I am having trouble understanding the behavior of a circuit I am using to apply gain and offset to a triangle wave.
The circuit in question is here:
My triangle wave output is 4 +/- 1.6 V, and I want 2.55 +/- 2.45 V out of the gain stage. (0.1 - 5 Vpp)
To find the topology and calculate the resistor values I used the TI offset and gain calculator located at:
My circuit is AC-coupled, but I can only get it to work properly if I shunt the input signal to ground, between the cap and gain resistor Rg=1370, with a low- value resistor (60 or so). This effectively creates a voltage divider with Rg and sets the DC part of Vi to 0.04 V or so. But it also loads the previous output stage...which I don't want.
If I remove the shunt resistor, the output is centered at about 1.5 V instead of
2.5 V, so my output signal dips a volt or so below ground.What I don't understand is why I need the shunt resistor in this inverting topology. All the op amp gain/offset diagrams I have ever seen show the input from the cap straight to the gain resistor, with no shunt. How could that ever work? The input's DC level, after the cap, seems to be set by V- (and hence by V
+), instead of being 0 as the design tool assumes.I'm confused! Am I making a mistake with the gain calculator or am I wrong to assume that an AC-coupled signal into an inverting amp should not need a shunt resistor to force its DC to 0?
Thanks,
doug