See page 18:
https://www.mouser.com/catalog/specs...02135187-1.pdf
truclacicr
Badcaps Veteran
Last Activity: Today, 05:35 PM
Joined: 04-21-2019
Location: brisbane
-
They appear to be ESD protection diodes. The circuit should work without them, but you won't have continued protection from ESD.Leave a comment:
-
Does the Fault pin of the IC remain low (latch) after the fault occurs?...Leave a comment:
-
If the third IC is an OR gate, then its output would be indeterminate when its external input is low. It would make some sense if it was a unity gain voltage follower op amp, but then there would be no purpose for two LEDs, as they would be displaying the same state....Leave a comment:
-
Page 18 of the datasheet shows the internal feedback divider resistors. The FB voltage is 0.65V. Reducing the output voltage would involve adding an external resistor between VOS and FB. What is your target Vout? Is the resistor on pin #7 your own modification? What is the present Vout?Leave a comment:
-
https://www.ametherm.com/datasheetspdf/SL222R510.pdf
The part marking is obscure. I can't understand how anyone could make the connection, especially since the datasheet is devoid of this information.
Nice find, by the way....Leave a comment:
-
The marking is cLASB.
https://monitor.espec.ws/section41/topic325493.html
This board has cLBHA in the same position:
https://old.reddit.com/r/AskElectronics/comments/txyaxs/can_anybody_identify_this_part/
cLxyz = SY6287CABC (Low Loss Power Distribution Switch with Programmable Current limit)
https://uploadcdn.oneyac.com/attachm...SY6287CABC.pdf...Last edited by truclacicr; 08-26-2025, 04:05 PM.Leave a comment:
-
Does the resistance increase or decrease as you warm it up?
"2407" is probably a YYWW date code.
I would have thought that "005" would be the resistance in ohms at 25C....Leave a comment:
-
It's not the same part, but it may have the same pinout:
https://www.monolithicpower.com/en/d...n/sku/MP86934/Leave a comment:
-
The batteries must be charged by the panel ... somehow. However, there is something missing from your circuit.
Measure the battery voltage in the dark, then do the same in bright sunlight. Alternatively, disconnect the battery and see whether the IC is getting any supply voltage from the panel.
Disconnect the panel and measure the battery voltage at the IC. This will tell you which ICs pins are connected to the panel and to the batteryLast edited by truclacicr; 08-18-2025, 12:52 PM.Leave a comment:
-
AFAICS, those resistors are too large for any charging current. Instead, they appear to be bias resistors.Leave a comment:
-
The circuit diagram is busting my brain. I presume that DC+ and DC- are the terminals of a pump motor, in which case it makes no sense to connect Q5 as in your circuit. Instead, it would make more sense to connect the emitter to ground.
Another problem is the absence of a path for the charging current from Solar+ to Bat+ via the IC. How is the battery being charged?
I can't understand what D1 is doing.Leave a comment:
-
I think the R15 and R16 current sharing resistors should be R510, 5W.
P = I^2 x R = 2.5^2 x 0.510 = 3.2WLeave a comment:
No activity results to display
Leave a comment: