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Old 11-06-2020, 05:35 PM #1
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How can I change the 110v to use 400watts all the time?

I'm just wondering if we can permanently use the 110v 400watts all the time while driving.

the reason for this is that i have a 180watts power adapter for a computer and during driving, i wanted to charge my computer. unfortunately the current setting is that when you are driving, the 110v only puts out 100watts but when parked, that's the only time you can use the 400watts.
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Old 11-06-2020, 06:33 PM #2
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What does your computer require for charging? My 13" Macbook Pro only takes about 80W. You might be ok without changing anything.
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Old 11-06-2020, 06:41 PM #3
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buy a 12V DC charger for your laptop

Last edited by jhguth; 11-06-2020 at 06:43 PM.
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Old 11-06-2020, 08:41 PM #4
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Quote:
Originally Posted by jhguth View Post
buy a 12V DC charger for your laptop
This is the correct answer. You're wasting a ton of power going DC -> AC -> DC and will kill your battery way faster than just using a 12VDC laptop charger.
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Old 11-07-2020, 02:51 PM #5
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Quote:
Originally Posted by jhguth View Post
buy a 12V DC charger for your laptop
The problem with most 12V DC charger adapters is they plug into the cigarette lighter socket which is limited to 10 Amps or about 120 to 140 watts. You don't dare put in a larger fuse for that may over heat the wiring.

My suggestion is to get a stand alone power inverter. I have a Coleman 800 watt (1600 watt peak) inverter that attaches directly to the battery. Yes you will have to run some wires if you want to use it while driving, but beats trying to modify the built in inverter which will void your warranty and possibly damage wiring.
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Old 11-07-2020, 04:51 PM #6
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Have you tried it with your current setup? I doubt the laptop will pull more than 100 watts unless its under heavy load.
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Old 11-08-2020, 09:52 AM #7
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Takes literally 5 minutes to do the mod... It is literally just cutting one wire and grounding it

400W Inverter Mod?
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Old 11-08-2020, 04:43 PM #8
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The watt rating for laptop chargers is what they can deliver at the charge port, so at 5-20v that USB or USB-C PD is rated for.

It’s not 100 or 160 watts at 120v but rather at the voltage delivered at the output of the charger.

100w at 120v ac is plenty to support even 160w at 20v charging of electronics.
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Old 11-08-2020, 06:17 PM #9
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Quote:
Originally Posted by jp.vegas View Post
The watt rating for laptop chargers is what they can deliver at the charge port, so at 5-20v that USB or USB-C PD is rated for.

It’s not 100 or 160 watts at 120v but rather at the voltage delivered at the output of the charger.

100w at 120v ac is plenty to support even 160w at 20v charging of electronics.
I believe you are confusing amps and watts....

Watts = Voltage * Amperage

160 watts = 120v * 1.33A
160 watts = 12v * 13.33A

Thus the inverter will be drawing around 13.3A @12v, yet outputting approximately 1.33A @ 120v... Of course, this is assuming it was 100% efficient, which it isn't, assuming ~87% efficiency to produce an output of 160w @ 120v, it would need to pull enough in to produce 183w, or approximately 15.3A @ 12v
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Old 11-08-2020, 07:10 PM #10
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Quote:
Originally Posted by Anand View Post
I believe you are confusing amps and watts....

Watts = Voltage * Amperage

160 watts = 120v * 1.33A
160 watts = 12v * 13.33A

Thus the inverter will be drawing around 13.3A @12v, yet outputting approximately 1.33A @ 120v... Of course, this is assuming it was 100% efficient, which it isn't, assuming ~87% efficiency to produce an output of 160w @ 120v, it would need to pull enough in to produce 183w, or approximately 15.3A @ 12v
I'm doing no such thing.

People are talking about their chargers being 100w or more so they may need the 400w portion of the inverter to run them. A 100w USB charger doesn't draw 100w at 120v, but rather can supply up to 100w to the device being charged. In the case of USB C IF, that's 100w at 20v or 5a at 20v. That is less than an amp or less than 100w at 120v ac.

They're confusing wattage at the supply voltage with what the charger draws at the wall, that's all I'm pointing out.
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Old 11-08-2020, 10:28 PM #11
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i have 180w power adapter.

Quote:
Originally Posted by flyrv9 View Post
What does your computer require for charging? My 13" Macbook Pro only takes about 80W. You might be ok without changing anything.
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Old 11-08-2020, 10:54 PM #12
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Anand, just what i needed. i'm going to make this mod at some point.

Quote:
Originally Posted by Anand View Post
Takes literally 5 minutes to do the mod... It is literally just cutting one wire and grounding it

400W Inverter Mod?
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Old 11-09-2020, 08:34 AM #13
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Quote:
Originally Posted by jp.vegas View Post
I'm doing no such thing.

People are talking about their chargers being 100w or more so they may need the 400w portion of the inverter to run them. A 100w USB charger doesn't draw 100w at 120v, but rather can supply up to 100w to the device being charged. In the case of USB C IF, that's 100w at 20v or 5a at 20v. That is less than an amp or less than 100w at 120v ac.

They're confusing wattage at the supply voltage with what the charger draws at the wall, that's all I'm pointing out.
P=VI, watts (power) equals voltage (potential) times amps (flow)

You can't just create or destroy power because you change the voltage (conservation of energy), if the voltage changes the current will change, and vice versa, but power will be constant.

Quote:
Originally Posted by fkheath View Post
The problem with most 12V DC charger adapters is they plug into the cigarette lighter socket which is limited to 10 Amps or about 120 to 140 watts. You don't dare put in a larger fuse for that may over heat the wiring.

My suggestion is to get a stand alone power inverter. I have a Coleman 800 watt (1600 watt peak) inverter that attaches directly to the battery. Yes you will have to run some wires if you want to use it while driving, but beats trying to modify the built in inverter which will void your warranty and possibly damage wiring.
inverters are inefficient and totally unnecessary considering the laptop runs on a DC power supply. if it's actually drawing 180W then worst case spend $15 and install a dedicated 12v power outlet or just hardware the charger.
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Old 11-09-2020, 12:48 PM #14
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Quote:
Originally Posted by jhguth View Post
P=VI, watts (power) equals voltage (potential) times amps (flow)

You can't just create or destroy power because you change the voltage (conservation of energy), if the voltage changes the current will change, and vice versa, but power will be constant.


inverters are inefficient and totally unnecessary considering the laptop runs on a DC power supply. if it's actually drawing 180W then worst case spend $15 and install a dedicated 12v power outlet or just hardware the charger.
Sigh. OK. So changing the voltage in your equation doesn't change the wattage? Gotcha
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Old 11-09-2020, 04:53 PM #15
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Quote:
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So changing the voltage in your equation doesn't change the wattage? Gotcha
Correct, because amperage will change

Power is work (energy) per unit of time, time isn't changing and since you can't just destroy or create energy from nothing that means your power is also constant.

If something is using 180 watts of power it is transferring 180 joules or 180 newton meters of energy every second, that's it's measure of work per unit of time. it doesn't create less or more work because you use a different voltage. if you increase both the voltage and amperage the power increases exponentially, so doubling both the voltage and amperage would increase the power 4X.

watts is joules per second (energy per time), volt is joules per coulomb (energy per charge, a potential), and amps is coulomb per second (charge per time)

if something can do a certain amount of energy per time -> if you decrease the amount of energy for each charge then you have to increase the amount of charges per time to still provide the same amount of energy per time. each charge has less energy, so you need to give it charges at a faster rate to keep the same amount of energy per time.

https://www.electronics-tutorials.ws...its/dcp_2.html


so if the 180W is at 19V

so P=VI; 180W = 19V * I --> 180W / 19V = I = 9.47 A

and 180W at 14V is 180W = 14V * I --> 180W / 14V = 12.86 A

or at 5V is 180W = 5V * I --> 180W / 5V = 36 A

you'll notice that as voltage decreases the amperage increases because of that relationship described by P=VI, one practical application of this related to vehicles is that you can use smaller wiring on 24V systems because of the lower amperage.

if you look at a power supply you will notice that the amperages are different on the different input and output voltages, if what you were saying is correct the amperages would be the same (the difference in the product of the voltage and amperage for input and output depends on the efficiency).
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