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USB to 7.5 volts adapter

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#1 Jerry Lodriguss

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Posted 30 June 2017 - 12:42 AM

Has anyone tried one of these USB to 7.5 volt adapters on a DSLR?

 

B&H lists Canon's ACK-E6 120 volt adapter as outputting 8v at 3amps, but the DC coupler (fake battery) as 7.6 volts.

 

The actual LP-E8 battery that goes in the camera says 7.2 volts.

 

I think Nikon's battery for the D5300 is 7.4 volts.

 

I've put a meter on the camera and I don't ever see it drawing 3 amps.

 

I was thinking of just plugging one into my powered USB hub on the top of the scope to power the camera.

 

Not knowing anything much about electricity, I though I would ask for some expert help here.

 

Jerry



#2 james7ca

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Posted 30 June 2017 - 02:22 AM

Well, a standard USB2 port is rated as 500mA @ 5V. That's 2.5 watts which is a long way from the 24 watts suggested by that 120V adapter (3A @ 8V). Thus, while the voltage might be about right the total power requirement for the camera could exceed what could be given over a USB2 port.

 

Having said that, I have a Nikon D5100 that takes a battery that is rated at just over 1Ahr at 7.4V and I know that I can get about three hours of operation on that one battery (when doing astrophotography). So, I guess the camera is averaging about 0.3A over those three hours which would indicate 2.2W and that's pretty close to the 2.5W that could be had from a USB port. But, those 3 hours of operation don't including using the LCD/liveview which can run the battery down much more quickly. Given that I wouldn't be surprised to find that the maximum power draw of the camera is greater than the 2.5W that could come from a standard USB port.


Edited by james7ca, 30 June 2017 - 03:30 AM.


#3 otocycle

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Posted 30 June 2017 - 06:32 AM

Like a lot of things....it depends.   What is your powered USB hub supply rated at and how many other devices are drawing current at the same time ?   I would guess that a Canon DSLR's biggest task load is charging the flash, which isn't used for AP unless Bigfoot shows up at a dark site.   So figure 500mA nominal for the camera's functions based on published numbers.

 

But you don't want to be operating at the margin for your USB hub's power supply or the 5.0V > 7.5V DC/DC converter.   If you are running a USB3 powered hub, the power specs were increased and it should be rated for bigger loads.

 

This is a neat idea if it works reliably.   I use old Sony PlayStation 1 car adapters (from ebay) for my Canon DSLRs in the field, (12V > 7.5V) into the dummy battery and it is rated at 3 amps output.   Probably overkill but cheap and reliable.

 

The only caution I would give is for the build quality of the USB DC converter for rated load.   If it fails it probably won't damage the camera at 5 volts in, but you would need a backup if far away from home.



#4 Ketut

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Posted 30 June 2017 - 06:41 AM

Jerry, I believe it will work.

Let's say I can running my 60D+LP-E6+BYEOS about 5hrs for imaging.

So 60D draws about -> 1865mA x 80% : 5hrs = 298mA

Which means, 500mA from the USB will be possible to power the camera on.

And, at 7.5 volts you will have higher ampere output anyway...more than 500mA I think.



#5 StuartJPP

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Posted 30 June 2017 - 09:11 AM

The link points to a product that powers a weighing scale whose power requirements is 100mA @ 5V to 9V so I would say no.

 

 

Stepping down DC voltages (Buck conversion) is always easier than stepping up (Boost conversion), using simple switchers anyway.

 

So technically it is possible, but the cheap switchers you get off eBay are quite rubbish, especially the step up ones, most use fake chips as well making wild claims (95% efficient). A properly designed boost regulator boosting 5V to 7.2V at 500mA and 90% efficiency is not difficult using modern components, but it is difficult finding an off-the-shelf solution.

 

Note that 500mA @ 7.2V = 3.6W which then equates to 720mA @ 5V assuming it is 100% efficient, which it won't be.

 

I am not sure about the peak power requirements of the Nikon but for the Canon 6D I measured circa 500mA with LiveView and about 300mA when exposing (at 7.2V). Since there is no flash there aren't any huge gulps charging up the flash capacitors so that helps (as mentioned above).

 

The other thing about USB hubs is that if it complies correctly with the USB specification it will only supply the higher power (500mA or more) after negotiation (which your wire won't provide), however some/most powered hubs will gladly supply much more power without negotiation. Modern hubs come with "charger ports" to overcome this.

 

Since I had 12V as well as the powered hub providing 5V, I opted for the 12V down to 7.2V route instead.



#6 Ketut

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Posted 30 June 2017 - 10:30 AM

Yup confirmed 1.05A output -> here

More than enough to power your camera on, Jerry :D



#7 Jerry Lodriguss

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Posted 30 June 2017 - 12:08 PM

Thanks guys for your replies. 

 

Now I know that yes, I can use it, and no I can't. :-)

 

Jerry



#8 james7ca

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Posted 30 June 2017 - 06:16 PM

I found another reference here on CN where a user measured the power draw on his Canon T3i and he reported 0.5A at at just over 7VDC (with liveview active, but obviously not while charging the flash). So, that means at least 3.5W which is more than a typical USB2 port can handle. In any case, if your USB port can't deliver that amount of power it makes no difference what the conversion device is rated at, so saying that the conversion device can output 1.05A doesn't help (also, the specs for the 7.5V version of the USB power converter says "5V USB to 7.5V 0.84A converter," not 1.05A).

 

There would be three ways to handle this:

 

1.) Use a hub that has a high current charging port (very often from 1A to 1.5A at 5V).
2.) Use a USB3 hub that can deliver the full USB3 output (0.9A at 5V).
3.) Tie the power output from two USB2 ports together in parallel (you may even be able to find a pre-made cable that would allow this, should give a total of 1A at 5V if the hub allows full power).

 

Further, note what Stuart said:

...The other thing about USB hubs is that if it complies correctly with the USB specification it will only supply the higher power (500mA or more) after negotiation (which your wire won't provide), however some/most powered hubs will gladly supply much more power without negotiation. Modern hubs come with "charger ports" to overcome this...


Edited by james7ca, 30 June 2017 - 06:23 PM.


#9 james7ca

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Posted 30 June 2017 - 06:25 PM

Jerry, I believe it will work.

Let's say I can running my 60D+LP-E6+BYEOS about 5hrs for imaging.

So 60D draws about -> 1865mA x 80% : 5hrs = 298mA

Which means, 500mA from the USB will be possible to power the camera on.

 

And, at 7.5 volts you will have higher ampere output anyway...more than 500mA I think.

Actually, it's just the opposite. At 7.5V you'll have less current because it's the power draw that makes the difference. So, if the USB port can deliver 2.5W at 5V that means at best the maximum current at 7.5V would be 2.5W / 7.5V = 0.33A or 330mA (since 7.5V x 0.33A is 2.5W).



#10 Ketut

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Posted 30 June 2017 - 08:50 PM

Thanks guys for your replies. 

 

Now I know that yes, I can use it, and no I can't. :-)

 

Jerry

It's only $9.99 worth to try grin.gif

Anyway, you will use it on your powered usb hub which is commonly deliver @900mA, not on a laptop @500mA grin.gif

 

 

 

Jerry, I believe it will work.

Let's say I can running my 60D+LP-E6+BYEOS about 5hrs for imaging.

So 60D draws about -> 1865mA x 80% : 5hrs = 298mA

Which means, 500mA from the USB will be possible to power the camera on.

 

And, at 7.5 volts you will have higher ampere output anyway...more than 500mA I think.

Actually, it's just the opposite. At 7.5V you'll have less current because it's the power draw that makes the difference. So, if the USB port can deliver 2.5W at 5V that means at best the maximum current at 7.5V would be 2.5W / 7.5V = 0.33A or 330mA (since 7.5V x 0.33A is 2.5W).

 

As you can see, at 7.5v it provides 0.84A and not 0.33A.

What is the math behind this?



#11 Phil Sherman

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Posted 30 June 2017 - 09:27 PM

If your mount runs on 12V, the simplest solution is to build/buy a splitter cable and use a "buck" regulator set to 7.2V to power your camera. I've been using a similar solution with the buck regulator connected directly to the battery and running a separate power cable to a salvaged battery eliminator for my Canon camera.

 

If you have power problems at the mount after doing this, you can make a new power cable for it using heavier wire than the one that came with the mount. Mount power cables usually have 20 gauge or smaller wire in them. Using one size larger wire would eliminate any wire losses between the battery and the mount caused by the additional load imposed by the camera.



#12 james7ca

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Posted 30 June 2017 - 09:36 PM

 

 

Jerry, I believe it will work.

Let's say I can running my 60D+LP-E6+BYEOS about 5hrs for imaging.

So 60D draws about -> 1865mA x 80% : 5hrs = 298mA

Which means, 500mA from the USB will be possible to power the camera on.

 

And, at 7.5 volts you will have higher ampere output anyway...more than 500mA I think.

Actually, it's just the opposite. At 7.5V you'll have less current because it's the power draw that makes the difference. So, if the USB port can deliver 2.5W at 5V that means at best the maximum current at 7.5V would be 2.5W / 7.5V = 0.33A or 330mA (since 7.5V x 0.33A is 2.5W).

 

As you can see, at 7.5v it provides 0.84A and not 0.33A.

What is the math behind this?

 

The math? Pretty simple, the adapter (by itself) can't create power, it depends upon the power source which is the USB port. Just because a conversion device is rated at a certain power output that doesn't mean it can output any level of power, it that were possible then you could power your entire home on a single USB port since you'd just need a converter that was rated at a high enough power output.

 

The fact that the converter can handle a certain amount of amperage doesn't reduce the power requirement from the actual source of the power (which in this case is limited by the wattage given by the USB port).

 

Think of a voltage requirement as a kind of pressure. If the voltage requirement goes up something has to give and that will be the current that can be delivered at that higher voltage.

 

Of course, if the power source can deliver more wattage that what is available with a standard USB2 port then that converter could handle up to 0.84 amps (which would mean a consumption of 0.84A x 7.5V or 6.3 watts of power, more than can be delivered by a standard USB2 port).



#13 Ketut

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Posted 30 June 2017 - 10:48 PM

The math? Pretty simple, the adapter (by itself) can't create power, it depends upon the power source which is the USB port. Just because a conversion device is rated at a certain power output that doesn't mean it can output any level of power, it that were possible then you could power your entire home on a single USB port since you'd just need a converter that was rated at a high enough power output.

 

 

 

 

 

The fact that the converter can handle a certain amount of amperage doesn't reduce the power requirement from the actual source of the power (which in this case is limited by the wattage given by the USB port).

 

Think of a voltage requirement as a kind of pressure. If the voltage requirement goes up something has to give and that will be the current that can be delivered at that higher voltage.

 

Of course, if the power source can deliver more wattage that what is available with a standard USB2 port then that converter could handle up to 0.84 amps (which would mean a consumption of 0.84A x 7.5V or 6.3 watts of power, more than can be delivered by a standard USB2 port).

 

Thank you, James.

This is my conclusion:

- Use on USB2 5V @500mA -> we get 7.5V @500mA

- Use on USB3 5V @900mA -> we get 7.5V @840mA

- Use on a powered hub 5V @900mA -> we get 7.5V @840mA

- Use on a power bank 5V @1A or @2.1A -> we get 7.5V @840mA

Good news for you Jerry grin.gif

 

@Jerry, be careful because some powered usb hub has 12V output voltage grin.gif



#14 james7ca

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Posted 30 June 2017 - 11:25 PM

Thank you, James.

This is my conclusion:

- Use on USB2 5V @500mA -> we get 7.5V @500mA

- Use on USB3 5V @900mA -> we get 7.5V @840mA

- Use on a powered hub 5V @900mA -> we get 7.5V @840mA

- Use on a power bank 5V @1A or @2.1A -> we get 7.5V @840mA

Good news for you Jerry grin.gif

 

@Jerry, be careful because some powered usb hub has 12V output voltage grin.gif

 

I'm not sure how you are calculating those numbers, but here is what could be given from a converter that outputs 7.5V and operates at 100% efficiency (the latter not likely).

 

For USB2 5V @500mA -> maximum 7.5V @333mA (perhaps not enough given a requirement for powering liveview)
For USB3 5V @900mA ->  maximum 7.5V @600mA (may or may not work depending upon the efficiency of the converter and the exact power draw from the camera)

For dual USB2 port 5V @ 1A (2x500mA) -> maximum 7.5V @667mA (perhaps a bit better than just one USB3 port)

For High Output Charging Port on Hub 5V @ 2.1A -> maximum 7.5V @ 1.4A (should work if the charging circuit actually provides a constant 2.1A, some -- few -- might not since they try to be "smart" about the charging rate).

 

Of course, this assumes that the USB power converter can actually provide up to its advertised 0.84A at 7.5V which might be the limit before the device cuts off from its overcurrent protection. So, you could hook the converter to a high output charging port and the maximum you could draw would likely be around that 0.84A limit (which should power the camera, as best as can be estimated).

 

Here is a $7 cable that will allow you to pool the power from two USB2 ports (could allow up to 5V @ 1A):  http://a.co/h2rTJbj


Edited by james7ca, 30 June 2017 - 11:56 PM.


#15 Ketut

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Posted 01 July 2017 - 01:57 AM

I'm not sure how you are calculating those numbers, but here is what could be given from a converter that outputs 7.5V and operates at 100% efficiency (the latter not likely).

 

 

 

For USB2 5V @500mA -> maximum 7.5V @333mA (perhaps not enough given a requirement for powering liveview)
For USB3 5V @900mA ->  maximum 7.5V @600mA (may or may not work depending upon the efficiency of the converter and the exact power draw from the camera)

For dual USB2 port 5V @ 1A (2x500mA) -> maximum 7.5V @667mA (perhaps a bit better than just one USB3 port)

For High Output Charging Port on Hub 5V @ 2.1A -> maximum 7.5V @ 1.4A (should work if the charging circuit actually provides a constant 2.1A, some -- few -- might not since they try to be "smart" about the charging rate).

 

Of course, this assumes that the USB power converter can actually provide up to its advertised 0.84A at 7.5V which might be the limit before the device cuts off from its overcurrent protection. So, you could hook the converter to a high output charging port and the maximum you could draw would likely be around that 0.84A limit (which should power the camera, as best as can be estimated).

 

Here is a $7 cable that will allow you to pool the power from two USB2 ports (could allow up to 5V @ 1A):  http://a.co/h2rTJbj

 

Thank you for the correction James.

Even though we use it on high output charging, I think it will provide only max 0.84A as advertised, and not 1.4A as your last paragraph.

But still, worth to try on USB3 grin.gif




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