11-26-2012 05:14 AM
I am simulate a circuit using LM358N op-amp to amplify the signal from the IC timer NE555P and then using that amplified signal to switch the LED by a transistor switch.
In theory, I think that when the switch key is off, there must be no signal go through LM358 and no signal go to the B-pin of the BJT, so the LED will be on due to BJT switch but in practice, when I switch off the key, the voltage at the B-pin of BJT is still 1.58V and it causes the LED on.
I attached the image and my simulation here, please take a look and any help will be appreciated.
Thanks in advance.
Solved! Go to Solution.
11-26-2012 06:56 AM
If you use your probe tool and measure the unterminated non-inverting input pin of U3A, you will see ~800mV which is the input offset voltage. This is being amplified and is why you are seeing ~1.6V. Best engineering practice is to not leave input unterminated so add a 1Kohm resistor to ground at the non-inverting input pin of U3A. Now measure the input pin and see the difference.
08-04-2017 12:28 AM
pin 4 need to be VEE, not VCC. -10.0 V that's all