Remove positive feedback and bias Vin(+) with 1 or 2 diodes with pullup to Vcc. or R ratio.
For RS232 the threshold is between this but close to 2 diode drops or 1.3V
It depends on slew rate of Vcc and noise level and resolution of short bursts you may want to stretch.
Obviously my logic must be wrong since the LED is always on.
I guess you did not read the datasheet for the LMV358 opamp. Its maximum output sourcing current (when its output goes high) is only 5mA which will make an LED appear fairly dim.
I agree than your 100% positive feedback is causing your circuit not to work:
1) When the output is high then the (+) input connected to it is also high.
2) Then for the output to go low the (-) input must be higher than the (+) input voltage that is not allowed with the opamp and does not happen with your input signal.
For that matter a good 1.8V AND buffer chip can drive the LED direct with Anode to ground or simply a 1.5V FET with suitable RdsOn.
Just recognised it was SunnySkyguy, not Brad. Sorry.But for your circuit i agree with Brad:
I can see I have to draw the answers for you.
Your circuit looks like an inverting voltage follower but wrong. It's non-inverting and that wont work.
To decide when to enable the LED, you must know the logic threshold , which for LV logic is ~1.2V so a Reference voltage is needed.
THese are 3 ways to do the same thing.
R choose for only 5 to 10mA depending on colour, which for a small indicator is all you need.
Using Ohm's Law [Vcc - Vf(LED)]/R= I_led
OUT is clearly shown to be the output of the opamp, not ground. Why would you short the output to ground?
Thank you for the very detailed response...that makes much better sense now. Only question I have is where would the output be connected? Im guessing for the non inverting op amp it would have to be ground, but would this not be bad during the LED off state since output will be at 5v and will have a short circuit to ground?
In the non-inverting opamp, it is obvious that the anode of the LED is connected to +5V through a 330 ohm current-limiting resistor and when the output of the opamp goes low it drives the cathode of the LED almost to ground to make it light.
Maybe you are confusing 0V with ground. They are the same, but ground is simply a common 0V it is not earth's ground.
You said it backwards:
1) IN- is shown to be +1.2V and it NEVER goes to 0V.
2) The input IN+ is never +1.2V, it is either 0V or +logic high.
Your baud rate is much too fast for the very slow LMV358 opamp. Make a digital data detector that drives the opamp with a DC voltage whenever there is data.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?