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I don't understand. The boost converter is rated the following:
1.Electrical characteristics
1). The maximum output current: 2 a
2). The input voltage: 2 v ~ 24 v
3). The maximum output voltage: 28 v
4). Efficiency: ＞93%
Does it mean that the lock is drawing too much current causing the boost converter to shut down? When it shuts down, the output voltage will drop to 1V and stay there forever until I remove the input battery supply?
Somehow its working now after I changed the battery. But when the solenoid is switched on, i hear a high pitch sound coming from the boost converter. Its disappears when the solenoid is turned off.
I will measure the DC resistance tomorrow in the lab and let you know.
I think your battery can not support the load.
You want 6.5V at 800mA to get that you must pull 1.6A from 3.25V battery. When the voltage goes 1:2 the current must also go 2:1.
With a boost supply, if the output is stuck at 1V then there is a good chance that the battery voltage has dropped to 2V. (please measure)
I found more information on the supply. I thing it is good for 800mA but not a lot more.
If your battery voltage drops much the input current will go up more. You might need 2A from the battery to do this job. You might decease the output voltage some. That will reduce the current from the battery.
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