[SOLVED] voltage cut off using LM324

Status
Not open for further replies.

thannara123

Advanced Member level 5
Joined
Jan 7, 2010
Messages
1,602
Helped
122
Reputation
244
Reaction score
116
Trophy points
1,353
Visit site
Activity points
10,626
i need make a low voltage cut off using LM324 .
the reference voltage given to the inverting pin .

can i directly connect the refferance voltage and the sensing voltage to the IC
is there any design ? if yes how ...
 

yes I am trying to that ,... but how to design that .



Yes there any other passive components needed ?

why dont simulate LM324 in proteus ? any solution ?
or any other simulator ?
 
Last edited:

First specify the performance you expect.
- Vin off to on range, VL
- Vin on to off range, VH
- Hysteresis is required by % positive feedback ratio of Vcc
- Vref determines middle of VH,
- potentiometer voltage range relative to Vref, determines if you want fixed R on either side of pot.

IOL = relay coil current, open collector drive.

I assume you want LOW Side switch for relay with reverse diode clamp across coil so + in to activate relay Vol ( V_out_low) at Iol (I_out_low) which switches relay from LOW side.

Please design this way like any data sheet. Then schematic is easier to create from your specs.
 
Why don't you look at the datasheet for the LM324? It has 4 opamps. The LM358 has only 2 of exactly the same opamps and is in a smaller 8-pins package.

The datasheet shows that its output high current might be high enough to destroy itself and/or the LED since you have no series current-limiting resistor.
The datasheet also says that its inputs DO NOT WORK if they have a voltage more than 1.5V less than the supply voltage but they work all the way down to 0V.
The maximum allowed input voltage is the positive power supply voltage.

An LM10 is an opamp and an adjustable reference voltage in an 8-pins package and will do what you want.
 
I am using proteus/ multisim how to get.
What considerations are needed for designing a simple voltage monitor using LM324.

any example .
 

Look at post #6 again. If you do not add hysteresis then the opamp will amplify its own noise when the input voltages are almost the same then the output voltage will swing all over the place.
The datasheet for the opamp shows you everything you should consider. Keep the input voltages 1.5V or more less than the positive supply voltage. Do not overload the output.
 
is this circuit is correct ?
working well ....
made from my practical knowledge .not a theoretical design
 
Last edited:

thannara123 said:
is this circuit is correct?
Click to expand...
You need to learn the basics about electronics.
1) The zener diode is upside down so it acts like an ordinary 0.7V diode. A zener diode should be reverse biased.
2) The circuit and the opamp have no positive supply voltage.
3) The opamp does not have any hysteresis.
4) The LED is missing a series current-limiting resistor.
 


Sorry that zener is wrongly connected , when it drawn ..
i given positive supply to the input that is RV1 (in practical it is VCC)
This is testing circuit my VCC is only 5volt .
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…