at 48V. Coil voltage is system voltage, rated at 48V. I'm actually currently running the coils in series as an experiment, so that if one fails or the voltage goes too low they'll click off.
 
This is getting a bit complicated if I understand what you want...

- 62v supply (54v nominal) (minimum undefined
- capacitor bank size not defined (farads)
- at power-up
  • battery is disconnected from motor until capacitor bank charges
  • Current flow is limited to capacitor bank and charges cap bank until some % (90%?) of battery charge level
  • When capacitor bank is at threshold level
    • connects cap bank to motor (or to motor speed controller)
    • Relay connects battery to motor (or to motor speed controller)
If all of that is correct, what happens if you're using the motor and high power demand causes the cap bank falls below your threshold level (90% of supply for example?). Should it continue working or should the motor cut out again until the cap bank charges again? I hope not.​

how long from power up until the cap is ready? 2, 5, 10 seconds? Longer?​
 
This is getting a bit complicated if I understand what you want...

- 62v supply (54v nominal) (minimum undefined
- capacitor bank size not defined (farads)
- at power-up
  • battery is disconnected from motor until capacitor bank charges
  • Current flow is limited to capacitor bank and charges cap bank until some % (90%?) of battery charge level
  • When capacitor bank is at threshold level
    • connects cap bank to motor (or to motor speed controller)
    • Relay connects battery to motor (or to motor speed controller)
If all of that is correct, what happens if you're using the motor and high power demand causes the cap bank falls below your threshold level (90% of supply for example?). Should it continue working or should the motor cut out again until the cap bank charges again? I hope not.​
how long from power up until the cap is ready? 2, 5, 10 seconds? Longer?​
Let's simplify it. There's only one part I really need help with, I think I can figure the rest out myself. I need something that can in some way detect DC voltage or DC current flow, and trigger a relay depending on it's reading, while functioning up to 62V.
 
I need something that can in some way detect DC voltage or DC current flow, and trigger a relay depending on it's reading, while functioning up to 62V.
The simplest would be a Zener in series with a resistor to the base of an NPN transistor that turns on the relay.
The voltage value of the Zener depends upon at what capacitor voltage you want the series resistor to be removed from the circuit.
If you can tell us that we can proceed.
 
An even simpler way is to rely on the pull-in voltage of the relay coil being close to the 54v operating voltage, and use one pair of it's contacts to self-latch, and the other pair shorts out the charging resistor. Coil is connected across the big cap. Go up to 3 sets of contacts and it can also toggle the motor connection. I'm guessing you'd need a relay with around a 60v coil to get pull-in at around 50v. Of course, I may have picked the most expensive way to do it, here.
 
An even simpler way is to rely on the pull-in voltage of the relay coil being close to the 54v operating voltage, and use one pair of it's contacts to self-latch, and the other pair shorts out the charging resistor. Coil is connected across the big cap. Go up to 3 sets of contacts and it can also toggle the motor connection. I'm guessing you'd need a relay with around a 60v coil to get pull-in at around 50v. Of course, I may have picked the most expensive way to do it, here.
So wait, are you saying use the resistance of the coils? The ones I have are about 1.2KOhms resistance each. So If I use them in series (or parallel, whichever is better) I could use their N/O contacts to charge the cap, then when the cap charges my voltage will equalize, which will turn them off and switch to N/C contacts, switching over to the unimpeded power. Is that your idea?

EDIT: Wait, you're saying coil is parallel to the big cap? I don't think I have any meaningful way to do that (except maybe adding a diode somewhere), because the cap is internal to my motor controller. I just have +/- contacts going to the the controller.

The simplest would be a Zener in series with a resistor to the base of an NPN transistor that turns on the relay.
The voltage value of the Zener depends upon at what capacitor voltage you want the series resistor to be removed from the circuit.
If you can tell us that we can proceed.
Would this re-trigger every time that the voltage drops below the threshold? If so, probably something like 45V due to voltage drop under load. If this would switch over only once (or could be easily set up to only trigger once, which I think I know how to do that...), then probably 50V give or take would be good enough to reduce the massive inrush.
 
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I was thinking of the well known soft start circuit. You charge the capacitor via a low value current-limiting resistor, with the relay coil connected across the capacitor. At some point the relay coil will have sufficient voltage across it to be able to close it's contacts - this is the pull-in voltage (I think the proper term is the "must close" voltage) which is somewhat below the rated coil voltage. You want it to be high enough that the capacitor is mostly charged when it activates, but also low enough that it doesn't cut out at the low end of the supply voltage range

You arrange the relay contacts so that one set will short-circuit the resistor when the relay closes them. I was mistaken earlier, you don't need another set to keep the coil connected, So for the simplest version of the circuit you only need 1 set of normally open contacts. if you are also switching the motor out of circuit until the cap is charged, you need another set.
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The relay will stay energised as long as there is enough voltage across the cap to keep it so - this will be somewhat a lower value than the pull-in voltage.

The circuit that crutschow suggests will work equally well and provides a much better defined activation voltage. It will drop out at approximately the same point as it activates, but can easily be made to have a lower release voltage, like the bare relay does.

Given that you only have access to the + and - terminals of the controller, well, it's safe to assume that the capacitors you refer to are connected across the supply terminals. If that is true then you can still use the relay, just treat the controller input as if it's the capacitor itself. No need for diodes or anything fancy. If it's not arranged this way then you have two options that I can see. One is to use a timer, which really only needs to be a capacitor and resistor. It can be connected in exactly the same way, with the advantage that you only need a much lower voltage relay if you size the resistor appropriately. The caveat is that it's a timer, not something governed by local conditions. The other alternative is to use a current sensing circuit, but this would require more components and probably is rather excessive for what you want.
 
Would this re-trigger every time that the voltage drops below the threshold?
You could add a diode and a capacitor to delay the relay opening when going below the threshold voltage.
Or would the time it goes below the threshold be quite long?
 
You could add a diode and a capacitor to delay the relay opening when going below the threshold voltage.
Or would the time it goes below the threshold be quite long?
Probably not long at all, just during hard acceleration or when the batteries are low.

I was thinking of the well known soft start circuit. You charge the capacitor via a low value current-limiting resistor, with the relay coil connected across the capacitor. At some point the relay coil will have sufficient voltage across it to be able to close it's contacts - this is the pull-in voltage (I think the proper term is the "must close" voltage) which is somewhat below the rated coil voltage. You want it to be high enough that the capacitor is mostly charged when it activates, but also low enough that it doesn't cut out at the low end of the supply voltage range

You arrange the relay contacts so that one set will short-circuit the resistor when the relay closes them. I was mistaken earlier, you don't need another set to keep the coil connected, So for the simplest version of the circuit you only need 1 set of normally open contacts. if you are also switching the motor out of circuit until the cap is charged, you need another set.
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The relay will stay energised as long as there is enough voltage across the cap to keep it so - this will be somewhat a lower value than the pull-in voltage.

The circuit that crutschow suggests will work equally well and provides a much better defined activation voltage. It will drop out at approximately the same point as it activates, but can easily be made to have a lower release voltage, like the bare relay does.

Given that you only have access to the + and - terminals of the controller, well, it's safe to assume that the capacitors you refer to are connected across the supply terminals. If that is true then you can still use the relay, just treat the controller input as if it's the capacitor itself. No need for diodes or anything fancy. If it's not arranged this way then you have two options that I can see. One is to use a timer, which really only needs to be a capacitor and resistor. It can be connected in exactly the same way, with the advantage that you only need a much lower voltage relay if you size the resistor appropriately. The caveat is that it's a timer, not something governed by local conditions. The other alternative is to use a current sensing circuit, but this would require more components and probably is rather excessive for what you want.
I think I understand now. The capacitor in parallel would reduce the voltage across the coil until it's charged sufficiently, I see what you're saying. Once the relay triggers, it'll bypass the resistor. I'm gonna try that first I think as I have everything I need already to test that.

My best option is probably dual switching, a key to start precharge, then a push start button for the relays (also gives me an easy kill switch option too). That would be my quickest option, requiring no parts that I don't already have.
 
Consider also a 3 position switch. Less holes to make, that way :)
 
Consider also a 3 position switch. Less holes to make, that way :)
I suppose. But that would make it much easier for someone to just turn it all the way practically totally bypassing the precharge circuit. Plus it would feel cooler to have some extra switches, like starting up a rocket.
 
What If I use a transistor (maybe a MOSFET for higher currents just in case) with a large resistor in parallel to the cap, and when it triggers it'll activate the primary relays which will then use their own N/O contacts to bypass the whole precharge circuit? That way once on, it won't shut off during high current draw (read: significant voltage drop). Would that work the way I think?

Edit: I think this is an accurate diagram (sans some specific values I suppose). And thinking about it now, would I even need the transistor at all, actually?
Edit2: the capacitor and [Loads] are functionally equivalent in terms of ultimate connections.
 

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