Photo Eye Power: Current, Voltage, & Resistance Explained

[A J]

I was reading the specs on a photo eye and it states its current consumption

65 mA at 24v
120 mA at 12v

both situations end up with roughly 1.56 VA of power...how does it draw more current with less voltage applied. If V=IR, the only way i see it is the resistance of the system must lower in order to draw more current to maintain similar power. How does the resistance lower?

 

[Averagesupernova]

I think you are misinterpreting the spec. Are there some light/dark specs in there?

 

[anorlunda]

Lots of devices are nonlinear. They are not collections of R, L, and C. Take a transistor for example. You can only apply Ohms law logic to linear circuits.

 

[A J]

Lots of devices are nonlinear. They are not collections of R, L, and C. Take a transistor for example. You can only apply Ohms law logic to linear circuits.

I don't know much about non-linear circuits, can you elaborate on how current can go up as voltage goes down? voltage is the "pressure" and Current is "flow"

 

[A J]

I think you are misinterpreting the spec. Are there some light/dark specs in there?

I guess you mean does it show different specs if its signalling or not?

 

[dlgoff]

I guess you mean does it show different specs if its signalling or not?

I'm sure Average means Dark Current.

 

[meBigGuy]

Please post the spec sheet and we can help you interpret it. Why would you ask questions about a specification we can't see. The context of those numbers is totally unknown.

 

[rbelli1]

Switching power supplies will also cause a device to only draw a basically fixed amount of power no matter the input voltage. This only applies if you are within the normal operating range that it is designed for.

Also if the load is very light (<10% or so typically) efficiency often suffers, sometimes dramatically. Some supplies will lose some efficiency at the top end as well.

The various supplies that have the ability to output a voltage higher or lower than the input (buck-boost, SEPIC, ZETA, etc) frequently have varying efficiencies depending if the voltage is higher or lower than the output.

BoB

 

[256bits]

I was reading the specs on a photo eye and it states its current consumption

65 mA at 24v
120 mA at 12v

both situations end up with roughly 1.56 VA of power...how does it draw more current with less voltage applied. If V=IR, the only way i see it is the resistance of the system must lower in order to draw more current to maintain similar power. How does the resistance lower?

Are you stating a spec under maximum power dissipation before the device blows up?
Photo eye is a vague term - it could be one of the types of light sensors, or a complete circuit package.
As in post #7, spec sheet please.

 

[meBigGuy]

Ahhh... as rbelli1 said:
.065ma X 24V = 1.44 watts

120ma @ 12V = 1.44 watts

So, the load is constant power (which is possible)

Assume, for example that an 100% efficient switching power supply (as an example) is supplying a 1.44 watt load at 5V (that would be 288ma ). Then, the input power drain would be as indicated. (you can tweak the numbers for what the actual load would be if the switcher were 80% efficient)

 

[A J]

I was reading the specs on a photo eye and it states its current consumption

65 mA at 24v
120 mA at 12v

both situations end up with roughly 1.56 VA of power...how does it draw more current with less voltage applied. If V=IR, the only way i see it is the resistance of the system must lower in order to draw more current to maintain similar power. How does the resistance lower?

For those who are asking for the spec sheet. http://www.keyence.com/products/sensor/photoelectric/lr-w/specs/index.jsp

 

[A J]

Ahhh... as rbelli1 said:
.065ma X 24V = 1.44 watts

120ma @ 12V = 1.44 watts

So, the load is constant power (which is possible)

Assume, for example that an 100% efficient switching power supply (as an example) is supplying a 1.44 watt load at 5V (that would be 288ma ). Then, the input power drain would be as indicated. (you can tweak the numbers for what the actual load would be if the switcher were 80% efficient)

Sure, so the photo eye requires that power +/- inefficiencies. I guess i am hung up on how it does this - Where is the reduction of resistance in order to allow more electrons to flow (higher amps). what component of the photo eye allows this to happen.

 

[A J]

Switching power supplies will also cause a device to only draw a basically fixed amount of power no matter the input voltage. This only applies if you are within the normal operating range that it is designed for.

Also if the load is very light (<10% or so typically) efficiency often suffers, sometimes dramatically. Some supplies will lose some efficiency at the top end as well.

The various supplies that have the ability to output a voltage higher or lower than the input (buck-boost, SEPIC, ZETA, etc) frequently have varying efficiencies depending if the voltage is higher or lower than the output.

BoB

Keeping in mind ohms law, how does it draw a fixed amount of power? what's changing to allow current to flow more freely?

 

[meBigGuy]

https://en.wikipedia.org/wiki/Switched-mode_power_supply

Please read about switching regulators. They are essentially power converters. They convert the input power from a wide range of voltages to an output power at a FIXED voltage. So, their current draw from the source voltage is different for different voltages. Please reread rbelli1's post and my post with that in mind.

whats changing to allow current to flow more freely?

A switching regulator draws pulses of energy from the source as needed to supply energy to the output. It is a dynamic process and everything is smoothed out so the current looks pretty constant.

EDIT:
Note that switching power supplies are used in nearly all electronic devices, from every PC power supply, to every wall wart, to every television or any other electronic device that plugs into the mains.

 

[rbelli1]

every

Well most everything from the last couple of decades at least.
I do come across transformers with linear regulators for some of the really cheap electrical devices I buy. I guess it depends on the price of copper at the time of manufacture.

BoB

 

[sophiecentaur]

voltage is the "pressure" and Current is "flow"

It is not hard to design a valve which will shut off when there's a lot of pressure but it will let fluid flow through at low pressure. So it is with some non linear electrical devices. You are making assumptions in your argument about how things should be working.

 

[dlgoff]

[Hijack]

... to every wall wart, to every television or any other electronic device that plugs into the mains.

And a whole lot of these:
http://www.howardcountymd.gov/uploadedImages/Home/Environment/Environmental_Services/CFL%20bulb.jpg
[/End Hijack]