Needing help powering up a Sanyo ICC-801D

[sislasl]

Hi,

I have found a Sanyo ICC-801D calculator without an AC adaptor. I want to test the calculator's functionality and therefore need to power up the calculator. The AC Adaptors which work would be a Sanyo CU-82 or Dictaphone CU-1680 but I found no place where to get one.
I have read that you can also power it up by connecting 6.6 VDC.
Now I would like to know how to do this.
Since I have never done something like this before I assume I have to open the calculator's case and connect the cable of some generic AC Adaptor with variable VDC to some spot inside the case. Can anybody tell me how to do this exactly and what kind of AC adaptor I need?
Thanks in advance

[Ren]

This website says it takes 6-7.3 Volts DC external.

http://mycalcdb.free.fr/main.php?l=0&id=5040

I have not seen a photo of it's power jack,
Does the label have any indication of the polarity of the jack?

[AndiGer]

The power plug seems to be on top side. If I look at some pictures of the machine I guess it has a rechargeable battery inside. You can imagine if you look at the photos on this site
https://blog.goo.ne.jp/vrc-tezuka/e/9599...7652602867
where the plugs from the battery go to. (And yes you see where better don't touch due to high voltage for the nixies :-) )

On this photograph you can see the plugs but battery not included:
https://auction-antique.com/marketplace/...i_icc-801d

Good luck
Andi

[sislasl]

(05-04-2021 02:12 PM)Ren Wrote:  This website says it takes 6-7.3 Volts DC external.

http://mycalcdb.free.fr/main.php?l=0&id=5040

I have not seen a photo of it's power jack,
Does the label have any indication of the polarity of the jack?

The owner of vintage calculators inc gave me the info about 6.6 VDC.
There is no indication of the polarity of the jack. I have atttached an image of the jack.

[sislasl]

(05-04-2021 05:06 PM)AndiGer Wrote:  The power plug seems to be on top side. If I look at some pictures of the machine I guess it has a rechargeable battery inside. You can imagine if you look at the photos on this site
https://blog.goo.ne.jp/vrc-tezuka/e/9599...7652602867
where the plugs from the battery go to. (And yes you see where better don't touch due to high voltage for the nixies :-) )

On this photograph you can see the plugs but battery not included:
https://auction-antique.com/marketplace/...i_icc-801d

Good luck
Andi

Yes, it has a rechargeable battery inside (which is probably dead).

So, just to be sure: I need anadjustable DC power supply like this and connect it to the plugs inside.
How do I know which cable I have to connect to which plug (I have never done something like this before)?
The plug on the outside of the case has 3 contacts, but I only need to connect the black and the red cable in the inside, right?

The calculator has 3 Watts at 6-7.3 VDC so I need 0.41 - 0.5 Ampere, right?

Thank you for your help

[Nigel (UK)]

Here's a link to the sale of this model of calculator on an auction site: Sanyo ICC-801D.

In the seventh picture you can see an image of the mains adapter. Here's a snip:

You can see that the power supply supplies two voltages, 6.6V and 7.3V. My guess is that the 6.6V powers the calculator, and the 7.3V charges the battery pack (which is 5 Ni-Cd cells, so nominally 6V). If this is correct it would do no harm to connect both of the pins to 6.6V.

Looking at the "jack" picture, there are three pins: 0V, +/-6.6V and +/-7.3V in some order. However, we still don't know what the polarity should be. Given that the calculator apparently uses PMOS technology, I wouldn't be surprised if the common pin was positive, not negative. Don't trust me on this!

A way around this is to open the case and get at the battery pack. Looking at the pictures of the interior (e.g., here), we see a red wire and black wire coming from the area on the right where the battery pack would be. Red is positive; black is negative, with close to 100% certainty. So open the case, remove the battery pack (probably not in great shape), clean things up, and connect your 6.6V supply to the battery pack terminals, with polarity according to wire colour.

(If your power supply allows it, limit current to 0.5A.)

Once you're inside you can use an ohmmeter to see whether any of the pins in the power supply socket are directly connected to plus (red battery pack wire) or minus (black wire). I would expect only one direct connection, and that would be the 0V pin. If it's connected to the red wire, connect the 0V pin to +6.6V and the other pins to 0V; if the direct connection is to the black wire, connect the 0V pin to 0V and the other pins to +6.6V.

Be really careful! If you don't want to open the case I would wait around in case some information on the adapter polarity turns up.

Nigel (UK)

[Ren]

Well, you might need a dual output adjustable power supply like this:

https://www.ebay.com/itm/144021599984?ep...SwEthgh7fD

I've done some looking, but I think a better price can be found (Ali-Express?).

Or, maybe just a single output power supply capable of 6.6 Volts and tie the two inputs together like was mentioned earlier.

[sislasl]

(05-07-2021 08:21 AM)Nigel (UK) Wrote:  Here's a link to the sale of this model of calculator on an auction site: Sanyo ICC-801D.

In the seventh picture you can see an image of the mains adapter. Here's a snip:

You can see that the power supply supplies two voltages, 6.6V and 7.3V. My guess is that the 6.6V powers the calculator, and the 7.3V charges the battery pack (which is 5 Ni-Cd cells, so nominally 6V). If this is correct it would do no harm to connect both of the pins to 6.6V.

Looking at the "jack" picture, there are three pins: 0V, +/-6.6V and +/-7.3V in some order. However, we still don't know what the polarity should be. Given that the calculator apparently uses PMOS technology, I wouldn't be surprised if the common pin was positive, not negative. Don't trust me on this!

A way around this is to open the case and get at the battery pack. Looking at the pictures of the interior (e.g., here), we see a red wire and black wire coming from the area on the right where the battery pack would be. Red is positive; black is negative, with close to 100% certainty. So open the case, remove the battery pack (probably not in great shape), clean things up, and connect your 6.6V supply to the battery pack terminals, with polarity according to wire colour.

(If your power supply allows it, limit current to 0.5A.)

Once you're inside you can use an ohmmeter to see whether any of the pins in the power supply socket are directly connected to plus (red battery pack wire) or minus (black wire). I would expect only one direct connection, and that would be the 0V pin. If it's connected to the red wire, connect the 0V pin to +6.6V and the other pins to 0V; if the direct connection is to the black wire, connect the 0V pin to 0V and the other pins to +6.6V.

Be really careful! If you don't want to open the case I would wait around in case some information on the adapter polarity turns up.

Nigel (UK)

Thank you for these very detailed instructions, Nigel!!! I appreciate them a lot and will definitely try it as soon as I have a DC power Supply

[AndiGer]

Any news? Could you power up the machine - and if please tell us what you did. Thank you

[sislasl]

(07-04-2021 06:04 PM)AndiGer Wrote:  Any news? Could you power up the machine - and if please tell us what you did. Thank you

I just got my hands on an adjustable DC power supply two days ago but was very hesistant about opening the case because it mostlikely would have damaged the rubbers covering the screw holes. So I asked the user vintagecalculatorsinc on ebay if he knew the order of the 3 pins. Yesterday he sent me a message with the correct order. In regard to the picture of the pins in post #4 it is Neg, +6,6V, +7,3V.
When I first attached the clips from the DC power supply to the pins of the calculator and tried to power it up nothing happened.
Then I found out that there is some kind of tongue at the bottom of the connector which is pushed downwards when the jack is inserted.
As I used a wooden chopstick to push this tongue down while powering up the calculator it finally came back to life.
A big thanks to all contributors here - the mission is finally accomplished

[megatesla]

Well done getting it powered up!

I have the similar ICC-82D and recently refurbished it, including the battery pack.

The early chipsets used by Sanyo had some bugs/quirks.

Try this: CA 0 ÷ 0.01 = CK