Discussion in "Project Help" started by    nagufever    Sep 13, 2010.
Tue Feb 22 2011, 02:10 am


how do i reduce the current from 1.2A to 0.6A without reducing the voltage which is 12V from the Ac power adapter?

nagufever


Why do you want to do this ?

I think you have mixed up the operation of the LDR circuits you have been given,
and come up with a circuit that will not do what you want.

Can you explain what your LDR circuit is for?
Wed Feb 23 2011, 12:05 am
Hi majoka,
If i add a 10k resistor to the switch does this configuration will result an active-low input. When the button is being pressed, reading of I/O pin will be in logic 0, while when the button is not pressed, reading of that I/O pin will be logic 1?iit means it will work the opposite way am i right? why u say it will damage the component if the Vcc is 5volt?
ExperimenterUK:Pin 9 of ULN2803 is connected to ground and 10 to power of relay.When controller pin at high logic transistor inside the uln turn on the relay. When controller pin at low logic, logic transistor inside the uln turn off the relay
Wed Feb 23 2011, 12:28 am


ExperimenterUK:Pin 9 of ULN2803 is connected to ground and 10 to power of relay.When controller pin at high logic transistor inside the uln turn on the relay. When controller pin at low logic, logic transistor inside the uln turn off the relay

nagufever


Why do you need a relay ?
Wed Feb 23 2011, 08:27 am
becuase it is light activated circuit.when it enabled, the it will turn the dc motor forward and vice versa.
Thu Feb 24 2011, 12:12 am

If i add a 10k resistor to the switch does this configuration will result an active-low input. When the button is being pressed, reading of I/O pin will be in logic 0, while when the button is not pressed, reading of that I/O pin will be logic 1?iit means it will work the opposite way am i right? why u say it will damage the component if the Vcc is 5volt?


yes ur right and i say for damaging because in ur jpg file resistor was not there so i mention this
Thu Feb 24 2011, 02:40 am



Why do you need a relay ?

ExperimenterUK


becuase it is light activated circuit.when it enabled, the it will turn the dc motor forward and vice versa.

nagufever



As you have an ADC converter in your PIC, the proper way is to read the LDR as an
analogue value, then use the program to control the motor.

What has a light got to do with opening or closing a curtain ?
Fri Feb 25 2011, 04:45 am
ExperimenterUK:I am using sun light as the light source for the LDR input.
Majoka:If i modify ur L298HN circuit like the attached file is it ok?
Attachment
Fri Feb 25 2011, 10:53 pm
yes u can do it
it is ok
Thu Mar 03 2011, 03:39 am
i have tested the receiver in a protoboard but there seems no current/voltage in Vcc and ground.Can u check my circuit if there is any mistake?
Thu Mar 03 2011, 04:44 am


there seems no current/voltage in Vcc and ground.
Can u check my circuit if there is any mistake?

nagufever


Just to the right of R1, Vcc is connected to ground.

That can often cause problems.

Get Social

Information

Powered by e107 Forum System

Downloads

Comments

Clydehet
Wed May 01 2024, 06:44 pm
Davidoried
Wed May 01 2024, 06:11 pm
KevinTab
Sun Apr 28 2024, 05:35 am
Tumergix
Sun Apr 28 2024, 12:59 am
StevenDrulk
Sat Apr 27 2024, 08:47 pm
StephenHauct
Sat Apr 27 2024, 09:38 am
Adamsaf
Sat Apr 27 2024, 07:12 am
Robertphype
Sat Apr 27 2024, 12:23 am