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DC in series with AC

ebrahim120

Aug 18, 2012
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Hi,
I need to add an DC offset to my AC voltage.I have a function generator for ac and source measurement unit (SMU) for DC. I have tried to see the signal on oscilloscope but I can't.anybody knows how to do that ?:confused:
 

Electrobrains

Jan 2, 2012
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Decouple the AC signal through a (or two) capacitor(s). Then you can add or subtract a DC voltage on the output side.
Important: The capacitors must be able to handle the current and polarity that will result.

p.s. If it's only a question of "seeing the signal" on your oscilloscope, then why not switching the input selector to AC on the oscilloscope? If you need to see both signals together on one channel and your DC is pushing the ray out of the screen, then you can attenuate the input signal to the oscilloscope by adding a resistor to the probe input. For instance if your probe has 1M Ohm input impedance, by adding 1M in series with the input, you will get double the voltage/div on your screen.
By doing this, high frequency signals might be slightly distorted, as the probes are calibrated without that resistor.
 
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ebrahim120

Aug 18, 2012
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thanks. but I am using Stanford resarech function generator DS345 in series with power supply, and in principle it should work. the problem is when I did according to this http://www.google.com/url?sa=t&rct=...lGl_mskDJbK4-BIDA&sig2=pI0xj8vmOGDjPI4vg-NXog
it doesn't work. while Instad of flotaing the function genartor I grounded it worked.
the problem now is when I measerd the dc offset with osilloscope it half of the value of the dc power supply, but when I measuerd with multimeter the offset it is correct. so I totally confused ...
 

Harald Kapp

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You may have introduced a ground loop. Many Oscilloscopes are grounded. If you ground the DC-source, you can create a loop that changes your measurements. There is also no obvious reason why a floating Dc source should not work.

Can you show us a drawing of your test setup? From the textual description alone it is hard to figure what may be wrong.
 

ebrahim120

Aug 18, 2012
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I attached the drawing of my design. Indeed I found the function generator is floated, so in principle my circuit is similar to the one of Agilent which I gave the link in previous comment. The only difference is instead if resistor I am using Oscilloscope which had 1 Mega ohm internal resistor.
And as I mentioned when I measured with multimeter the offset is right while oscilloscope shows the half of that.
 

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Harald Kapp

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The circuit looks right. How do you measure the offset on the scope?
 

ebrahim120

Aug 18, 2012
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I put the oscilloscope in DC mode and I saw the shifting when I applied the dc, but it is the half value which I am applying.
 

Harald Kapp

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That's illogical. What happens when you measure only the DC source (no AC connected)?
Does your scope give you a direct measurement or do you read the voltage from the trace on the screen? Check the settings (offset, gain) of the scope.
 

superam

Aug 22, 2012
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If the internal resistance of Oscilloscope and the function generator is the same, then half the voltage may fall across the function generator!
R{o} = R{fn}

----- youmeusandabeautifullife.wordpress.com
 

Harald Kapp

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Right, but:
I am using Oscilloscope which had 1 Mega ohm internal resistor
That would mean the function generator has 1MOhm internal resistance. Very unlikely. A likely scenario could be if the scope was set to 50 Ohm - contrary to the posted setting. But then again the multimeter should have shown the halfed value, too. Unless, that is, if the scope was not connected while measuring with the multimeter.
 

KrisBlueNZ

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Nov 28, 2011
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If it's an analogue scope, make sure the "variable" vertical gain adjustment is turned fully clockwise. This is usually a little rotary control in the middle of the input range selector. It should click when it's fully clockwise.
 

superam

Aug 22, 2012
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Right, but:
That would mean the function generator has 1MOhm internal resistance. Very unlikely. A likely scenario could be if the scope was set to 50 Ohm - contrary to the posted setting. But then again the multimeter should have shown the halfed value, too. Unless, that is, if the scope was not connected while measuring with the multimeter.


I thought internal resistance is something different than output impedance. :confused:
 

Harald Kapp

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What else should it be?

For the generator it is the series resistance of the output, assuming an otherwise (almost) ideal voltage source within the generator.
For the scope it is the parallel resistance of the input, assuming an otherwise (almost) ideal input stage of the scope's vertical amplifier.
 
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