View Full Version : asking for parts info

Craig Swensson
09-15-2013, 06:17 PM
yes i know its a scanner question, however all i require is info on the correct Ac adapter to use for Canon 8800f.
output voltage, output current,power rating, i have a large box of adapters and i need to correctly match the adapter to the scanner.
anyone able to provide info, thanking you in advance

09-29-2013, 12:22 PM
Hi Craig, I've just dived under my desk to check.
My AC adapter is model no. K30286
Input 100-240V 0.45A
Output 32V 0.7A

I couldn't see any indication of the polarity of the connector into the scanner.

Good luck.

Craig Swensson
10-01-2013, 12:19 AM
tokam, thanx for that info. I had forgotten all about this thread. After much research it seems that the power adapter is hard to come by.My Solution -buy another scanner:)
i`m going to keep the 8800 and see if I can find a fix, either that our destroy it trying

10-01-2013, 12:33 AM
There is nothing that looks like this on the adapter or near where it goes into the scanner?

Craig Swensson
10-01-2013, 03:33 AM
no Truzi. I took the scanner to a technician, he just rolled his eyes and said'canon'..?
There is an HP adapter with matching output and a 3 pin plug, though the plug has square fitting and the canon round.However I have found some connectors from inside an older computer that match the canon fitting.So perhaps there is a fix eventually.

10-01-2013, 10:21 AM
You certainly can splice any connector you need once you find the polarity.

Perhaps tokam can take a VOM to his adapter to find the polarity, though the specs he posted do seem a bit hard to come by.

Peter Simpson
10-01-2013, 12:15 PM
EE here. 32V is 32V. Any power supply with an output of 32V DC and a current rating of over 0.7A will work fine. Do what you need to to get the connector to fit. If it's one of the simple coaxial ones, you can get them at Digikey.com. No need to buy a new scanner unless you want to upgrade.

eBay has several 32V supplies available at under $10: http://www.ebay.com/bhp/32v-power-supply

OK, with a little more Googling, I have found that the Canon K30286 seems to be a dual output supply: 24V @ 0.5A and 32V @ 0.7A.
The used/refurb places seem to be capitalizing on these supplies being hard to find, and offering them for sale with "price upon request".

Not necessary at all. The solution is to buy both a 24V and a 32V supply. Each should be available on eBay at under $10 used. You want a "1A output, regulated switching supply". Connect the negative leads of both supplies together, now you have GND, +24 and +32. If you can get the owner (tokam?) of a working scanner to apply a meter to his power connector, you'll know which lead connects to which pin and you're off to the races.

Craig Swensson
10-01-2013, 04:51 PM
thank you peter.Yes i have seen the supplies you mention on ebay. I will try your solution, even though i am not the sharpest tool in the box when it comes to power i can work out how your solution is supposed to work. There is another thread concerning this problem on flickr, some guys have listed the polarity, though there is some confusion over the correct numbers so i may message tokam.I have actually brought another scanner[8400] which is working fine, would still like to get a 'win' on the 8800f.If it all works out i will post your fix to flickr as well.

Peter Simpson
10-01-2013, 07:29 PM
The trick with figuring out the connector is to be methodical. Take the negative lead of the meter and place it in the first contact. Next, use the red lead to measure the voltage on the other two contacts in turn. You will get two readings -- if they are 24 and 32, you are done, and the contact with the black lead is GND. If the voltages are something else, move the black lead to the next contact and repeat the above procedure. You'll only have to do it a maximum of 3 times :-)

10-02-2013, 02:06 AM
I just pulled the plug from scanner to check polarity and its not going to be that simple. Canon, (Bless 'em!!!), have use a very proprietory looking three pin flat connector.
See attached.



10-02-2013, 02:13 AM
I've just seen the eBay link from Peter and the power connectors on the HP power supplies look very similar. I'll borrow a meter tonight and try and determine what power is on each pin. At $7 they could be the answer, even if you have to cut and splice the lead to get correct polarities. (Quite likely they will be the same).

10-02-2013, 10:26 AM
I just pulled the plug from scanner to check polarity and its not going to be that simple. Canon, (Bless 'em!!!), have use a very proprietory looking three pin flat connector.
See attached.



Paper clips... very carefully.
If the probes/leads on your meter are too big, get a bit of metal that will fit inside the holes on the connector. They will be nothing more than "pins" sticking out for you. Perhaps wrap electrical tape around them so you don't short anything. Do this before plugging it in to make sure you will not accidentally short any leads.

Then touch the VOM's probes to the "pins" you made.

Peter Simpson
10-02-2013, 12:15 PM
Another tip: you'll find you need three hands to do the voltage measurement. Get a hold of one of those hobbyist bench vises (any vise will work, but in the US, there's a brand called PANAVISE that's my favorite). Clamp the connector in the vise, pins pointing up. Plug in the AC cord and enjoy the (almost) effortless probing! :-)

(profuse apologies for this being an electrical debug thread and not a film camera thread; though, hopefully, the strategy and techniques will be of interest to anyone repairing stuff)

Craig Swensson
10-02-2013, 03:11 PM
yeah, i could see from the research that this was not going to be really straight forward. I have a brother in law who is very good with this stuff - sometimes..haha.
Don`t worry about apologies for this becoming an electrical de bug thread on a film camera site, I posted the same in DPUG - no replies.All great info from everyone, well done and cheers

10-03-2013, 03:21 AM
OK, I've located a cheap meter, probably only good for continuity checking but it gives me the polarity. Also found my vise, last used when I replaced two caps on the power board of a Samsung monitor last Christmas. The voltages per my meter are are low but from left to right on the attached picture are + gnd, -12V, -27V.


Peter Simpson
10-03-2013, 08:15 AM
Well, *that* doesn't make much sense...seems like Mr. Murphy has stepped in to make life interesting.
Thank you, tokam, for making those measurements!

Someone in the Flickr conversation says he took the scanner apart and claims the center pin is GND.
But I'm at a loss to figure out how that matches up with the voltages you measured.

A possible next step would be to take the scanner apart and try to understand the circuitry connected
to the pins, and figure out what voltages it wants to see. That's possible, and I have done it, but it
takes a lot of experience and not a small bit of luck.

Another possibility is to measure the voltages with the scanner connected and powered on.
Often, these switching supplies do not regulate their outputs well until a load is connected.
Again, this is a probably an exercise for "advanced students".

I believe this scanner has an LED light source. These are sometimes powered by current-regulated
supplies, and this could be the purpose of the third pin. If this is the case, you would not see a
stable voltage on that pin, it would be whatever voltage is necessary to get the specified current
to flow through the LEDs. We just don't know if this is the case.

It may be that this is beyond us. It's probably a good thing that Craig has purchased another scanner :-)
I'm sorry I couldn't have been more help. I can promise you that if I ever run across one of these scanners,
I will have an irresistable desire to disassemble it, and post my findings on the web somewhere!

Craig Swensson
10-05-2013, 03:59 AM
thank you all.Yes, it seems canon really made this one as difficult as possible.Probably beyond my capabilities so fortunate that i had another scanner available.I will put this particular project aside in the hope that perhaps a solution is found or i come across someone local that likes a challenge.Thank you all for your input and excellent clear replies - all much easier understood and clearer than the flickr info.Peter, i guess you will be on the lookout for one of these scanners, just so you can solve the problem:D: