Simple circuit to convert Bi-polar Square wave to Unipolar

Hi everyone. I want to convert my square wave signal generator to produce only a positive going waveform without just chopping off the -ve signal.

The signal generator has variable output from 0-10V (20Vpeaktopeak or

-10V to +10V)).

I came up with a simple circuit that nearly works: The hot output of the signal generator goes to the cathode of a diode, the anode connects to the -ve of a small value electrolytic capacitor and the +ve goes to the ground output. My resulting signal is taken either side of the diode.

Works well, but I get the signal going a fraction of a volt negative. If I use a shottky diode the negative offset is reduced but still there.

It is important that the output does not go negative at all - in fact I need it about a quarter of a volt positive offset. I have power supply rails that I can access from the signal generator, but connecting a "pull-up" resistor does not work - I think because of the cap.

Any ideas - keeping it simple? I know I could achieve this with an op-amp but want to avoid that and try and keep it relatively passive. If I have to I could use one of two transistors.

Is what I ask possible? Thanks

Reply to
rylmp2k1
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How about an isolation transformer?

Reply to
Greg Neill

I need to avoid passing the signal through a transformer or DC blocking cap as it screws the square waveform up. I need a good square waveform all the way from 1Hz to 2MHz - but thanks for taking the trouble of replying.

Reply to
rylmp2k1

This is a classic "DC restore" circuit where the baseline is set by a DC power supply:

in---------cap--------+----------------out | gnd | gnd k diode a | | | + dc supply - | | | | gnd

Set the DC supply to roughly +1 volt to get a small positive baseline on the output. You'll need a big cap to get to 1 Hz into most loads.

You could actually derive the DC supply from the sig gen output. That would be cute.

John

Reply to
John Larkin

If you don't mind inverting the signal..

Drive a common emitter inverting circuit via a bias R so not to over drive the NPN transistor..

Something in the line of a 4.7k sounds good.

Put a SI diode from base to common so that the (-)input gets redirected to common and not cause the transistor to go into zener mode.

Select a sufficient load R for the collector from a fixed DC source some where..

The collector will be your output.

When the transistor goes into saturation state via a (+10) through the bias R, you'll get ~ .25 volts on most commonly used switching transistors. Maybe less.

When input goes (-10), it'll get sucked up in the SI diode and force the NPN to be off, which will allow your load R to give you what you need.

You'll get a .2 to What ever DC source you select as your square wave.

If you don't want it inverted, just pass it through another inverter stage..

Reply to
Jamie

A clamp? Simply a diode? You can cut off the negative part because it is a square wave. It will only reduce the peak to peak amplitude.

You can use a diode to simply clip the negative half. If you don't like that use a diode clamp or an active diode.

Reply to
Jon Slaughter

You state the output is variable 0 to 10v. Can't you adjust it as so? Does it have an offset control? Care to share more details that you are not telling us yet? regards, al

Reply to
mickgeyver

Another method might be an optoisolator(the bidir kind)...

Reply to
Jon Slaughter

a resistive divider?

try this: .-------------------------------------------------------------. | This is an ascii schematic, if the diagram appears garbled | | try switching to a fixed-pitch font (courier works well) | | pasting it into notepad works well on ms-windows. | | or in google groups "show original" (in "more options") | `-------------------------------------------------------------' + ---[1000]-----. 10V | supply | - ---. | | | gnd | | -------[1100]---+---- sig gen out --. +---- | | gnd gnd

Reply to
Jasen Betts

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