From what I've been seeing online, they reduce the Amps by 10, and add 12v, and that tells you if you can use it or not for your DC system. Just wondering if I'm doing this right just for future reference?
That sounds like internet folklore, not any kind of substantiated science. I'd say forget it. You should derate the switch (because it's being used with an inductive load, the mower motor), but not by that weird-sounding scheme.
It's a little cheaper, and he also stated that, some ways to check if the switch will not work is if the switch is getting very hot, or if it stops working obviously?
I think I'll try this one first then use the one you linked to which we "know" will work. I just really like the extra Off position
That sounds like internet folklore, not any kind of substantiated science. I'd say forget it. You should derate the switch (because it's being used with an inductive load, the mower motor), but not by that weird-sounding scheme.
lol. Okay, I was seeing that word being used as well, "derate" which I don't understand how I'm suppose to do that? Can you please elaborate the process?
Does this circuit look right to you for what I'm trying to do, as it seems like something may be missing or I may have hooked up something wrong: **broken link removed**
*Since multisim doesn't have the on-off-on switch in my library I had to just use the on-on dpdt switch.
I'm thinking a diode should go in front of the AC chargers + connection...or maybe not.
Digi-Key says "part not found". Copy the URL correctly?
It's a little cheaper, and he also stated that, some ways to check if the switch will not work is if the switch is getting very hot, or if it stops working obviously?
I think I'll try this one first then use the one you linked to which we "know" will work. I just really like the extra Off position
lol. Okay, I was seeing that word being used as well, "derate" which I don't understand how I'm suppose to do that? Can you please elaborate the process?
It just means reducing the rated value of something to give a safety margin. There are rules for how to derate various things (voltage, current, on time, etc.), but it can be done by a seat-of-the-pants approach as well. For instance, reduce the rated current (of the switch) by 75%. It may not be totally technically accurate, but it is a step in the right direction at least.
Does this circuit look right to you for what I'm trying to do, as it seems like something may be missing or I may have hooked up something wrong: **broken link removed**
It just means reducing the rated value of something to give a safety margin. There are rules for how to derate various things (voltage, current, on time, etc.), but it can be done by a seat-of-the-pants approach as well. For instance, reduce the rated current (of the switch) by 75%. It may not be totally technically accurate, but it is a step in the right direction at least.
Yeah I understood the basic meaning of it, however, I wasn't sure if I was already doing that in my circuit. In other words, do I need to buy something else for this circuit so that I'm correctly derating the voltage/current like a resistor or something?
Two basic questions here: 1. What happens if you hook up the diode the wrong way, what signs will you see? 2. I understand when using induction motors people recommend these diodes, however, I'm not fully clear on the overall picture of what they help protect that the motor can possibly destroy. Does the diode help reduce the large starting current or something else? I'll probably google this a bit more later though...
Lastly, what's your address so I can send you a chocolate cake! lol Seriously, you and some others here are so darn helpful it's amazing! Thanks a ton for your help thus far.
The diode will keep the switch from arcing due to the inductance of the motor. The switch won't last long without it.
If your motor draws 25 amps I would get a bigger diode that the 4007. If you put it in backwards it will smoke.
I'd suggest this switch from Digi-Key instead. It's rated for 30 amps. Even so, it probably won't last forever. Your application is a hard life for a switch.
I'd suggest this switch from Digi-Key instead. It's rated for 30 amps. Even so, it probably won't last forever. Your application is a hard life for a switch.
Well according to, ronv, the diode (4007) should help protect to switch? Is there another component I can place in the circuit to help extend the life of the switch.
And I know in my original post that on the switch on the lawn mower it showed 25A, however, that's most likely the max that the mower will ever reach, but more than likely the amps will range from 5A - 15A tops. Only time will tell once I test it and hook up an Amp meter.
BTW: I'm about to buy this, anyone want a couple: **broken link removed** ??
lol amazing how cheap these things are!
Well according to, ronv, the diode (4007) should help protect to switch? Is there another component I can place in the circuit to help extend the life of the switch.
And I know in my original post that on the switch on the lawn mower it showed 25A, however, that's most likely the max that the mower will ever reach, but more than likely the amps will range from 5A - 15A tops. Only time will tell once I test it and hook up an Amp meter.
The catch diode protects the switch against reverse EMF pulses, but it doesn't protect against the high inrush current of the motor when it starts. Nothing really you can do about that. The switch will last, well, as long as it lasts. You can always get another one ...
Regarding that derating, is my current circuit taking care of that? I'm guessing when you/others say derating regarding this AC switch which is being used for DC, to pick a AC switch that is rated high enough to handle the DC voltage/current your circuit will have? Or am I still not getting the big picture?
I've also been searching around trying to find articles that explain how to reduce sparks from batteries but haven't found much of anything yet.
Basically my concept would be, when the contacts of the switch close, some how the voltage gradually builds up to the full 36v thus reducing the large spark. This would result in the motor starting out spinning slowly and gradually getting faster and faster until it's at it's normal peak.
I remember talking about this stuff in school some but don't remember what was needed to make this happen, as think in the industrial field for very large power systems they use something like this?
---- Update: After digging some more, I saw some people suggested the following devices: soft starters, inverters (not sure what type), wye something...connection (something you build yourself), or another component which I can't remember now.
Anyway, I looked up soft starters and ran into this: **broken link removed**
It allows you to adjust the speed of the motor with a pot which I don't care to do , but the most important thing is the automatic soft starter and that it can handle my voltage/amps. I need to contact them though to see if it will do what I'm wanting it to do which is prevent the large sparks.
Can anyone shed some light on the device above regarding if it'll probably work or not?
Okay I received the switch in the mail today, still waiting on the diodes though, but I am curious about what connector types fit on the end, and where can I buy them locally (Autozone, walmart, etc?)?
I also bought one of these **broken link removed**, and wondering if I can turn it off with a simple radio shack 125VAC 16A switch while it's connected across the 36v battery? I would think so since it does not draw much current, I was just unsure if the motor in the circuit would damage it some how or is it isolated from that side?
The voltmeter draws very little current, so use any kind of switch. Doesn't say on that eBay page, but it probably draws something in the tens of mA neighborhood. A tiny little slide or toggle switch would work just fine.
Unless you really need a full-time voltage display, another way to do this would be to get one of those $3 digital multimeters from Harbor Freight and connect it across your battery. Of course, you'd have to turn it on and off by hand.
Haha, no, I just don't want to take a risk since I'd have to wait another week if I screwed it up. I love having stuff confirmed by experts like yourself even if I'm 95% sure a regular switch will do.
Anyway, I know you are busy, but regarding the derating, do I need to place something else in the circuit? That's basically what I'm worried about before I hook up this switch, as you and others said it should be derated.
Are you talking about this new switch, the one to turn the voltmeter on and off? Why would you think it needs to be derated? All it's doing is controlling the tiny bit of current the voltmeter draws. Doesn't matter if you hooked it up to a 10 billion-amp bank of batteries: all it'll ever see going through it will be the 20-30-40 mA of current the voltmeter uses.
Are you talking about this new switch, the one to turn the voltmeter on and off? Why would you think it needs to be derated? All it's doing is controlling the tiny bit of current the voltmeter draws. Doesn't matter if you hooked it up to a 10 billion-amp bank of batteries: all it'll ever see going through it will be the 20-30-40 mA of current the voltmeter uses.
Nope talking about the original switch (dpdt) which will be controlling the AC charger and power to the motor. I know this new switch from radio shack doesn't need to be derated .
So what do you recommend about that one since it will be having more amps going through it?
No, going back to your Post# 21, you said the following: You should derate the switch (because it's being used with an inductive load, the mower motor).
And I'm asking, or been asking what do I need to properly derate the switch? Or is my current circuit already doing that with the diode? I just want to make sure I'm not missing something in the circuit.