Its been while for me, and it sure feels great to be back,
I sure need some help I am trying to design a clipping circuit from 12 volts input clipped down to 5 volts using lm-324 OP-AMP.
thanks for stopping by, what I am looking for is to I have 12 volts input dc signal, and I want the op-amp to clip it down to 5 volts. I don't know if it is the correct term "clipping", please correct me if I am wrong.
Clipping typically refers to the distortion of an AC waveform that occurs when the input signal to an amplifier is of too great an amplitude, so it isn't clear what your goal is here.
Is the 5V to be used as a power supply for a subsequent circuit or perhaps an input to an ADC? Monitor the voltage of a car battery? Perhaps you're seeking a voltage divider or scaling amp? More info is needed!
what I am looking for is to I have 12 volts input dc signal, and I want the op-amp to clip it down to 5 volts. I don't know if it is the correct term "clipping", please correct me if I am wrong.Best Regards,
The Sniper (+)
it took me a while to come back,
yes pranam77, what I want is to drop the 12 volts to 5 volts, not to cut off the supply at 12 volts, its like what baysidebecca said, what I am looking for is a divider apparently and the term clip is incorrect in this meaning!
You may use a LM7805 to do the said job. Only three terminal need to connect to 12 volts Gnd and take the output 5volts from the third pin. Use filters of 1000Mfd /25v to the input and 470mfd /25v to the output. Reading the datasheet of LM7805 will explain you better. Good luck.
if you have a 12volt DC (no signal variation you are trying to preserve) and simply want to drop it to 5 VDC, as said, a linear voltage regulator is a good choice.
However, a linear regulator is inefficient, as the entire voltage drop times the load current is turned into heat.
Your options:
linear voltage regulator (inefficient but accurate)
a series 7 volt zener diode of correct wattage (inefficient, not accurate, but cheap)
A switching voltage regulator (efficient, accurate, some ripple, more parts count)
A series dropping resistor of the correct wattage (inefficient, very in-accurate, very cheap)
Use the voltage drop for some other component, like take up the 7 volt drop with two series connected led diode indcator lights, etc.
it took me a while to come back,
yes pranam77, what I want is to drop the 12 volts to 5 volts, not to cut off the supply at 12 volts, its like what baysidebecca said, what I am looking for is a divider apparently and the term clip is incorrect in this meaning!
The easiest way to do this is the circuit submitted by eecs4ever above. A very simple voltage divider + voltage follower buffer, it will provide Vo = 5Vin/12.
To be precise, note that you need to use 1% resistors, and the closest standard 1% values are 6.98kΩ and 4.99kΩ. This should yield an output accuracy of ≈±2% or better.