Op-Amp Confusion

[logictom]

Hi, I'm trying to build a circuit which amplifies a small pulse with a large dc offset. I understand the principle of what I need to do but I'm having trouble implementing it.
I have found a circuit in an appnote which works in ltspice but I'm unclear on a few of the details.

As I understand it, the signal is ac coupled by C4,  R2 & R5 are a voltage divider to offset the signal above ground to ~0.6v.
I'm unsure about R1, is this to minimise the loading of the voltage divider? Or is it working with C4 as a high pass filter?
R4 & R3 are the standard gain setting resistors, C2 is linked to the gain but I'm unsure how or why.

I've been reading through the TI single supply opamp circuit collection which is coming in very handy but I'm having trouble relating this circuit to any of those mentioned.

It's been some time since I learnt about opamps and I'm very rusty, am I on the right track or am I barking up the wrong tree?

[ejeffrey]

As I understand it, the signal is ac coupled by C4,  R2 & R5 are a voltage divider to offset the signal above ground to ~0.6v.

Yes.

I'm unsure about R1, is this to minimise the loading of the voltage divider? Or is it working with C4 as a high pass filter?

R1 does two things.  It sets the input impedance high so that the circuit doesn't load the source, and it forms a high-pass filter with C4 for AC coupling.

R4 & R3 are the standard gain setting resistors, C2 is linked to the gain but I'm unsure how or why.

C2 essentially AC couples the gain network.   At low frequency C2 becomes an open circuit and the op-amp becomes a unity gain follower.  The reason it is there is to avoid amplifying the op-amps DC offset.

[logictom]

Thanks for the quick reply.

In the 'standard' non-inverting setup R3 & R4 are connected to Vcc/2, in this case we connect to ground, through C2, because at high frequencies C2 looks like a short? That's what is still throwing me, why in this case is it ground and not Vcc/2?

[logictom]

Just been doing a little more poking around in ltspice, does this amplification work due to a phase delay caused by C2 charging and it's the difference between the phase that the opamp is amplifying? Or am I talking absolute rubbish?

[ejeffrey]

Thanks for the quick reply.

In the 'standard' non-inverting setup R3 & R4 are connected to Vcc/2, in this case we connect to ground, through C2, because at high frequencies C2 looks like a short? That's what is still throwing me, why in this case is it ground and not Vcc/2?

Since it is capacitively coupled the DC voltage doesn't matter -- ground is just as good as Vcc/2.  All that matters is that the capacitor goes to a low-impedance node with a constant voltage. What will happen is that the capacitor will charge up to the operating point set by the voltage divider bias.  It has to do this because in the steady state C2 makes sure no current can flow through the feedback network, meaning all three terminals of the op-amp must be at the same potential.

[logictom]

Thanks for clarifying that, the circuit makes a lot more sense now :)