# REPOSTED Ohms law & watts law



## Morix (4/4/19)

Credit :

JustSayNoToDiacetyl
Found this on reddit hope that it will be helpful.


Most people still seem to be living in the mech mod days, where in order to calculate the current (amps) being drawn from the battery, you only need to know the resistance of the coil and the nominal voltage of the battery. This makes sense on a mech because the resistance of the coil is the ONLY thing you have control over (well that and the charge of the battery). However, if you are using a regulated mod, this does not apply. Resistance of the coil means absolutely zero on a regulated mod.

Let me give a brief explanation on why this is. (NOTE: I am not an electrical engineer, but I did stay in the honeymoon suite of a Holiday Inn Express last night).

Regulated mods typically use a DC-DC converter (probably in most cases a switched-mode converter). That is, they separate the input and output voltage (in other words, they separate the battery from the atomizer). So, just because you have 3.7v going in from your battery doesn't mean this is what will be hitting the atomizer. On a mech mod, yes, that is what happens because there is nothing in between the atomizer and the battery. On a regulated mod, there is a voltage regulator in between the battery and the atomizer.

On a mech mod, as your battery drains, you have less voltage (and thus less power) hitting the atty. This means the vapor production diminishes over time. We know from Ohm's law that Power = voltage X current. As you can see from this simple equation, as the voltage drops, this necessarily means less power (watts).

The circuitry in a regulated mod stops this from happening. The regulator will swap voltage for current in order to achieve the power (watts) you have your mod set at. Again, P = I * V. As the voltage on the right side of the equation drops, the power also must drop. So, looking at this equation, how can we keep the power constant throughout the charge of the battery? Yep, we need to increase the "I" (current) to compensate for the battery being drained. This will allow you to keep your desired power setting all the way through the battery's charge.

So let's cut to the brass tacks. If you want to know the current being drawn from your battery on a regulated mod, you need to solve the equation for I (since the value of I is what you want). The equation for current is: I = P/V. That is, Current = Power/Voltage. Voltage will be a fixed parameter (and it will depend on the charge of your battery) and power is easy enough to see on your mod's screen. (Some mods also show the remaining battery charge in volts, which comes in handy for this. If not, you will have to guess, but 3.7v is a good "guess" for a battery that's not low).

Example: Let's say you have your mod set at 50 watts and your battery has about a 3.7v charge on it. Thus: 50/3.7 = 13.5 amps. If you have your mod set to 50 watts and you have a (relatively) fresh battery, you are going to be pulling 13.5 amps no matter what the resistance of your coil is.

But, for the sake of argument, let's say your coil is a 0.5Ω kanthal build. What most people would do would be to go to an online calculator and input the resistance of their coil and their power (watts). If you did this, the current draw would be shown as 10 amps (which is incorrect). Why is that incorrect? Because the voltage required to make that work is 5v. None of our batteries can output 5v (unless they are in series) or unless the regulator is increasing the amp draw to make those 5 volts. And the regulator will have to increase the amp draw to make that work, so in reality you are drawing 13.5 amps (as I showed in my original calculation), not 10 amps. The same result will happen if you input the output voltage and resistance. You will get back 10 amps, which is incorrect.

I think the biggest bit of confusion comes from the fact that people don't know the difference in input voltage (what comes directly from the battery and varies based on charge level) and output voltage (what the regulator puts to the atomizer to achieve your desired power level). And the mods we use don't help the matter. A lot of these mods show the power, the resistance of the coil and the applied voltage on the screen at the same time. So, a lot of people assume this is the voltage value you plug into an Ohm's law equation. Using our .5Ω example, if you had it set to 50 watts, the screen would show 5v. So if you plugged in 5v and 0.5Ω into a calculator, you would once again get back the incorrect result of 10 amps.

I saw a guy earlier who was talking about his build on a regulated mod. He said "I am running 0.3Ω at 80 watts. This equals 16 amps, so I am well within the safe limits of my 20 amp battery." Well, he committed the cardinal sin of using the output voltage in his calculation (or using Power and atomizer resistance to calculate current -- neither are correct). Let's do his calculation properly. If his battery is fully charged (around 4v) and he is running at 80 watts, then:

I = P/V

80/4 = 20 amps.

In reality he is pulling 20 amps on a full charge, not 16 as he thinks. But that's not all. Since the battery voltage drops during use, the regulator will have to increase the amperage drawn to keep him at 80 watts. So, let's say his battery is near dead and is at 3.2v.

80/3.2 = 25 amps

Now, since most batteries are 20 amps continuous, we might be getting into some danger territory (possibly). Meanwhile this guy is vaping happy thinking he is still only pulling 16 amps from the battery.

Multiple Battery mods:

Some mods will run multiple batteries in series or parallel. First we need to explain the difference:

A series connection will treat the two batteries as one. The advantage of a series connection is the voltage doubles (from 3.7v nominal to 7.4v nominal). This means you will, in a sense, be putting less stress on the battery. For instance, if you were to run 100w with a 3.7v battery, you are drawing 27 amps on a fresh battery. If, however, you have two batteries in series, you will only be drawing 13 amps. Remember, in series, the two batteries behave as one. Therefore, the amp limit of one battery is the amp limit of both batteries (25R's in series still have a 20 amp limit). However, with the doubled voltage, you don't need as much current to achieve the desired power. So, even though the amp limit hasn't magically doubled, the need for those amps has decreased.

Parallel configurations do not double the voltage. Two 3.7v batteries in parallel still only provide 3.7v. However, the capacity of the cells (amp hours) is doubled. This also means that the amp limit will also double (from 20 amps for one battery to 40 amps for two, and so on). So, if you want 100 watts with parallel batteries, you will still only be applying 3.7v to achieve it, which means you are drawing 27 amps on a fresh battery. But since you have two batteries with a 20 amp limit, the 27 amps is well within that 40 amp margin of safety.

Essentially, as far as current draw is concerned, series and parallel achieve the same thing. It's just the way they achieve it is different. Series achieves it with more voltage (which decreases the need for more current), while parallel achieves it by doubling the amp hours (increasing the available current).

Example: You are running your Sigelei 150 at 150 watts and you want to know the amp draw on your batteries.

I = 150/ 7.4v = 20 amps

Since your two batteries become a single battery in series, the nominal voltage effectively doubles. This means you don't need as much current to hit that 150 watts.

Don't confuse output voltage for input voltage on regulated mods. If you want to determine your current draw from the battery on a regulated mod, here is the only correct way to do it: I = P/V. That means your current will equal your watt level divided by how much charge you have on your battery. If you don't know the charge, then just plug in 3.7v (as that's the nominal rating). Atomizer resistance has nothing to do with the current being drawn from your battery on a regulated mod.

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## Silver (4/4/19)

Wow @Morix 

That was an excellent explanation


Thank you!

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## DougP (4/4/19)

Dam that was informative

Thank you for that and the effort it must have taken to write it

And that guys is why I love this forum 

Sent from my LYA-L09 using Tapatalk

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## Dela Rey Steyn (4/4/19)

Thanks @Morix you are a champ! Now I can just direct people here instead of trying to explain it all everytime! You are much better with putting it in writing than me! After the 3rd puzzled look I get, I tend to lose hope

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## Dela Rey Steyn (4/4/19)

@Morix, would love to hear your thoughts on PWM mods!

I personally prefer PWM over DC-DC, it's more power efficient and knowing how to build correctly for your PWM mod gives you a much smoother delivery of power to the coils. I think there are still some regulated mods around that use a combo action of both, PWM for stepping down and DC-DC to go above.

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## Morix (5/4/19)

Dela Rey Steyn said:


> @Morix, would love to hear your thoughts on PWM mods!
> 
> I personally prefer PWM over DC-DC, it's more power efficient and knowing how to build correctly for your PWM mod gives you a much smoother delivery of power to the coils. I think there are still some regulated mods around that use a combo action of both, PWM for stepping down and DC-DC to go above.


 PWM vape is the simplest of the regulated type vapes that allows you to adjust how much power goes to your coil. While it's nice to be able to fine tune it does have it's disadvantages such as it may need adjusting as your batteries drop and any time you change your coil. But, it is the simplest design that lets you easily adjust the power you're using for vaping and can extend the life of your batter charge since it doesn't run the vape coil constantly.


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## Dela Rey Steyn (5/4/19)

Morix said:


> PWM vape is the simplest of the regulated type vapes that allows you to adjust how much power goes to your coil. While it's nice to be able to fine tune it does have it's disadvantages such as it may need adjusting as your batteries drop and any time you change your coil. But, it is the simplest design that lets you easily adjust the power you're using for vaping and can extend the life of your batter charge since it doesn't run the vape coil constantly.



It's beauty lies in the simplicity for me. Ordered a few OKR chips and a LMC555 CMOS timer, going to build my own PWM mods during the course of the year, really looking forward to it

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## Morix (5/4/19)

PWM works by repeatedly turning power off & on at the desired rate so that the average power can be a specific % of the available power. For PWM vapes that means if you have 8V batteries running with 50%, how have an average of 4V, 75% would be an average of 6V. By adjusting the on/off ratio up or down you can get almost any effective power. You just have to remember that the batteries and switch (usually a MOSFET for vapes) needs to be able to handle the maximum current of 100%

How fast you do the pulses makes a difference once you get practical. High speed pulses make the average smoother, but ,most high current MOSFET's can have limits to how fast they can switch. On top of that, the more often the MOSFET switches the hotter it will get because while it is turning off or on is when most of the heat is generated in PWM circuits. Another factor that comes up is that PWM vapes can make noise, from a low buzz to a high pitch whine that can irritate people. That means that personal preferences are also factors in PWM frequency.

Finally there are additional technical issues in specific designs caused by specifications and voltages involved that become additional factors, for example the lower the voltages you are working with the slower you should switch the MOSFET off & on to reduce heating because the parts take longer to turn off & on at lower voltages.

Most simple PWM systems used with vapes will have a simple adjustable resistor you change to set what the on/off percentage in. It can be a pot on or through the case or a more precises multi turn trimpot. There are advantages and disadvantages to both types of pots. Since the PWM's power is based on % of the voltage, as your batteries drop, you may find yourself adjusting the PWM, and you definitely will want to adjust it each time you change the vape coils.

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## Dela Rey Steyn (5/4/19)

yes, its also very efficient. Lol, yip, they can Buzz, but with a nice N-fet you can get rid of most buzzing. there are also some nice pre-made PWM boards available online from Mod makers in the US and the UK, but its a bit expensive shipping them here, rather building my own.


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## Morix (5/4/19)

Dela Rey Steyn said:


> yes, its also very efficient. Lol, yip, they can Buzz, but with a nice N-fet you can get rid of most buzzing. there are also some nice pre-made PWM boards available online from Mod makers in the US and the UK, but its a bit expensive shipping them here, rather building my own.


Start with a basic 555 timer. It's usually cheap to assemble.


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## Dela Rey Steyn (5/4/19)

Morix said:


> Start with a basic 555 timer. It's usually cheap to assemble.


yip, cant believe the local pricing on some components!


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## Silver (5/4/19)

Hi @Morix 

Firstly, thank you for highlighting and emphasising that in a regulated mod there are actually 2 circuits.

The way I understand it from your post is as follows:

The input circuit - which is the power source - i.e. connected to your battery
and
The output circuit - which is what is connected to your coil (the load)

I like how you explained it
As you dial in the power you want, it needs to draw more or less current from the input circuit to meet the power demand on the output circuit. So as the battery goes flatter (lower voltage) then it needs to draw more current to meet the power you have dialled in.

*I have a question relating to your original post.*

For a single battery DC-DC regulated mod, *is the Input Power and Output Power the same?*
(Other than efficiency losses)

I would like to just do an example of this with some calculations. Please let me know if my thinking and calculations are correct.

Suppose you have a *single battery *regulated mod.
You are using a coil that is *0.2 *ohms
You set the mod to power mode and set it to *70 Watts*

*Scenario 1 - battery is fresh at 4.0V*

Input Circuit
(no need to discuss resistance because the load is not connected here)
P=V*I
I=P/V=70/4=*17.5 amps*
Power = 80 Watts, V = 4.0 Volts 

Output Circuit
I am assuming that the Power of the output circuit is the same as the input circuit? i.e. 70 Watts (lets ignore efficiency loss for now)

How do you calculate the V and the I? I am assuming you need to use an ohms law equation that includes the Resistance R (0.2 ohms) and the Power? Because the resistance is connected to the output circuit?

I assume then as follows:
P=Vsquared/R, solve for V, so V=sqrt(P*R)
V=sqrt (70*0.2)
V=sqrt (14)
V=*3.74 Volts*

Solving for I
I=P/V
I=70/3.74
I=*18.7 amps*

So on the output circuit, the volts are 3.74 volts (it's less than the freshly charged battery on the input circuit) and the current is 18.7 amps in this circuit - a bit more than on the input circuit. 

All seems good and well.

Now lets consider what happens when the battery goes flat and you still keep it dialled in at 70 Watts.

*Scenario 2 - battery is flatter at 3.2V*

Input Circuit
P=V*I
I=P/V=70/3.2=*21.9 amps (i.e. it has to draw more to reach the 70 Watts)*
Power = 80 Watts, V = 3.2 Volts 

Output Circuit
Using the same assumptions as above:
P=Vsquared/R, solve for V, so V=sqrt(P*R)
V=sqrt (70*0.2)
V=sqrt (14)
V=*3.74 Volts *

Solving for I
I=P/V
I=70/3.74
I=*18.7 amps*

I.e. the output circuit's voltage and current are the same when the battery is flatter. (Makes sense because you have the same power requirement dialled in and the same coil resistance) It just had to draw more current on the input circuit to get to this power from the flatter 3.2 V battery.

Does this all sound right to you?

The issue being then that if you were using a 20A max continuous rated battery (20A CDR) you would be exceeding that 20 amp draw on the input circuit even though the numbers on the output circuit look okay. 

Follow up questions if I may

When a single battery regulated mod shows you the voltage and current on the screen, is this on the input circuit or output circuit? Or does it change depending on the manufacturer?
When they say a regulated mod has a certain current limit (eg 30A), is that the limit on the output circuit or the input circuit?

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## Dela Rey Steyn (5/4/19)

@Silver , I found this article by Mooch explains it quite well (but unfortunately it is for multiple battery mods):

Calculating the current being drawn from the batteries in a regulated device can be very confusing. You can't do it the same way as you would for a mechanical/unregulated device and there are so many different battery configurations; single, dual parallel, dual series, triple series, etc.

The way I keep it all sorted out is to remember that, in a regulated mod, the coil isn't connected to the battery. The regulator is. To calculate the current being drawn from each battery when using variable-wattage (VW) mode you need to calculate the maximum wattage each battery supplies.

Here's how I do it...
As an example, the RX200 has a maximum wattage rating of 200W. Since it uses three batteries that means each battery supplies 200W / 3 = 67W. For dual parallel or series 150W devices each battery supplies 150W / 2 = 75W. You use this method for series or parallel devices, it doesn't matter.

Once you have the maximum wattage for each battery then you can use the following formula to determine the maximum amount of current that can be drawn from each battery...

Max Amps Per Battery = Max Wattage Per Battery / Minimum Voltage Per Battery

For the RX200 the minimum possible cutoff voltage is 9.0V, which is 3.0V per battery (unless you set the cutoff higher). For most other devices the minimum is 3.2V or 3.1V per battery. Let's use the Sigelei 150W TC device as an example. This device has a minimum battery voltage of 6.4V, which is 3.2V per battery...

Max Amps Per Battery = 75W / 3.2V = 23.4A

So you want a battery that can safely supply 23.4A of current if you're using the mod at its maximum rating of 150W.

I should add that to get as close as possible to calculating the max current being pulled from your batteries you should add an additional 10%. This will account for the inefficiency of the regulator. For example, if your device draws 23.4A then add 2.34A for a total of 25.74A. Not a big difference, but it's there. That changes the equation to...

Max Amps Per Battery = (Max Wattage Per Battery / Minimum Voltage Per Battery) / 0.9

If you know you will not be exceeding a particular wattage that is less than the maximum then you can use that wattage in the equation instead. This often means you're able to use a higher capacity battery like the HG2 or 30Q instead of a high current rated, but lower capacity, battery like the VTC4 or HB6. It's worth doing the math to find out.

This works for series or parallel devices. It does not matter how they are connected as we are already taking that into account when we calculate the max power for each battery.

It takes much longer to explain all this than it does to actually calculate the amount of current being drawn from your batteries. I hope this helps make the very confusing process of determining how much current is being drawn a little bit easier.

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## Dela Rey Steyn (5/4/19)

@Silver , also there is this:

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## Silver (5/4/19)

Thanks @Dela Rey Steyn 

That is very useful indeed!

In my post above I am focusing on a single battery mod however.

I get the idea that its not straightforward to calculate how much current is being drawn from the battery in the input circuit - but my main question relates to how to work out what is going on in the output circuit and to confirm whether the input and output circuits have the same power?


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## Silver (5/4/19)

Dela Rey Steyn said:


> @Silver , also there is this:
> 
> View attachment 162758



Thanks @Dela Rey Steyn 

But that doesn't make sense to me. 

Surely the input circuit will pull whatever current it can from the batteries to make the required power. So if it needs 22 amps and your batteries are only 20A continuous rated - it will pull the 22 amps. I don't see how the processor will know that its only a 20A CDR battery in there. Presumably a 20A CDR battery can drain a bit more than that - and that's where the trouble comes in. Unless it is monitoring the temperature of the battery and checking if it goes above a certain temp then it must stop. 

Who wrote that post if I may ask?


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## Dela Rey Steyn (5/4/19)

Silver said:


> Thanks @Dela Rey Steyn
> 
> But that doesn't make sense to me.
> 
> ...



Here is a couple of threads i found relating to these regulated mods and battery usage:

https://www.e-cigarette-forum.com/threads/confusion-with-voltage-output-for-box-mods.732652/
https://www.e-cigarette-forum.com/b...attery-current-draw-for-a-regulated-mod.7532/

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## Dela Rey Steyn (5/4/19)

also found this interesting, From Evolve, datasheet on a DNA60 chip:

https://downloads.evolvapor.com/dna60.pdf

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## Morix (5/4/19)

@Silver

Check this out. Learned alot from this professor.

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## Silver (5/4/19)

Morix said:


> @Silver
> 
> Check this out. Learned alot from this professor.




Thanks @Morix 
But can you check my calculations above and let me know if they look right to you?


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## Dela Rey Steyn (5/4/19)

Silver said:


> Thanks @Morix
> But can you check my calculations above and let me know if they look right to you?



your equations looks correct @Silver
I don't think we can successfully calculate the input/output of a circuit without knowing how that circuit is setup unfortunately. Different chips handle the power draw and output differently. I don't think a universal equation for this is applicable.

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## Silver (5/4/19)

Dela Rey Steyn said:


> your equations looks correct @Silver
> I don't think we can successfully calculate the input/output of a circuit without knowing how that circuit is setup unfortunately. Different chips handle the power draw and output differently. I don't think a universal equation for this is applicable.



Ok thanks @Dela Rey Steyn 

So then the next question is that when you set the power on your regulated mod (single batt mod), is that the power on the output circuit (where the coil is) or the input circuit (where the battery is)

Or are you saying that it depends on how the mod works?


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## Dela Rey Steyn (5/4/19)

the only constant is Ohm's law, how much power is pulled from the battery will be defined by the chip, and most are setup to not draw more than what can be drawn ie. a mod letting you know "Ohm's to low" or "Battery too weak"



Silver said:


> Ok thanks @Dela Rey Steyn
> 
> So then the next question is that when you set the power on your regulated mod (single batt mod), is that the power on the output circuit (where the coil is) or the input circuit (where the battery is)
> 
> Or are you saying that it depends on how the mod works?



My understanding would be that what is displayed on the screen is what you ideally want the output to be, and not necessarily what is being pulled from the battery. for example, i have a 30q in an Aegis, at 100w, it wont let me fire a 0.22 coil and at 70w it shows an output of 14Amp and fires no problem, though I doubt i can sustain that 70w for very long

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## Dela Rey Steyn (5/4/19)



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## Dela Rey Steyn (5/4/19)

its unfortunately the only regulated mod i have on me now, so can't explore other chipsets right now.


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## Dela Rey Steyn (5/4/19)

@Silver, if you look at the below images, the 70w for 14amp drawn doesn't make sense normally, but the chip makes it work:


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## Alex (5/4/19)

Morix said:


> Most people still seem to be living in the mech mod days, where in order to calculate the current (amps) being drawn from the battery, you only need to know the resistance of the coil and the nominal voltage of the battery. This makes sense on a mech because the resistance of the coil is the ONLY thing you have control over (well that and the charge of the battery). However, if you are using a regulated mod, this does not apply. Resistance of the coil means absolutely zero on a regulated mod.
> 
> Let me give a brief explanation on why this is. (NOTE: I am not an electrical engineer, but I did stay in the honeymoon suite of a Holiday Inn Express last night).
> 
> ...



Thanks for sharing, in future please make an effort to credit the original poster, with a link to the original post.




source

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## Morix (5/4/19)

Alex said:


> Thanks for sharing, in future please make an effort to credit the original poster, with a link to the original post.
> 
> View attachment 162772
> 
> ...



Yes @Silver has spoken to me. I was not aware reposting wasn't allowed. It will still be easier toe find it here for new users. Thank you for the direct attack. Very professional.

And even in reposting says in no way that i don't know what im talking about with my posts. I thought i would repost for those new people that have not seen or have no idea of watts law with a regulated device. 

Im new to the forum and was not aware that you can't repost to help others. 

Even though you @Alex reacted like a child to attack someone straight on for speading good things that someone might need.. it doesn't bother me. But that you. I know now to make "original posts. 

Im sure there are more reposted content. Try to handle those folks in a more professional manner. 

Have a good day.


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## Silver (5/4/19)

Hi @Morix 

Your reposting of other content is more than welcome.
Certainly not disallowed here.

Its just that the way you posted that original post, it looked like you wrote it and I even congratulated you on the excellent writeup.

So with this kind of thing its always best to mention that you found it somewhere else (and ideally provide a link to the source) and hopefully it can add value to members. 

Don't be upset - just try to credit your source when you are posting other content in future. And it helps for everyone to know where its from instead of thinking its you that wrote it.

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