Dual polarity square wave generator

[Venkadesh_M]

A combination of potentiometers can do this. One potentiometer conveys a portion of a 5V supply V.

The capacitor blocks DC from flowing back toward your input signal. However the capacitor can distort a square waveform.



BradtheRod this not seem to work in all frequencies...... Isnt it??


Try this

 

[Elex-factor]

A combination of potentiometers can do this. One potentiometer conveys a portion of a 5V supply V.

The capacitor blocks DC from flowing back toward your input signal. However the capacitor can distort a square waveform.

Screenshot:

Thanks for your help. I tried the above circuit and the voltage level shifted. However, the voltage range was much higher than 5 V even when the input was between -3 and 3 V. How could i shift down the output voltage? I am using 500 ohm potentiometer instead of 200 ohm. Also, what value should i keep the potentiometer to? Does the output voltage depends on potentiometer resistance? Please help.

 

[BradtheRad]

BradtheRod this not seem to work in all frequencies...... Isnt it??

Adding a battery is a method that could work at low frequencies. However at high frequencies it is likely to distort the waveform.

Screenshot of the simulation (theoretical):

Notice the output goes below 0V.

Also notice the battery receives a charging current of a few mA when the input is -4V. Some types of batteries cannot tolerate this.

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I modified the above by putting the diode in series.
This schematic could do the job:

The output will not reach 5V if:

a) the diode has too high a forward volt spec,
or
b) the following stage draws too much current.

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Try this

Yes, the resistors create a virtual ground.



If the load draws too much current, the output level will drop. This will require raising the supply V at the 555 IC.

There will be a question about whether the output is referenced correctly to the next stage's ground rail.

 

[Venkadesh_M]

Also notice the battery receives a charging current of a few mA when the input is -4V. Some types of batteries cannot tolerate this.

Ya the modified circuit could do the job perfectly if the 100k replaced by a 10k............ I just draw a modal which is easy to understand...

f the load draws too much current, the output level will drop. This will require raising the supply V at the 555 IC

I already said that. this circuit cant be used for higher loads.. and for the ground concern it can be used for different ground configurations....

- - - Updated - - -

But a capacitor in that ciruit could damage the sync ness in fast mode...........

 

[BradtheRad]

Elex-factor:
Thanks for your help. I tried the above circuit and the voltage level shifted. However, the voltage range was much higher than 5 V even when the input was between -3 and 3 V. How could i shift down the output voltage? I am using 500 ohm potentiometer instead of 200 ohm. Also, what value should i keep the potentiometer to? Does the output voltage depends on potentiometer resistance? Please help.

The pot between supply and ground needs to be a sufficiently low value, so that each of its arms can be a path for current through the capacitor. In other words, the capacitor needs to have a load attached to its output, in order to attenuate the incoming signal.

A 1k value is suitable. Just a mA or two goes through it.

As for the 200 ohm pot, yes it is fine if you replace it with a 500 or 1k, etc., whatever value gives you the voltage swing you need.

Success depends on fine-tuning the potentiometers, until the waveform looks right.

The same with adjusting the capacitor value. You can reduce it if your signal is at a higher frequency than 100 Hz. It acquires a DC charge, so it must be oriented correctly.

 

[Venkadesh_M]

do u still think cap can be useful.?




Reliable....

 

[BradtheRad]

do u still think cap can be useful.?


View attachment 92993

Reliable....

This makes use of the comparator inside a 555 IC. This is fine, provided that it is safe to apply a negative signal to the pins.

 

[Elex-factor]

Hi

I also wanted to know, if I want to reduce the voltage range from 0-10 V to 0-5 V, what kind of circuit should I use? Should I just use a resistor to drop down the voltage level or is there any better way to do that? Please help..

 

[rahdirs]

Depends on what you want to do with your output.You could use a simple resistive divider to bring 0-10 V to 0-5 V.

But what is it that you want to do with that 0-5 V output ?????

 

[Venkadesh_M]

You can use the same 555Ic cicuit(VCC must be 5V)....

 

[Elex-factor]

Depends on what you want to do with your output.You could use a simple resistive divider to bring 0-10 V to 0-5 V.

But what is it that you want to do with that 0-5 V output ?????

I want to input this signal to an ADC of a microcontroller. Is it ok if I just use a resistive divider for this purpose?

 

[Venkadesh_M]

this is the first thing you have to say...