Turning a small AC analog signal into DC

[HankMcSpank]

Hi there,

I wonder if you knowledgable guys can help me out here.

I'm running a small analogue signal through two oamp stages. The second stage has a variable gain, controlled by a JFET. The JFET is biased by the output from a PIC (essentially a varying DC level - more DC on the JFET Gate, the more it 'opens' & vice versa).

I need to take the output of this second opamp stage & rectify it into DC, thereby having DC reflect the 'signal level' at the opamp output. This DC will feed into the PIC, which will then decide if more/less gain needs applying to the second stage. It's really just AGC using a PIC.

This second opamp output stage will have a maximum a signal level of about 4.2V (peak to peak) & at it's lowest it'll be about 1.2V peak to peak.

My opamps have a single rail of about 5V & therefore I'm running with a virtual earth, therefore the AC signal sits on about 2.5V DC to start with.

Therefore, referencing this 2.5V, (& just focusing on the AC signal positive cycles here)...the maximum voltage the AC signal will 'hit' will be 4.2V divided by 2 = 2.1V + the residual virtual ground level (2.5V), therefore 4.6V max. The lowest will be 1.2V divided by 2 = 0.6V + the residual virtual ground level (2.5V)...therefore 3.1V.

So at it's lowest, the AC signal will be 3.1V...& at its highest 4.6V (referenced to 2.5V). This represents a theoretical peak to trough swing of about only 1.5V.

What's the best approach of extracting this small 'swing' into DC.

I can forsake a couple of hundred millivolts being 'lost' due to say germanium diode voltage drop, but probably not 0.7V! Also, I can tolerate a bit of ripple (therefore half wave recification probably ok...I'll use a larger smoothing cap & have the PIC average it a little).

I'm not clued up at all about diodes (ie the variants of...& which would be best for my needs)

I really need to keep the component count as low as possible, so extra opamp circuits are out for now.


Any ideas?

[Artie]

Wait for more answers from the more experienced, but it seems that you could just capacitively-couple, or transformer-couple the output, to remove the DC component, then run the AC portion through any simple opamp rectifier as shown in many opamp circuit collection PDF's. Such as this one:



AN-31

There's simple full-wave rectifiers and more complex precision rectifiers in that document.

[HankMcSpank]

I'm just thinking here that if I remove the DC component, it could send the results out of whack.

At the moment, when the signal is at max peak to peak , let's say it's 4V. Becuase I'm using a virtual earth sitting at 2.5V, the signal at max will see 4.5V at the top of it positive cycle & 0.5V at the bottom. This sits nicely within the confines of 0V & 5V rails.

If I remove the DC component, would the zero crossing point would then be 0V? (as opposed to 2.5V), if so, what I can't wrap my head around is - since the negative parts of the signal can't go below 0V, what happens to them when I remove the DC component....do i only get the positive cycles? Or do I get the full AC signal, but with the lowest part of the AC signal touching 0V.

perhaps I haven't put it across well - but it's a head trip for me!

Here's a circuit that I think might be appropriate.....




(taken from http://www.holmea.demon.co.uk/Spread/Spread.htm )

but I can't quite figure out what's happening with the D1, R3 & R5...& C2. Obviously, there's some rectifying/smoothing going on...but what's the technical explanation eg what conditions does D1 conduct with those resistors on its anode)

[AndrewT]

use a precision rectifier using two opamps.
or
use a peak hold circuit again using opamps.

[HankMcSpank]

Hi Andrew,

Thanks for your input. I really need to keep the component count low - what I've omitted to say is that there'll be six of these 2 x opamp chains ...I can't really afford the space to host yet another six opamps (1 rectifier per chain) to go the direction you suggest - the resulting small board will be going inside my guitar!

How do I arrange the half wave rectifier circuitry, so that when the signal is at it's lowest, it is barely conducting, but when the AC signal is at its max, it is passing through the full 2V (ie half the max 4V peak to peak) of signal peak cycles.

Do I need to use germaium diodes? (therefore only losing 0.2V across the diode)

I'm thinking if my opamp output has a residual 2.5V DC, then I'll need 2.5V DC on the other side of the diode too (to maintain the virtual earth)....but the other side of the diode must also be able to drift upward in DC as those large positive cycles come through.

I can then program my PIC to say that an incoming level of 2.5V DC = lowest AC signal ......& an incoming 5V = highest signal ....this is give a full 250 steps (which ties in reasonably well with the PICs 8 bit AtoD resolution of 256 steps)

[AndrewT]

I think you are wasting your time using a conventional rectifier (whether Schottky or germanium or silicon diode) on low level signals.

[wakibaki]

Originally posted by HankMcSpank
If I remove the DC component, would the zero crossing point then be 0V?

The zero crossing point is always 0V.

Originally posted by HankMcSpank
since the negative parts of the signal can't go below 0V, what happens to them when I remove the DC component....do i only get the positive cycles? Or do I get the full AC signal, but with the lowest part of the AC signal touching 0V.

The signal is a DC level with the AC superimposed. When you take away the DC level the AC remaining is symmetrical about 0V.

I know it seems weird but if you have a 5V DC with 2V peak-peak AC superimposed on it then the max value is 6V and the min is 4V. If you feed this signal to (thru) a capacitor you will see AC on the other side with max +1V and min -1V symmetrical about 0V when both sides share a common ground. Presuming no other losses... The way that you are generating the AC has implications only for the apparent impedance of the AC source.

Magic. Otherwise known as application of the superposition principle.

The superposition theorem for electrical circuits states that the response in any branch of a linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are replaced by their internal impedances.

AndrewT is right, you need a precision rectifier. I found these to be a bit of a bastard to get working right but it was probably just a bad day. There are a few true RMS chips out there for this function. You need a dual-rail power supply, which could be a problem.

w

[Nico Ras]

If the frequency is not too high, use the processing power of the PIC, you are using the analog port and all you need to do is detect when you are at a positive or negative peak then read the voltage.

Use dv/dt to determine the peak, pretty straight forward. There are many PICs that run at 40 MHz and multiplexing several ports and sampling should not pose a problem.

[HankMcSpank]

>> If the frequency is not too high, use the processing power of the PIC,

I had considered that, but this is ultimately going to be a battery powered device that is already going to have a fair amount of draw (there's other stuff following on in the circuit tha's a bit of a current hog). By using the latest most powerful PIC variants, the current draw will likely go way up.

I'm using a PIC from the 16 range....apparently it's not up t the job of extracting peak fpresented on six different input pins (max frequency is 2khz). there are some on the PIC forums that said this requirement is too big a task for even the latest PIC variants.

therefore I'm faced with using discreet (in part)

I think I can get the PIC to 'average out' the ripple contingent, so a simple half wave rectifier is likely all I need. Back to my conundrum - but I'm still not sure I understand how to apply the circuit.

So with the opamp outputting 4V peak to peak, with the DC level at 2.5V (virtual ground DC level), how do I implement a circuit to extract the DC content of the signal.

I don't care whether the extracted DC is 'added' to the base 2.5V, or whether it's referenced to ground...all I care about is getting a circuit that will act to extract the D C component of the signal.

what do I need here?

Opamp (AC signal sitting on 2.5V DC) -> Diode (in series)....

Then what? It's this other side of the diode that's causing me probs!