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| rylmp2k1... |
 Posted: Sat Oct 24, 2009 5:28 am |
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Guest
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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 |
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| rylmp2k1... |
| Posted: Sat Oct 24, 2009 6:31 am |
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Guest
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On Oct 24, 11:51 am, "Greg Neill" <gneil... at (no spam) MOVEsympatico.ca> wrote:
[quote]rylmp2k1 wrote:
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
How about an isolation transformer?
[/quote]
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. |
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| Greg Neill... |
| Posted: Sat Oct 24, 2009 9:51 am |
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Guest
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rylmp2k1 wrote:
[quote]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
[/quote]
How about an isolation transformer? |
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| John Larkin... |
| Posted: Sat Oct 24, 2009 10:57 am |
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Guest
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On Sat, 24 Oct 2009 08:28:30 -0700 (PDT), rylmp2k1
<royalmp2001 at (no spam) gmail.com> wrote:
[quote]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
[/quote]
This is a classic "DC restore" circuit where the baseline is set by a
DC power supply:
in---------cap--------+----------------out
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gnd | gnd
k diode
a
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+ dc supply
-
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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 |
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| mickgeyver... |
| Posted: Sat Oct 24, 2009 12:59 pm |
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On Oct 24, 11:28 am, rylmp2k1 <royalmp2... at (no spam) gmail.com> wrote:
[quote]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
[/quote]
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 |
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| Jamie... |
| Posted: Sat Oct 24, 2009 2:20 pm |
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Guest
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rylmp2k1 wrote:
[quote]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
If you don't mind inverting the signal..[/quote]
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.. |
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| Jon Slaughter... |
| Posted: Sat Oct 24, 2009 4:09 pm |
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Guest
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rylmp2k1 wrote:
[quote]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
[/quote]
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. |
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| Jon Slaughter... |
| Posted: Sat Oct 24, 2009 5:17 pm |
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Guest
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Jon Slaughter wrote:
[quote]rylmp2k1 wrote:
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
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.
[/quote]
Another method might be an optoisolator(the bidir kind)... |
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| Jasen Betts... |
| Posted: Sun Oct 25, 2009 1:40 am |
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Guest
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On 2009-10-24, rylmp2k1 <royalmp2001 at (no spam) gmail.com> wrote:
[quote]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
[/quote]
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 |
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-------[1100]---+----
sig
gen out
--. +----
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gnd gnd |
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