IR Demodulator Design
Answer rewritten due to new information from Jason:

Some other answers talk about using a custom IR received IC - which is what I would tend to do except where price was utterly crucial and I could make a cheaper design wit acceptable performance discreetly.

But, this answer is aimed at making the existing circuit work as it should, as requested. 

MCP601 op amp datasheet here
 This is single supply, rail to rail output, Vin = ground to Vdd-1.2V. So with Vdd = V5.0 = 5V Vin range = 0 - 3.8V.
 Set ideally set mid point at about range/2 = 1.9V or so but 2.5V used is OK. 

R31 to Vground = Vdd/2 provides DC level for opamp input and output. With no signal Vout = V5.0/2 = 2.5V so output of D5 will be about  2.7-Vd = 2.5-0.5 =! 2.2V. You ideally want Vout ~+ 0 at 0 input - see below. 

Filter looks to have potential for massive gain at some frequency ( ~~R17/R18 = 400:1 as noted) - whether this occurs anywhere depends on overall filter action. I'd be more comfortable if the basis for the design was described. 

C16/R31 provide a high pass at &lt;&lt; 1 kHz so well below IR frequency.
 You could increase this till it approaches IR frequency but far enough away for minimimal attenuatio bit that depends whether it was a formal part of your overall filter design - if it wasn't it should be. For best results te overall front end frequency response should be tailored to be a bandpass of designed characteristics.

Try this: 
_

Aim: AC couple op-amp output so
Vout = positive signal peaks from op amp - V_D5


Drive D5 with a capacitor = Cout = say 10 uF electrolytic so D5 is AC driven.   
Add D6 to ground with D6 cathode to Anode (D5 input) and D6 anode grounded.


This causes positive half cycles to be passed through D5 and to charge C15 and negative half cycles to be conducted through D6 to ground.
 This is potentially " a bit hard" on U1a and Cout = 1 uF may suffice.
 Cout > C15, to Cout >> C15 is desirable to stop the two dividing the output positive peaks by too much. 


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