Dual +ve edge

Discussion in 'Electronic Basics' started by [email protected], Jun 18, 2005.

1. Guest

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.

2. ChrisGuest

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
` |
` .-.
` | |
` | |
` '-'
` |
` ===
` GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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.

`
` VCC
` +
` |30K* VCC
` VCC .-. +
` + | | |
` 30K*| | | .-.
` .-. '-' VCC | |
` | | | + | |
` | | | |\| '-'
` Vin '-' o-----|-\ | Vout
` || ___ | | | >-----o-----o
` o---||--|___|-o-----|--o--|+/ |
` || 120K | | | |/ |
` 100pF .-. | | |
` 20K*| | | | |
` | | .-. | |
` '-' | |3.3K |
` | | | | |
` === '-' | |
` GND | | |\ |
` | '--|-\ |
` | | >-----'
` o-----|+/
` | |/|
` | ===
` .-. GND
` 20K*| |
` | |
` '-'
` |
` ===
` GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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.

Good luck
Chris

3. John FieldsGuest

---

+-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.

4. Guest

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??

5. BanGuest

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.