I need a low power very low offset voltage comparator opamp which will take 0-5v as supply voltage. It should take -5v to 5v as its input. Will the -5v input wth 0-5v supply burn the opamp??? PLease let me know about ny suitable opamp. I will use the opamp 4 the purpose described below. Any critical thoughts abut my plan is most welcome...
I need to produce +ve edges from each of the +ve and -ve edge of an square wave input. The simplest idea was to diffenciate the square wave and full wave rectify the output. I cannot do this b cos I'm restricted to a single power supply. Instead I designed the followin circuit: A capacitor and a resistor is connected in series.The other end of the cap is connected to square wave input. The other end of the resistor is grounded. The +ve input of one opamp is connected to the -ve input of another opamp and they r grounded. The other +ve and ive inputs of the opamps r also joined and they both r connected betn the cap and resistor. Each of the output of the opamps r connected togethe via rectifiers. THis is the output. PLEASE DRAW THIS SIMPLE SCHEMATIC AND ITS FUNCTION WOULD B COM LUCID.
Good morning, AlienOnEarth. Sorry about the delay. I can detect the note of panic in your post.
First off, I'm hearing you say you want to interface a -5V/5V square wave with a +5V only system and output a pulse both for the positive transition of the square wave and the negative transition. That shouldn't require comparators at all -- you simply use a diode and a resistor to rectify the input to a +4.3V/0V square wave. ` ` Vin Vout ` o->|-o----o ` | ` .-. ` | | ` | | ` '-' ` | ` =3D=3D=3D ` GND created by Andy=B4s ASCII-Circuit v1.24.140803 Beta
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You can then use standard digital logic to get your pulses (I'm assuming this is a digital logic interface).
But, if you want to use a dual comparator to do this, it becomes pretty easy with DC level shifting.
You can see by the diagram that you've established a DC level of 2V for the input signal. The input resistor and cap then cause approx. 1V excursions above and below that DC level with the +5V/-5V excursions of your input square wave. You then use two comparators as a window comparator to get pulses if the input signal goes above the higher divider voltage, or below the lower one. (Resistors with an asterisk are 1% or better).
I'm not sure why you'd need a low offset comparator for a square wave that slams 10V either way. With this setup, you don't have to worry about the differentiator GND. Low power is relative, and up to you. But the voltage dividers in the above take less than 0.5 mA, and any single supply dual comparator should do the job. An LM393 will only take another 1 mA or so with a 5V supply.
I hope this has been of help.
By the way, I'd like to add a few style notes. First, many respondents don't even bother answering posts that are written in cell phone text messaging style or all caps, so if you're actually posting from a cell phone, the emergency must be truly dire (all caps usually indicate either a hostage situation, a fire, or shouting). Please note this in the body of your text, but only after you call the police or the fire department. If not, plz uz std wrds -- r 2 hrd 2 rd.
Also, you might want to download Andy's ASCII circuits -- it's much easier to get a point across with a picture, even within the limitations of ASCII art. The program is incredibly easy to use, and is practically intuitive if you've ever done anything in CAD.
thanx a lot chris, Use net is a wonderful service. It's my 1st time in the groups so I was ignorant about its precepts.
Any way I will use "+5v/0v not +5v/-5v square wave".As u guessed, it will be a digital logic interface. You said I can use standard digital logic to get pulses from a rectified +5v/-5v squarewave. So it implies I can also get the desired pulses from +5v/0v input using digital logic. How can I do that??
--
+-IN>---[1N4148>]-------+--A
| Y--+------A
+--B | EXOR Y-->OUT
| [R] +--B
| | |
[1K] +---+
| |
| [C]
| |
GND>--------------------+--------+
Any gate will work at the junction of the diode and resistor since
its purpose is to provide a totem-pole output to source and sink
current to charge and discharge the cap.
You invert the digital pulse train and then with a small capacitor you can differentiate both outputs, so the pulse length is short compared to the original frequency. this signal you can feed into two comparators and or-them by two diodes or another gate.
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