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Converting +/-50V source to 0-5V via diff amp?

J

John

I'm just fed-the-@!#*()-up with these people jumping right into
electronics when they haven't even passed Electricity I.

Anyway- he doesn't need anything except a rotated T, assuming input
impedance >> 10K and forgetting about CMR which can usually be handled
with good layout, and there are no "negative" voltages in this circuit:
View in a fixed-width font such as
Courier.


.
.
.
. Va >------.
. |
. [120k]
. |
. +---------+----> IN
. | |
. [120k] [15k]
. | |
. Vb >------' |
. ---
. ///
.
.
.
. Va Vb IN
.
. 50 0 5
.
. 0 50 5


Thanks for the divider tip...I'll check it out.

I happened to have made a basic EE mistake (something we all do once
in a while), but I made it very publicly.

Though I do not have your level of knowledge, I am most certainly not
an EE noob and sincerely appreciate your patience in answering my
original post.

And I learned something new today...always a good thing. :)

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J

John

Where does the 0V in the first diagram and the 50V in the second diagram
Yea, instant poof. :)
In my first paragraph (in OP) I had discounted using the voltage
divider and was leaning towards a diff amp but hadn't used them
before...lots of questions resulted and the adventure began.
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F

Fred Bloggs

John said:
Anyway- he doesn't need anything except a rotated T, assuming input
impedance >> 10K and forgetting about CMR which can usually be handled
with good layout, and there are no "negative" voltages in this circuit:
View in a fixed-width font such as
Courier.


.
.
.
. Va >------.
. |
. [120k]
. |
. +---------+----> IN
. | |
. [120k] [15k]
. | |
. Vb >------' |
. ---
. ///
.
.
.
. Va Vb IN
.
. 50 0 5
.
. 0 50 5


Thanks for the divider tip...I'll check it out.

I happened to have made a basic EE mistake (something we all do once
in a while), but I made it very publicly.

Though I do not have your level of knowledge, I am most certainly not
an EE noob and sincerely appreciate your patience in answering my
original post.

And I learned something new today...always a good thing. :)

-- remove SPAMMENOT for e-mail responses --

Okay- well it's really not viewed as a divider per se, it is two parts.
The 2 x 120K form a common mode summer, meaning the voltage at their
junction is (Va+Vb)/2=Vcm- their common mode. Then from the perspective
of the 15K node, this looks like a Thevenin source of 120K/2=60k output
impedance with open circuit voltage Vcm. The 15K then loads this down to
Vcm*15/(15+60)=Vcm/5=(Va+Vb)/10. So it doesn't make any difference
which of Va or Vb is connected to GND or the source you are monitoring,
the IN node will read |Voltage|/10. This assumes your circuit loading of
the IN node is small compared to 10K, an easy thing to do. There will be
two main types of common interference into your measurement. The
conducted type due to currents circulated between your measurement
source common and the PIC GND will be negligible because that current
divides in the ratio of approximately 120K to probably milli-ohms on
frame or whatever, so this will be down 160dB or something ridiculous.
The second major interference will be electric field coupling injection
currents into your connecting wiring. Generally the effective coupling
capacitance is on the the order of 10's pF, meaning only high frequency
fields will have any kind of measurable effect. Shunt the 15K with 0.1u
to kill these by 80dB. Simply twisting the Va/Vb wiring buys immunity
from magnetic induction coupling into the circuit. You would have to use
that in any case because that type of interference is differential.
 
J

John

. Va >------.
. |
. [120k]
. |
. +---------+----> IN
. | |
. [120k] [15k]
. | |
. Vb >------' |
. ---
. ///
.
.
.
. Va Vb IN
.
. 50 0 5
.
. 0 50 5

Thanks Fred. Good tips on reducing the noise too.

But doesn't the circuit require that the negative terminal for the
SOURCE and the ground for the PIC's power supply be connected? Then I
can swap the SOURCE's +++ and --- leads with the Va/Vb connection
points and the output is always +5V (with respect to PIC ground).
Otherwise, if the SOURCE neg term. and the PIC ground are not
connected, no current flows thru the 15K resistor and I never get my
5V.

The SOURCE pos. and neg. connections to the PIC circuit might be made
either way so the two grounds might not be connected together. And if
I connect PIC ground to Va or Vb, IN is +5V or -5V (with respect to
PIC ground), depending on how the SOURCE is connected.

Am I missing something? I breadboarded the circuit but still had the
same results. Thanks!
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J

John

Can't use a polarized connector so I'm leaning towards the inline
Just a FYI....
I tested the schottky diodes I have (11DQ05) over the temperature
range they might be subjected to during use (25-70 degrees C) and the
voltage dropped over 0.1 volts as the temperature rose.

While a fixed voltage drop is no problem (as long as it's less than
0.4 or so) and can be compensated for in software, a varying voltage
is a problem and prevents me from using an inline diode.

A fuse and a beefy reversed schottky across the input connections is
looking better and better.


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T

Tony Williams

John said:
I tested the schottky diodes I have (11DQ05) over the temperature
range they might be subjected to during use (25-70 degrees C) and
the voltage dropped over 0.1 volts as the temperature rose.
[snip]

Try this?
R4 R5
+5v---------+---/\/\--+--/\/\----+-->Vout
| | |
\ | +5v |
R3/ | __ | |
\ +--|- \| | R1= 10*R.
R1 | |Op >---+
Vin---/\/\---+------------|+_/| R2= 10*R/9.
|\ |
\ \_Voltage, V1. | R3= R.
R2/ | R4= R.
\ | R5= R.
| |
0v----------+----------------+------0v

When Vin= +50, V1=5. When Vin= -50, V1=0.

The opamp has to be rail-rail input and output.
 
J

John

Try this?
Thanks Tony,
Been crazy busy this week but will try it this weekend. If I can swap
the leads from a 50V source on Vin and 0V and have a single-ended
Vout, I'll be a very happy man. :)

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T

Tony Williams

[/QUOTE]
Thanks Tony, Been crazy busy this week but will try it this
weekend. If I can swap the leads from a 50V source on Vin and 0V
and have a single-ended Vout, I'll be a very happy man. :)

Just to clarify.

Vout= Vin/10 for Vin= 0 to +50V.

Vout= 0V for Vin= 0 to -50V, but no damage to the opamp.

There could/will be damage to the opamp if Vin is
connected when the opamp is not powered. Protected by
the addition of two Schottky diodes, across R2 and R3,
both pointing upwards. Protection also helped by a
large value for R1, say 100k or greater.
 
J

John Woodgate

In message <[email protected]>, dated Fri, 8
Sep 2006 said:
I think the U3 and U1 functions can be combined into a single stage,
but I don't have time to play with it right now.

What are you playing with instead? Put it DOWN, boy!(;=)
 
F

Fred Bloggs

Jim said:
Or do it right....

http://analog-innovations.com/SED/ConvertUnknownPolarity50V-OutputZeroTo+5V.pdf

I think the U3 and U1 functions can be combined into a single stage,
but I don't have time to play with it right now.

...Jim Thompson

This is what I mean about your SED circuits. You /can/ combine it all
into one stage, did it several days ago, but with those small input
currents, 10uA on the low end of Vin=1V, you need low leakage diodes,
maybe even JFETs diode connected. I gave up on this OP, he says one
thing but then goes on to display zeroid ability. TW's circuit is good
but I think it's too risky for a zero ohm connection to an input lead.
 
J

Jim Thompson

This is what I mean about your SED circuits. You /can/ combine it all
into one stage, did it several days ago,

Show me. Message ID ?:)
but with those small input
currents, 10uA on the low end of Vin=1V, you need low leakage diodes,
maybe even JFETs diode connected. I gave up on this OP, he says one
thing but then goes on to display zeroid ability. TW's circuit is good
but I think it's too risky for a zero ohm connection to an input lead.


...Jim Thompson
 
J

John

I gave up on this OP, he says one
Shucks, and I was really hoping to learn something here. :)
You don't even want to tell me why I didn't get the results I thought
I was going to with your circuit? Was I supposed to know something
about your circuit that you didn't mention?

Zeroid ability would mean I couln't even breadboard your circuit to
try it out. Or worse yet, wouldn't even want to. But I did want to
and did take the time to test your circuit.

Politely asking questions of those who know more than I and being
willing to take the time to test out recommended circuits, that's a
lot more than zeroid ability. I never deserved a bit of the sarcasm
and brushing off I've gotten from you and was sincerely hoping you
could help me.

I'm more than willing to do the leg work here, just needed some
pointers. I'm damn sure you once knew less than you do now and that
you have asked many, many questions about EE stuff in your lifetime.
I'm doing nothing more than you have already done and would hope that
you could give me the same basic respect that you deserved when you
asked questions from those who knew more.
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J

John

Thanks Tony, Been crazy busy this week but will try it this
Thank you Tony. I've been thinking about the circuits I've been
playing with, and yours, and realized that Vout would stay at 0V.
This is definitely no problem. :)


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T

Tony Williams

John said:
Thank you Tony. I've been thinking about the circuits I've been
playing with, and yours, and realized that Vout would stay at 0V.
This is definitely no problem. :)

What exactly is your 1-50V voltage source that
can be accidentally wired in reverse?

Is it truly floating, (as my circuit requires), or
does it have a Gnd/0v connection somewhere, (as JT's
or Fred's differential circuits can cope with)?
 
J

Jim Thompson

What exactly is your 1-50V voltage source that
can be accidentally wired in reverse?

Is it truly floating, (as my circuit requires), or
does it have a Gnd/0v connection somewhere, (as JT's
or Fred's differential circuits can cope with)?

My version DOES handle a floating OR grounded source.

...Jim Thompson
 
F

Fred Bloggs

John wrote:

You do entirely too much talking and far too little actual work to be
worthy of any investment in time from others.
 
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