What's going on in this circuit

[Jim Barnes]

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...
http://blea.ch/wiki/index.php/PPM_Meter

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

 

[James Arthur]

stage 3 is a precision full wave rectifier.

Bob

Rectifier? Yes. Precision? Nope.

Cheers,
James Arthur

 

[Jim Barnes]

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

Thanks for your reply Bill.

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

 

[Jim Barnes]

A lot closer than just using diodes.

Bob

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

 

[Jim Thompson]

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...
http://blea.ch/wiki/index.php/PPM_Meter

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

http://www.analog-innovations.com/SED/FullWaveRectifier.pdf

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.

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

Any feed back is greatly appreciated.

As another poster has already opined, think _current_flow_.

Regards


Jim

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Stormy on the East Coast today... due to Bush's failed policies.

 

[James Arthur]

A lot closer than just using diodes.

Bob

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-|<-o----o-----o--|___|---o----|___|---.
| V2+R9 | | | | | R11 | R10 |
=== ~2k |a V - c |C1 .-. 220k | 220k |
GND | - V --- | |R8 | |
| | | --- | |22k | 13|\ |
| o-|<-o |.22 '-' '---|-\ 14 |
| | d | | uF | ___ 12| >-----o-->Vout
| '----)----O-----O--|___|--o----|+/
| | [2] R12 | |/ U1d
| | 220k |
| V+ | .-.
| 9|\| | | |R13 (effective)
'--|-\ | | |220k
10KHz 10| >-'8 '-'

----------|+/ U1c |

|/| ===
V- GND

(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)



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


------
FIG. 2
------

___ ___
.-|___|---o------|___|---.
| R | R |
| O--|<---. |
| | | | C
| | |\ | | .---||----.
Vin | '-|-\ | | ___ | ___ |
~ >-O | >--O--|<--o--|___|--o--|___|--O
| .-|+/ R/2 | 10R |
| | |/ ___ | |\ |
| === .--|___|--o-|-\ |
| | R | >----O--> Vout
| | .--|+/
'------------------------' | |/
===

(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)


The circuit above provides 20x gain and an integrator; even simpler
circuits are possible if you don't need those.

Cheers,
James Arthur

 

[James Arthur]

A lot closer than just using diodes.

Bob

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-|<-o----o-----o--|___|---o----|___|---.
| V2+R9 | | | | | R11 | R10 |
=== ~2k |a V - c |C1 .-. 220k | 220k |
GND | - V --- | |R8 | |
| | | --- | |22k | 13|\ |
| o-|<-o |.22 '-' '---|-\ 14 |
| | d | | uF | ___ 12| >-----o-->Vout
| '----)----O-----O--|___|--o----|+/
| | [2] R12 | |/ U1d
| | 220k |
| V+ | .-.
| 9|\| | | |R13 (effective)
'--|-\ | | |220k
10KHz 10| >-'8 '-'

----------|+/ U1c |

|/| ===
V- GND

(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)



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


------
FIG. 2
------

___ ___
.-|___|---o------|___|---.
| R | R |
| O--|<---. |
| | | | C
| | |\ | | .---||----.
Vin | '-|-\ | | ___ | ___ |
~ >-O | >--O--|<--o--|___|--o--|___|--O
| .-|+/ R/2 | 10R |
| | |/ ___ | |\ |
| === .--|___|--o-|-\ |
| | R | >----O--> Vout
| | .--|+/
'------------------------' | |/
===

(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)


The circuit above provides 20x gain and an integrator; even simpler
circuits are possible if you don't need those.

Cheers,
James Arthur

Whoops, 10x gain, that is.

 

[Jim Barnes]

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