Elex-factor
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@BradtheRad: By saying back to back diodes,are saying like the one attached so that you are capable of varying duty cycle by moving the wiper.
The first one with the series capacitor are you saying in the (0-5)V DC pulse,DC offset 2.5V would be blocked ???
Would the said two work at high frequency ???
Zeners back-to-back. The resulting amplitude is the zener value plus 0.6V.
Diodes (or led's) arranged anti-parallel.
It could be difficult to obtain exactly 2.5V amplitude. A low ohm resistor can be added to 'bend' the value a little.
As a matter of fact, if the op amp output is at an unchanging volt amplitude, a simple resistor divider could do the job without diodes.
Using a series capacitor to turn DC pulses into AC square waves. (The resistor to ground is necessary.)
Only as high as component specs permit. If we try to go faster than that, the waveforms will start to deteriorate, of course.
Could you use +2.5 and -2.5 supplies?
But do we need to change the capacitor and resistor value for different voltage input? For e.g. I tested using 12V input and got the square wave from -6 V to +6 V with the same capacitor and resistor value. If I give 15 V input, do I need to change the capacitor and resistor values?
Hi rahdirs,You'll use a resistive divider to convert bipolar to unipolar square wave ???? How do you intend to bring (-4 - 4) V to (0 - 5)V
What if we need to step up the voltage level from the range of -4 V to +4 V to a microcontroller compatible voltage level between 0 and 5 V? Can we just use a resistor circuit or is there any other better way? Please suggest me some solutions.
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