audiodc
Firstly, I'm a hobbyist and have a lot to learn.
I'm trying to get a 0-5V signal created by a microcontroller to +/-2V audio level.
I don't want to spend a lot and would like to keep the part count down, so I started experimenting using resistor dividers and I can get it into either the 0-2V or 0-4V range.
But I don't know how to make the signal go negative. What are some ways to do that?
Some rotary Pot's com in Linear, Audio and reverse Audio (where audio=log taper)
Actually it is "quasi"-log scale and not precision logarithm.
Notice it has a rapid transition from two different linear slopes. which gives you more sensitivity over a wider range of audio inputs rather than an apparent Off to ON effect if using a linear pot. If one adjusts audio equipment over a 30 dB range, that might occur in the 1st 5% of a Linear pot.
Many options may be selected when you order.
Precision Log pots would be mcuch more expensive than
Imagine two scenarios: an AC signal which has such a high frequency that the 4.7 nF capacitor appears as a short. And a signal whose frequency is so low (or DC) that the 4.7 nF capacitor is open.
Signal that is blocked by the capacitor (i.e. that is in your filter's low pass band) sees a gain of \$-8/3.4\$.
Signal that is shorted to ground by the capacitor sees even less gain, because the inverting stage is a current-to-voltage amplifier, and flow through the capacitor robs the input of current.
So basically your amplifier tops out at a gain of \$-8/3.4\$ gain.
You might be thinking that there is somehow a gain of 80 due to the ratio between the 8K resistor and the 100 ohms. But that's not how it works! When you consider the input to be at the 100 ohm resistor, then you have to regard the 3.3K on the left as a source impedance. That source impedance matters because it restricts the current that flows to the current-to-voltage amp. The gain formula for an inverting stage assumes zero source impedance!
Best Answer
I'm going to make an assumption here because you didn't provide enough information.
Specifically, I assume that this is an analog signal that ranges from 0V through +5V.
The solution is simple. All you need to do is to capacitively-couple the signal from the controller output network into your destination device.
I'm going to make a couple more assumptions here:
1) The output impedance from the controller and its output network is fairly low.
2) The input impedance of the device that you are trying to feed is fairly high.
simulate this circuit – Schematic created using CircuitLab
I'm showing a polarized capacitor because that is the most common. The (+) side of the capacitor goes to the node with the most positive (highest) average DC voltage. That would be from your microcontroller: the average voltage is about 2.5 Vdc assuming that the signal swings all the way from 0V to 5V.
It's really that simple!