# What's going on in this circuit

• posted

Hi All,

I was wondering if anyone might be able to give me a run down on what happening in stage 3 of the circuit on the following link...

The documentation shows that R8 has a constant 2V across it and I don't really understand how this is working. If it has a 2V drop across it that means that the current through it is a constant 91uA, how does this happen when the input varies? I don't think im too clear on how the bridge is working here.

Any feed back is greatly appreciated.

Regards

Jim

• posted

stage 3 is a precision full wave rectifier.

Bob

• posted

Hi Jim

We know with feedback, the voltages at pins 9 and 10 of UC1c have to be the same. The output will give whatever voltage it needs to make sure pin 9 adjusts to the voltage at pin10 (0.2V). The output ramps up until the bottom right diode in the bridge conducts. This passes a voltage to the bottom of R8. Current flows UP through R8 and through the top left diode of the bridge. This current then flows on through R9 and V2.

If you think current instead of voltage: the same current flows through R8 as R9 and V2. If R8 is 10x the value of (R9 + V2), you will get 10x the voltage across it.

Put another way, with feedback, we know the voltage at pin 9, so hence know the current flowing through (R9 +V2). If R8 is 10x bigger, with the same current, the voltage will be 10x bigger.

C3 smoothes the waveform into dc.

the next stage is a differential to single ended converter that maps the capacitors voltage to a ground referenced voltage to feed into the ADC.

Damn good interview question I think....

If you want the rest of the circuit explaining, let me know and I will talk you through it

(I have a sinewave osc kit to cater for the front end if it helps)

Best of luck

```--
Bill Naylor
www.electronworks.co.uk```
• posted

Rectifier? Yes. Precision? Nope.

Cheers, James Arthur

• posted

A lot closer than just using diodes.

Bob

• posted

Here is an explaination on how the sinewave bit works:

```--
Bill Naylor
www.electronworks.co.uk```
• posted

Your explanation seems to match my understanding of the circuit. What I don't understand is how the voltage across R8 is a constant 2V like it mentions in the documentation. As you point out, if there is a +0.1v-peak input and R9 and V2 = 2.2K then the gain would be 10 and the voltage across R8 would be 1V, when the input drops to +0.05V-peak the voltage across R8 would be 0.5V etc. Where does the constant 2V across R8 come from when the input varies?

I also think this would make a good interview question, has been driving me mad for a couple of hours.

Regards,

Jim

• posted

I don't think it is a precision rectifier because it only compensates for 1 of the 2 diodes (in conduction) in the bridge.

• posted

Hi Jim

This is where it starts getting complex and I would need SPICE to prove my comments. My last post was referring to dc conditions. In ac the following happens:

If pin 9 is constant any positive disturbance on pin 10 will cause the output to shoot high. This will put a spike on the bottom of C3. Because capacitors conduct ac, this spike will appear at the top of C3 and thus put a spike on pin 9 to bring it back into regulation with the voltage on pin

1. The same with a negative disturbance on pin 10, but with a corresponding negative spike.

The bridge recitifier charges the cap in the same polarity with both negative and positive spikes so over time although the capacitor is conducting the ac spikes, it will build up a dc voltage across it.

The voltage at pin 9 with respect to ground will be alternating, as will the voltages at either end of C3. However, the voltage ACROSS C3 will be smooth dc. This is why the circuit has a differential amplifier after it - to pick off the voltage ACROSS C3 and not just the voltage at one of its terminals.

Is this OK?

```--
Bill Naylor
www.electronworks.co.uk```
• posted

An unusual way to do a full-wave rectifier. More common would be as in...

I think the author originally meant for a given example input. It definitely varies with signal.

As another poster has already opined, think _current_flow_.

...Jim Thompson

```--
| James E.Thompson, P.E.                           |    mens     |
• posted

Yep, it's better than plain rectifiers, so 'precision' was probably fair, but relative. The circuit (Fig. 1) depends on diode-matching in the D3-quad, whose Vf's /won't/ match-- the loading due to R11 ensures significantly different diode currents (and in different diodes) for the two halves of the a.c. cycle. The error is particularly troublesome for low- level signals.

The delta.i also introduces an unnecessary temperature term.

------ FIG. 1

------ D3 ___ b [1] ___ ___ [3] .-|___|--o--o-|----------|+/ U1c | |/| === V- GND

(created by AACircuit v1.28.4 beta 13/12/04

All those errors could be eliminated, using a simpler full-wave rectifier circuit, e.g. this one:

------ FIG. 2

------

___ ___ .-|___|---o------|___|---. | R | R | | O--|

• posted

Whoops, 10x gain, that is.

• posted

HI Bill,

That does clear it up a bit for me, thanks so much for your efforts. Sorry for my delayed response, I have not been in the office for the past week.

Best regards

Jim

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