Winch voltage
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Most winches draw 300 - 400 amps on 12 volts. Running 24 volts will cut that in half. You need 2 batteries and a 100 amp diode to isolate the second battery, so it will charge but not discharge thru the 24 volt series jumper to second battery. Then you can run the 24volt winch. Your choice.
Main battery - ground with diode transferring to 2nd battery ground - positive (main battery) through diode transferring power to second battery positive terminal but not back - charges 2nd battery. Then another wire from main (+) to second battery (-) with a diode. Then connect ground of your winch to the main battery ground (-). Connect your winch (+) to the second battery positive = 24 volt winch. Don't think I would attempt it unless there was an electrician around. Could also do like a transfer switch set up...
or connect a second battery in parallel for 12 volt, with upsized wire, fuses, and alternator accordingly.
Charging for 24volt system will also need a bigger alternator, wire, fuses to first and second battery...
Main battery - ground with diode transferring to 2nd battery ground - positive (main battery) through diode transferring power to second battery positive terminal but not back - charges 2nd battery. Then another wire from main (+) to second battery (-) with a diode. Then connect ground of your winch to the main battery ground (-). Connect your winch (+) to the second battery positive = 24 volt winch. Don't think I would attempt it unless there was an electrician around. Could also do like a transfer switch set up...
or connect a second battery in parallel for 12 volt, with upsized wire, fuses, and alternator accordingly.
Charging for 24volt system will also need a bigger alternator, wire, fuses to first and second battery...
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202 Posts
try this,
So I have been thinking about this winch setup.
-main battery1 negative to second battery2 negative (use 350 amp wire.
-main battery1 positive to 350 amp diode -> to second battery2 (use 350 amp wire)
*what this does is charge the second battery but never draws on it through normal power. Second battery remains in fully charged state.
To Winch from second battery (normal setup)
-negative battery2 to negative winch (use 350 amp wire or whatever amp wire the winch draws)
-positive battery2 to positive winch (use 350 amp wire or winch draw amps is best for sizing the wire)
*So what happens is the second battery2(fully charged) is supplying power to the winch. But if the second battery2 voltage drops below the main battery1 voltage, then the main battery will supplement power to the second battery2 through the diode but not from the second battery2 to the main battery1.
-I think it would be best to use 2 ROCKFORD FOSGATE distribution terminals, have very large wire terminals and usually have gold or platinum plating (good connection).
Advanced - Power factor correction
*so everybody knows that resistance is a type of load, lets say it goes right ->, now capacitance leads resistance by 90 degrees, lets say it goes down !, and inductance (motor load) goes up ^. So if you have a large motor load going up, you can cancel this load out with the large capacitance load going down !, POWER FACTOR CORRECTION, basically free power. This will drop the winch amperage considerably. Electricians do it all the time for large motors (inductive loads). What you do not want is the same inductance as capacitance load, this will give you infinity amps - BOOM. Basically 350over0 - 350/0 or 1/0 to simplify means infinity. What you want is like a 90% capacitance load of the inductance. Now you have to be careful, some of your winch amps are resistance (heat). Lets say you have 50 amps resistance with a right vector 50> and your winch is 350 amps total hypotinouse, sorry my spelling sucks. Anyways you can figure out your inductive load.
a2+b2=c2 so c2(350squared) - a2(50squared)= b2(120,000) then you root it to get rid of the 2 so you have a single b=346.4amps inductive load.
figure out ohm I(amps)=E(volts)/R(resistance-ohms)
346.4 amps/14.4 volts = 24 ohms inductive load
-so you need a capacitor that is 90% of 24 ohms
24ohms x.9(90%)= 21 ohm
-so now we need to figure out farads(capacitance)
R(capacitance - ohms) going down remember !
R(ohms) = 1/2xpie(3.14) x frequency x farads(capacitance)
so we need to get farads by itself (algebra)
1/R= 2pieFC
so
C=1/R/2/pie/F
C(farads)=1/21/2/3.14/20,000hertz
C(farads)=.000 000 3789 farads
*the frequency of a series motor varies depending on the speed, how many windings in the armature... I used 20,000Hz because variable frequency drives run DC motors well at 20,000 hz, almost a flat line it is so fast.
-so our desired capacitor is a 350 amp
C(farads)=.38 micro farads to correct the power factor to 90%
-also the one we can't have is the 24 ohm capacitor, so lets figure that out too so we don't get it.
C=1/R/2/pie/F
=1/24ohms/2/3.14/20,000Hz
=.000 000 015 789 198 1 farads
=.0158 micro farads
*so to keep from blowing up we need a cap bigger than .0158 micro farads
-2 ROCKFORD FOSGATE RFDB1 sells 2 micro farad capacitors for sterios (inductive load correction) This would work well if the load rating is above the load draw of the winch (350amps)
*so what happens is the winch draws an inductive load of 350amps up ^ we will say. The cap needs to discharge 350 amps down !to stabilize the voltage. By having a larger cap 2 micro farads instead of the .38 micro farad the cap is correcting the draw on the battery (cancelling it out). This will minimize the pulse draw on the batterie(s).
If I had just the cap and one battery setup, I would just use the winch for short pulses. Maybe a few seconds, depends on how fast your cap drains. Then let it charge and pulse it again(turn on winch)
-from positive battery2 through the cap +in>+our to the winch posative. There might be a ground you need to connect on the cap to the ground of the battery too, size should be indicated by the capacitor instructions.
There you go, the best 2 battery setup with capacitor for power factor correction for a winch.
Designed by a Red Seal Industrial Electrician/Instrumentation Tech.
So I have been thinking about this winch setup.
-main battery1 negative to second battery2 negative (use 350 amp wire.
-main battery1 positive to 350 amp diode -> to second battery2 (use 350 amp wire)
*what this does is charge the second battery but never draws on it through normal power. Second battery remains in fully charged state.
To Winch from second battery (normal setup)
-negative battery2 to negative winch (use 350 amp wire or whatever amp wire the winch draws)
-positive battery2 to positive winch (use 350 amp wire or winch draw amps is best for sizing the wire)
*So what happens is the second battery2(fully charged) is supplying power to the winch. But if the second battery2 voltage drops below the main battery1 voltage, then the main battery will supplement power to the second battery2 through the diode but not from the second battery2 to the main battery1.
-I think it would be best to use 2 ROCKFORD FOSGATE distribution terminals, have very large wire terminals and usually have gold or platinum plating (good connection).
Advanced - Power factor correction
*so everybody knows that resistance is a type of load, lets say it goes right ->, now capacitance leads resistance by 90 degrees, lets say it goes down !, and inductance (motor load) goes up ^. So if you have a large motor load going up, you can cancel this load out with the large capacitance load going down !, POWER FACTOR CORRECTION, basically free power. This will drop the winch amperage considerably. Electricians do it all the time for large motors (inductive loads). What you do not want is the same inductance as capacitance load, this will give you infinity amps - BOOM. Basically 350over0 - 350/0 or 1/0 to simplify means infinity. What you want is like a 90% capacitance load of the inductance. Now you have to be careful, some of your winch amps are resistance (heat). Lets say you have 50 amps resistance with a right vector 50> and your winch is 350 amps total hypotinouse, sorry my spelling sucks. Anyways you can figure out your inductive load.
a2+b2=c2 so c2(350squared) - a2(50squared)= b2(120,000) then you root it to get rid of the 2 so you have a single b=346.4amps inductive load.
figure out ohm I(amps)=E(volts)/R(resistance-ohms)
346.4 amps/14.4 volts = 24 ohms inductive load
-so you need a capacitor that is 90% of 24 ohms
24ohms x.9(90%)= 21 ohm
-so now we need to figure out farads(capacitance)
R(capacitance - ohms) going down remember !
R(ohms) = 1/2xpie(3.14) x frequency x farads(capacitance)
so we need to get farads by itself (algebra)
1/R= 2pieFC
so
C=1/R/2/pie/F
C(farads)=1/21/2/3.14/20,000hertz
C(farads)=.000 000 3789 farads
*the frequency of a series motor varies depending on the speed, how many windings in the armature... I used 20,000Hz because variable frequency drives run DC motors well at 20,000 hz, almost a flat line it is so fast.
-so our desired capacitor is a 350 amp
C(farads)=.38 micro farads to correct the power factor to 90%
-also the one we can't have is the 24 ohm capacitor, so lets figure that out too so we don't get it.
C=1/R/2/pie/F
=1/24ohms/2/3.14/20,000Hz
=.000 000 015 789 198 1 farads
=.0158 micro farads
*so to keep from blowing up we need a cap bigger than .0158 micro farads
-2 ROCKFORD FOSGATE RFDB1 sells 2 micro farad capacitors for sterios (inductive load correction) This would work well if the load rating is above the load draw of the winch (350amps)
*so what happens is the winch draws an inductive load of 350amps up ^ we will say. The cap needs to discharge 350 amps down !to stabilize the voltage. By having a larger cap 2 micro farads instead of the .38 micro farad the cap is correcting the draw on the battery (cancelling it out). This will minimize the pulse draw on the batterie(s).
If I had just the cap and one battery setup, I would just use the winch for short pulses. Maybe a few seconds, depends on how fast your cap drains. Then let it charge and pulse it again(turn on winch)
-from positive battery2 through the cap +in>+our to the winch posative. There might be a ground you need to connect on the cap to the ground of the battery too, size should be indicated by the capacitor instructions.
There you go, the best 2 battery setup with capacitor for power factor correction for a winch.
Designed by a Red Seal Industrial Electrician/Instrumentation Tech.
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Stick with 12v which is the native voltage for everything on your Jeep and modern civilian cars and trucks. Forget all the above 24v malarkey which is more common to large military vehicles. .
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I have an 8000 lb 12 volt unit and it does the job well :bop:
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umm 12 will get you 12 and 24 will get you 24:lmao:
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