Is there a common way to generate relatively accurate microvolt (like 5uVpp, 0.5uV resolution) waveforms with bandwidth under 4kHz, output impedance around 1k? I can't think of much more than resistors divider on a bigger signal.
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You better ask yourself - what are you going to do with such a signal? – Eugene Sh. Jul 23 '15 at 20:06
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1Why, playing around is not an option? – Jul 23 '15 at 20:15
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You are asking of "common way", right? There is no common way for playing around. Common ways are usually for practical things. – Eugene Sh. Jul 23 '15 at 20:17
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1low current is easier to generate, and precision resistors are easy to find (sort of), why not use a precision shunt resistor and a precision current-steering waveform generator over that resistor. – KyranF Jul 23 '15 at 20:19
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another issue is trying to buffer that signal, you would need zero input-offset (chopper stabilized?) op-amp to voltage buffer the output? – KyranF Jul 23 '15 at 20:20
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@KyranF, thanks! This is more or less what i thought about. In fact, maybe it doesn't even have to be a current sense resistor. It could be a 10R in series with 10000R and millivolt waveform. Measurement of that thing should be easier, i guess. At least they told us to use instrumentation amps with high gain, right? :) – Jul 23 '15 at 20:28
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With a very low output you will probably not need buffering as the bottom resistor in the divider would be only a few ohms which is then the effective output impedance.
I normally generate the signal at a few volts p-p so you can easily display it on a scope then divide it down with resistors that are in the kilohm range. You may need to divide it in more than one stage to avoid the wire resistance being an issue. The following diagram shows a 1000:1 attenuator. To get 1 million to 1 you could replace the upper resistor with a 1megohm resistor or put two attenuators in cascade. For low frequencies a single stage would be ok but for frequencies above a few 10's of kilohertz two stages would avoid capacitative parasitics being a problem. I wouldn't advise going below a few ohms for the lower resistor or wire resistance could cause errors.
simulate this circuit – Schematic created using CircuitLab
Rather than making the attenuator you may be able to use off the shelf Audio or RF attenuators to do the job, they will often go down to DC. A 1,000 to 1 ratio is 60dB, 1,000,000 to 1 ratio is equivalent to 120dB.
For example you could generate the signal at 5v p-p then attenuate it by 120dB to get 5uV p-p. Be careful about shielding and grounding or you may compromise the accuracy of the output.
Kevin White
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Cool, thank you! So if you did such circuit, is there any chance you could spare a waveform screenshot of something like that? – Jul 23 '15 at 20:37
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This is actually pretty easy.
simulate this circuit – Schematic created using CircuitLab
Use 0.1% resistors for the voltage divider, and your accuracy will be about the same.
Since the output resistance from the divider is the Thevenin equivalent resistance (about 1 ohm) R3 is added to bring the output resistance up to 1 k, and you can modify as necessary.
The thing to be careful about is stray resistance on your ground path and between R1 and R2. Since R2 is only 1 ohm, .01 ohm between R2 and the source voltage, or between R2 and R1, will reduce the signal output by 1%. If you really need the accuracy, you can increase R1 to 1M and R2 to 10 ohms. Or, as Kevin White suggests, use 2 or more divider stages in succession, with larger values in each stage. A starting point would be two 100k/1k stages, which will give an ouput impedance of a bit less than 900 ohms without a separate output resistor. The only difficulty with this is that stray capacitance becomes a potential issue if the frequency starts to rise.
At audio frequencies, stray capacitance will not be a problem, even though R1 is high, since it is counteracted by the very low R2.
WhatRoughBeast
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