SASM wrote:There´s a manual available on my website for the SASM80 module.
User LED wrote a nice summery. Dont´even try to charge the watch under a light. Riehl must have gotten a bunch of watches back and that´s why we now see all these inner parts from Barry popping up on ebay.
But let us take a look at the solar cell: The output of a modern silicium cell with the dimension of 100cm² (~40"x 40" inches) is round about 2 Ampere at 0,5 Volts in direct sunlight at 1000Watts/meter². But that would mean the solar cell is placed in a 90° angle to the sunbeams and there´s no red plastic cover above the cells and no sync-goo. Since P=VxA the output is 1Watt or 0,5W if the angle is not right with red top and so on. Now the solar cell has been produced in the 70s and has aged quite a bit, so I would say if you would use this 70s cell with a dimension of 100cm² (same as above), the output would be 0,25W at best. However the sync cell is just 20x25mm, so you have to multiply 0,25W with (20x25/100x100) which is 0,0125W. So the current flow in Ampere at 3Volts (a few cells are tied together in a row to archieve a 3V voltage) is... P/V=A..or 0,0125W/3V=0,0042A or 4,2mA. But we do not charge with 3V but just with 2,7V, because there´s a schottky diode between batteries and solar cell, so that the batteries cannot discharge through the solar cell. This diode takes 0,3V of the total voltage. The original module had a silicium diode in place (because these new schottkys weren´t available) so we would have a voltage drop of 0,7V.
Anyway let´s calculate the current flow with the schottky diode for the SASM80. We have 4,2mA x 2,7V/3V, thus 3,8mA.
Two V80 batteries (in case they are new) have an output of 80mAh at 2,8V which means you can power a circuit which takes 80mA at 2,8V for one hour. Let´s say the batteries are not completely drained an stll have an output of 30mAh at 2,8V , so that we would need to charge 50mAh.
We now search the time t until the batteries are full again. We can charge with 3,8mA at 2,7V. Now we look at the output P: P cell x h: 50mAh x 2,8V; P charge x h = 3,8mA x t [h] x 2,7V. Set P cell = P charge and calculate t.
t= 50mAh x 2,8V / (3,8mA x 2,7V) = 13,65h (in the sun!). Keep in mind that this a theoretical value with new batteries and no other losses. I recommend to charge the watch for at least 3 days in a row every month.
Hanno.
I found this post as I wanted to compare the performance of my solar powered nixie watch with the Synchronar watch. I think the mentioned 3.8mA is still on the high side (I reckon it's only half of that). My panels are 31x11mm each and produce about 2mA @ 4.5V. I am charging 3.7V lipos. The required battery current to display the time in my watch is between 6mA (total darkness) and 200mA (any ambient light over 2,000 lux). Does anyone know how much current the Synchronar consumes in broad daylight? I estimated that with so many LED segments, it must be close to 80mA.
The mentioned series diode doesn't really matter, the open voltage of the solar panels is much higher than the rated 3V (probably close to 4.5V), so a 0.7V voltage drop over the series diode is neglectable. My panels are rated 4.5V but produce 7V in direct sunlight.
It's interesting that the Synchronar advertisements do not mention how much light is required to keep the battery charged. I am quite convinced that if you check the time about 30 times per day, you can't keep the battery charged. Does anyone have more information about that?
Best regards,
Michel