Hi,

I'm pretty new to electronics, having skimmed though the AAE course notes over the weekend. I'm making a buffer so I can use my line in as an oscilloscope.

This is the circuit I'm will hope to use for each channel. http://www.sciencetronics.com/geocities/electronics/projects/soundcard_osci.html
The reason why I chose this circuit was because it was the simplest one I could find.

Even though the author says it's a voltage follower, what I would like to know is what is the gain of the op amp. Would it be unity? Or does the resistor on the negative feedback loop reduce the gain?

I'm assuming is does reduce the voltage, but I'm not sure how to calculate the voltage wrt gnd at the inverting input. (Unless it's a question of current. However the op amp is not a current-feedback op amp.)


Richard

 

The amp is unity gain, the 2k feedback resistor has no effect on gain as it is the only connection to the negative input ignoring the cap for now. As op amp inputs are very high impedance then the resulting current in the 2k is tiny and hence the voltage drop is small. Maybe somethng like 20nA which equals 40uV.

 

It's a voltage follower as described. The op-amp will work to keep the voltage at the "-" input the same as whatever it's seeing on the "+" input.

One thing to contemplate with this design is what is ground. If it's battery powered and floating relative to your sound card, fine. But if there's another path to true ground somewhere, you might fall victim to a ground loop. Always measure voltages before hooking up different power supplies.

 

The gain of the opamp is unity because it has 100% negative feedback. Usually the 2k resistor is a piece of wire and the 100nF capacitor is not used.

 

There is no example on the datasheet that explains the resistor and capacitor in the feedback loop. I think they could be replaced with a wire. Feel free to try it that way.

 

The datasheet for the "oddball" opamp shows the RC in the negative feedback and explains why.

 

So, this circuit is bascially a buffer. It doesn't actually bring down voltages to a safe level for my soundcard?

 

Thanks. I have never thought about an opamp commiting suicide by shooting itself in the input transistor.

Edit: Right. This circuit changes nothing about the voltage level.
If your voltage levels are all that high, a volume control can be attached to the input of the CA3140 circuit.

 

The output capacitor of the opamp causes the output of the capacitor to swing as high as +4.5V and -4.5V. What is the maximum voltage rating of the input of your sound card?

 

The output capacitor of the opamp causes the output of the capacitor to swing as high as +4.5V and -4.5V. What is the maximum voltage rating of the input of your sound card?

That's slightly worrying as the peak voltage on a line in is 0.45 V.

Can I just remove the 20uF capacitors?

 

If you remove the output coupling capacitor then the output of the opamp will swing a maximum from 0V to +9V and might destroy your sound card. With low levels then the output will swing a little above and a little below +4.5V.

 

Oh, I didn't see the thread go over to page 2.

I see, the coupling capacitor stops the 9V DC going into my line in. But isn't the coupling also meant to stop the all DC bias? So why isn't +4.5 V offset blocked? Or am I missing something?

 

The output of the opamp swings from 0V to +9V so its average is +4.5V.
If you add an output coupling capacitor to block the +4.5V then the signal swings from -4.5V to +4.5V.

I don't know if the input of your sound card is for a microphone or for line level. The 9Vp-p signal might destroy it.

 

Oh right, I see what you mean.

Yeah, I'm planning to add a clipper circuit after. Something along the lines of this:

http://www.eeweb.com/electronics-quiz/design-this-clipper-circuit