Find AC current from DC Load

I'm working on a project where I need to use a switch to control AC power to a laptop power supply (roughly 19VDC 3.5A). I need to make sure I'm not exceeding the max current on the switch, and I don't want to start cutting the power supply cable to measure it.

To figure out the current, I'd assume I find the DC power (Amps*Volts) then divide by 120 VAC. Something tells me the true conversion isn't so simple/ideal. I estimated that I'll be using about 40-45 watts DC (mostly at 12VDC after going through a buck converter). Can someone tell me if this is correct or point me to the right conversion if it is not? Thanks

I need to use a switch to control AC power to a laptop power supply

Use a SSR (SOLID STATE RELAY)
http://www.ebay.com/itm/IG1NC-210D-Solid-State-Relay-10A-Output-0-220VAC-1pcs-/360614748118

I'm using the electricity I'm switching to power my Arduino, so that won't work.

I don't understand the question about the switch. Why can't you just buy a toggle switch to switch the AC ? If the AC you are switching is to power your arduino, then you can't use the arduino to turn it's own power on so what is it you are trying to find out. ?
Is there something about the "switch" that you are not telling us ? What exactly do you mean by "switch" ? You can't just buy an in-line switch equipped extension cord. (rocker switch in line , built into the cord) ?

Hi,
I am a bit lost too. You want to switch the AC power to the power supply, then you say you can’t switch it because it is powering your Arduino. Did I misunderstand something?

Anyway,
As far as the supply current goes, your approach is probably a little too simple, though in some cases it might be near enough (depending how precise you need to be).
Your power supply is rated for 19V at 3.5A, i.e. for 66W.
You need about 45W, but this is at 12V, after a buck converter. What is the efficiency of the buck converter at 45W? This I don’t think you will know, so lets estimate it to be about 75% (I see computer power supply efficiencies from about 75% to 90%+ from a quick Google search… and will use the same figures for your buck converter… the lower end to be conservative).
So your 45W at 12V becomes 45/0.75=60W at the laptop power supply. Nearly at its rated output.
Now what is the efficiency of the laptop power supply? Say we use 75% again, to get 60W out, we need to put 80W in.
So at 120VAC(rms), you need 0.67A to get 80W. But that assumes the power factor is unity. It’s been a long time since I looked at the input current wave shape of a laptop power supply, but if it’s distorted (and I expect it might be… high in third harmonic), then your power factor will drop and the rms current goes up. I would expect it could go up to, say, 1 amp.
In fact, with all these very approximate assumptions, I would allow for a couple of amps.

You could use a pcb mount relay something like this:

but the current draw of the coil is greater than the output capabilities of the arduino i/o pins. You could parallel a couple together… I think.

Eric

Which is why Raschemmel suggested a SOLID STATE RELAY. They CAN be run directly from the Arduino.

d_vee:
I'm working on a project where I need to use a switch to control AC power to a laptop power supply (roughly 19VDC 3.5A). I need to make sure I'm not exceeding the max current on the switch, and I don't want to start cutting the power supply cable to measure it.

To figure out the current, I'd assume I find the DC power (Amps*Volts) then divide by 120 VAC. Something tells me the true conversion isn't so simple/ideal. I estimated that I'll be using about 40-45 watts DC (mostly at 12VDC after going through a buck converter). Can someone tell me if this is correct or point me to the right conversion if it is not? Thanks

Read the label on the laptop PSU, it's required to say the current draw of the device,
which gives you the answer without any work!

Mine says 19.5V 6.2A out, 100V -- 240V 1.2--0.7A in

Sorry, I forgot to come back to this. I believe monkey-man answered my question nicely, but I'm going to answer things for anyone who may come across this.

monkey-man:
I am a bit lost too. You want to switch the AC power to the power supply, then you say you can't switch it because it is powering your Arduino. Did I misunderstand something?

I was responding to the proposal to use a relay. The power I'm trying to switch will be powering the Arduino. If the Arduino has no power, I can't hope to switch it on with a relay. I have a manual switch that I'm using, and it has max values that I want to stay under.

MarkT:
Read the label on the laptop PSU, it's required to say the current draw of the device,
which gives you the answer without any work!

The max AC current on the power supply is 1.6 A. I don't intend to get too close to this value, and I wanted to make sure I have a good amount of headroom on the switch, so I wanted to know the AC current that would be pulled due to my DC loads.

First of all don't mess with the AC side, its harder and more dangerous especially if you don't do it correctly.

It is always preferable to switch the lower voltage side whenever possible. Its safer and you can use a smaller switch on the DC side.

120Vx1.6A = 192W
20V x 3.5A = 70W

The max AC current on the power supply is 1.6 A. I don’t intend to get too close to this value, and I wanted to make sure I have a good amount of headroom on the switch, so I wanted to know the AC current that would be pulled due to my DC loads.

First of all don’t mess with the AC side, its harder and more dangerous especially if you don’t do it correctly.

It is always preferable to switch the lower voltage side whenever possible. Its safer and you can use a smaller switch on the DC side.

If the switch is on the dc side, the power supply will still be powered and using electricity… This is not a problem, but if anything went
wrong it internally the only way to shut it down would be yank the plug;, not the best way to wire something up.

Almost any toggle switch you purchase will be rated for 10A @250vac so there is nothing to worry about there. Put the switch on the AC side but just MAKE SURE you solder it correctly. If you cannot guarantee that, then do not install the switch , period. Purchase a switchable extension cord with the inline rocker switch.

It is always preferable to switch the lower voltage side whenever possible. Its safer and you can use a smaller switch on the DC side.

120Vx1.6A = 192W
20V x 3.5A = 70W

That's sort of a half truth. If we were controlling an AC motor and it draws 10A at 240V, to keep the same power it would require 20A at 120V That will require a switch (or relay) with TWICE the contact rating thus a larger switch.

If we were switching things very quickly, the DC side is the side to switch since that little "wall wart" is a power supply and they don't take kindly to fast on/off cycles.

If you use a solid state relay none of your calculations are valid because it has a zero crossing detector and switches at the zero crossing point of the ac waveform. That is the beauty of the SSR. That and the fact that it works perfectly with a 5V signal from an arduino.

@Raschemmel: That does NOT invalidate the calculations. Where in the cycle it switches "ON" is moot. Your solid state relay will still have enough "Amps" and Volts capacity to sustain the load!

@Raschemmel: That does NOT invalidate the calculations. Where in the cycle it switches "ON" is moot. Your solid state relay will still have enough "Amps" and Volts capacity to sustain the load!

Ok. It may not invalidate the calculations but those calculations can't apply to the moment it turns on or off like it would with a conventional toggle switch or relay. With 33 years of experience you know very well that switching ac under load conditions ALWAYS
results in arcing across the contacts of the toggle switch or the relay. My point is that will not occur with an SSR, so while the calculations may apply to all other times, the one time they do not apply is at the moment of switching because AMPS and VOLTS as you put it are ZERO. (at the zero crossing point )
It comes down to the question "Do you want arcing to occur every time you turn it on or off ?"
If the answer is NO, then the solution is a solid state relay.

Which is exactly why I cringe whenever someone suggest controlling an inductive load with a relay. Those Arduino add-on boards that have 8 relays. Yikes.

Hi, I assume what you want to do is have a switch close to the power supply, rather than having to switch the supply on and off at the wall switch.

If that is the case why not buy a power board with switches on it, plug it into the wall and then plug the power supply into the board.
Position the board next to the power supply and there you have it, a switch next to the supply to turn it on and off.

Tom............ :slight_smile:

To figure out the current, I'd assume I find the DC power (Amps*Volts) then divide by 120 VAC. Something tells me the true conversion isn't so simple/ideal. I estimated that I'll be using about 40-45 watts DC (mostly at 12VDC after going through a buck converter). Can someone tell me if this is correct or point me to the right conversion if it is not? Thanks

Well . . . say that calculation is roughly correct; it gives you the RMS current value on the AC side. You have to take into account -in the "general case"- the voltage that the relay contacts support (sure much higher than 120 VAC).

On the other hand, although the current value specified for the contacts is RMS, I would choose (a relay with) at least double the value the calculation gives; that would prevent effects associated with conmmuting inductive (very small in this case) currents and other transformer open circuit currents (I dont remember all the names of them: Foucault, . . . .)

Regards

Not that it matters, the peak voltage is SQRT(2)*120v=170v PK.
Almost any toggle switch you buy (or even rocker sw) is rated for 250V /10A

I am lost... where is the schematic ?

sounds like the OP wants to plug in a wall wart and use a switch the DC on and off to the arduino ?

so, the OP wants to switch a DC load, but wants the AC load requirements ?

anyone have the data sheet for the power supply ? what is the load imposed by that device ?

what is truly unsettling is that my friend raschemmel got sucked into this without demanding a schematic !

It is always preferable to switch the lower voltage side whenever possible. Its safer and you can use a smaller switch on the DC side.

120Vx1.6A = 192W
20V x 3.5A = 70W

That's sort of a half truth. If we were controlling an AC motor and it draws 10A at 240V, to keep the same power it would require 20A at 120V That will require a switch (or relay) with TWICE the contact rating thus a larger switch.

If we were switching things very quickly, the DC side is the side to switch since that little "wall wart" is a power supply and they don't take kindly to fast on/off cycles.

But we aren't controlling a AC motor, nor did we mention anything about power requirements of such a motor. You rate your switch to match the load(s), whatever that might be (Amps) at whatever voltage your using.

You would never do any kind of fast switching or PWM on top of a AC line without a phase lock. Most wall adapters these days use switching regulators, your more or less free to do your own DC switching right from the output as they have there own ripple rejection on the inside. Although adding more capacitance on the adapters output is a very common practice.

Its not exactly more dangerous to put a on/off switch on the AC side but with non-professional modifications I would recommend to stay on the DC side. When you don't know the skill of the person doing the work, having bad wiring and/or insulation on a 120V AC line is far more dangerous then if it where on a low voltage DC line.