12 volt to 9 volt resistors?

I installed an actuator in my tailgate to automate the lock and tested it with a 9v battery. Once I tied it into the 12v system I noticed that the action was a lot more violent and louder than using 9v. Does anyone know of a way to step down the voltage from 12v to 9v? I found some products on Amazon and Ebay but are designed to have a dedicated positive and negative line. But actuators change positive and negative flow depending on if it locks or unlocks.

 

V=IR Ohms Law

 

Which translates into 9v = .5 amps x 18 ohms

So how would one get a 12v line down to .5 amps and 18 ohms?

 

https://www.google.com/search?q=res....1.69i57j0.19997j0j8&sourceid=chrome&ie=UTF-8

 

The question is how much current does the actuator draw from the +12V source when "actuating"? If a resistor is going in series with the actuator, the total current of the circuit will help you in figuring out the voltage drop across said resistor.

 

I bought a decent multi meter and the amps while actuating is 3.5 on the truck battery. However, on the little 9v battery it drew only 500 milli amps (the max the 9v battery can provide). So, does the resistor need to be 42 watts (12v x 3.5 amps) or 6 watts (12v x .5 amps) or 4.5 watts (9v x. 5 amps)? And, more importantly, what would the ohlm value be to drop the 3 volts and 3 amps? I get the ohm's law equation but which voltage is entered, 12v or 9v, and which amp rating, 3.5a or .5a)? Is this right?.. R=V/I or 9v / .5a = 18 ohms.. So the resitor needs to be 50 watts (worst case) with 18 ohms resistance?

 

First we need to get effective resistance of the actuator.
(NOTE, discovered below error in this post, due to 9v battery sagging its voltage. DO NOT USE THIS DATA. LEFT FOR HISTORICAL DOCUMENTATION OF INCOMPETENCE)
Ractuator=9v/.5amp = 18 ohm

Then figure total circuit resistance to maintain .5 amp at about 12.5 volt (assume engine not running)

Rtotal rqd= 12.5v/.5amp = 25 ohm

Rtotal rqd = 18 ohm + Resistor = 25 ohm

Resistor = 7 ohm

Resistor power = RxI^2
Power = 7 ohm x .5amp^2 = 1.75 watt.

Note, this assumes the actuator is a purely resistive load, which it absolutely isnt. But suggestion to TRY and TEST approximately 7 ohm resistor with 1.75 watt power rating. (I would do at least 5 watt)

 

I want to address some things in this thread. Some of you may get it, I didn't read everyone's response 100%.

Ohms law is V=IR

V=Volts
I=Current (amps)
R=Resistance (ohms)

watt
/wät/
noun
plural noun: watts

1. the SI unit of power, equivalent to one joule per second, corresponding to the power in an electric circuit in which the potential difference is one volt and the current one ampere.

Watts is the amount of heat your resistor will be able to dissipate.
Oversizing the resistors wattage rating will make it run cooler.
Always account for voltage ratings.
And always select the correct resistance.

Another tip is that if resistance stays the same and voltage drops; current has to increase. lets just make a simple example

V=IR ----> I=V/R

V=12
I=?
R=2

I=12/2=6 amps. This is 60W

If you drop the V down to 9...

I=9/2=4.5 amps. This is 40.5W

Undervolting can lead to overcurrent issues in some electronics. Be careful.

The over-engineered answer to this question is to get a switching or constant voltage power supply and use that through a relay to trigger the solenoid. When you're running you'll have around 14.4 volts, and when you've been parked you will have around 12 volts. It is impossible to have 9V in all circumstances unless you stabilize the voltage supply.

My real answer... It was designed to run at 12v. Put some sound deadening in the cavity of your tailgate and hook it up to power.

 

I like it.. One part of your work through I could use some clarification on though.. Why would the actuator's resistance be different on different volts? I.E. 9v/.5a = 18 ohm vs 12.5v/3.5a = 3.57 ohm. Does resistance drop as power increases? One other question, an actuator reverses polarity on lock to unlock. So, does it matter if the resistor is wired before or after the actuator? Thanks for your help!

 

The actuator resistance doesnt change.

I need to draw figures to explain this well, but on my phone here, so will just try text.
First calculation is for actuator only powered by your battery with an ammeter on it. First calculation I want to know how much effective resistance in just the actuator (by your test).

Second calculation for resistance is a series circuit, with the actuator and an added resistor in series, powered by the truck. The actuator resistance is assumed to be the same, but now there is an added dropping resistor in the circuit. Second calculation I want to know how much total resistance (actuator + resistor) with a higher voltage to keep current the same.

Now I have resistance for actuator and resistor. Subtract off the previously calculated actuator resistance to get the resistance of just the resistor. (Items in series, resistance is just added).

Doesnt matter in series where the resistor is.

BTW, didnt think I had 3.5 anywhere in my calculation. Not sure where you got 3.5?

 

I get what you are saying and I did end up installing a roll of U-seal which worked great. It just seems like this is something that would be doable and I like coming up with new ways of doing things.

 

I'm following.. However, the .5 amp was due mostly because that's all the 9v battery could push in the moment it took to fire. So could the actuator's resistance actually be 3.57 ohm (12.5v / 3.5a) but I want to lower the amps to .5? Yet, if that was the case then the resistor would need to be 21.43 ohm (25 ohm - 3.57)?

 

Think your right. Forgot your test of 3.5 amp at full battery voltage completely. (With your 9v battery test your voltage sagged way below 9v we assume therefore). My fail there.

But solenoid (firing especially) is pretty dynamic to be calculating static conditions here.


If it were i, get a nice power potentiometer from Amazon for 10-$15 (0-20 ohm) Toss it on there and adjust and see what makes a good solid operation even under load, but isnt as loud.

Or buy 2 x 10ohm power resistors. Try both in series, see if too weak, if so just go with 1 in series.... ($6 on amazon for a pair of 100w 10ohm)

Measure its resistance and replace with power resistor.

I think your calc is right. Pretty amazing if it is burning 3.5 amp but works fine at .5 amp...

 

There is definitely a dramatic difference. The 9v is nice and smooth and quiet while the 12v looks and sounds like there is a hammer smacking the lock open and closed. The actuator is designed to move old manual door locks so I can see it needing a lot of power to do it's job. But the tailgate's lock is just a couple plasitc pieces so the power is an overkill.

 

Skip the physics and math problems:

$8 DZS Elec 2PCS LM2596 DC-DC Step Down Variable Volt Regulator Input 3.2V-40V Output 1.25V-35V Adjustable Buck Converter Electronic Voltage Stabilizer Power Supply Module

https://www.amazon.com/gp/aw/d/B06XRN7NFQ/ref=sspa_mw_detail_4?ie=UTF8&psc=1

Turn the little screw until it works good.

 

Resistors are the best way to go, as these would ony let the lock go one way. You would need 2 and put some diodes in the supply side for as the OP mentions reverse polarity is needed when the lock closes and opens. But they are also only 2 amps output, the locks draw more if loaded.

 

Just get a new actuator in that case.

 

I seem to remember reading that a second resistor in series has a multiplying effect though right? Once the power goes through the first resistor then the volt/amp variables are lower so the second resistor has less work to do so ends up resisting more. Am I making any sence? E.G. a 10 ohm resistor followed by a 3 ohm resistor would result in a 20 ohm reduction (just an example to make the math simple).

 

For series circuit, unless some really wierd shit is going on, it is just purely additive.

Circuit with 10ohm and 3 ohm resistor in series will behave same as a 13 ohm resistor overall.

Circuit with 2x 10 ohm resistors (in series) will behave same as 1x20 ohm resistor.

Absolute voltage at the second resistor will be lower for the double resistor case, but voltage drop over each smaller resistor will just be their proportional amount of the voltage drop of the bigger resistor.

Power dissipated by smaller resistors will be half of the big resistor (if 2 half resistors used).

[This is much different than comparing what happens in the circuit with a single specific resistor, to what happens in the same circuit but adding an additional resistor in addition to the first. Ie comparing circuit with 1x10 ohm resistor to similar circuit except 2x10 ohm resistors]

 

So two 10 ohm resistors will get to the same voltage and amps as one 20 ohm?