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

[John]

I'd like to monitor a slowly changing 1V to 50V source via the ADC on
a PIC16F876A. I was originally going to use a 10:1 voltage divider
(0.1% resistors) to bring the voltage down to 0.1V-5V and apply that
to the ADC. Then I realized I would eventually accidently hook up the
source "backwards" and blow the PIC.

I was considering using a difference amp (AD628) for the job. Not
only would I not have to worry about GND problems, it would drop the
comon mode voltage that might be riding on the long leads from the
diff. amp. to the voltage source.

But, while using a 10V supply and a 2.5 Vref gives me an output of
2.6V-7.5V for a "correctly" connected source (SRC POS to +IN of amp
and SRC NEG to -IN of amp), which I would either divide-by-2 before
going to the PIC or change the amp's default 0.1 gain, what happens if
the source is connected "backwards"?

Does the amp just output zero volts (or whatever its min. voltage
swing is, IIRC, 30mV) because I'm using a unipolar supply for the amp?
Or do I end up zapping the amp?

I'd like to avoid using a bipolar supply for the amp as I don't need
the neg. voltage for anything else.

And while it would be great to be able to monitor the source no matter
how it was connected, I can't have the the PIC's ADC cover a +/-50V
source with high enough resolution. I need 10-bits available at the
DAC for a 50V range, minimum.

Adding a 12-bit ADC would do the job but I'd like to keep this as
simple as possible and I think I can if the diff. amp. setup described
above just drops to zero volts (almost) when the source is connected
"backwards". I could then use a amp output of less than 2.5V as an
indicator that the battery pack was connected backwards and sound a
klaxon, flash a strobe, send out a flare, whatever, to let me know.

Thanks for your help!
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[John]

But, while using a 10V supply and a 2.5 Vref gives me an output of
Ahhhh...OK, I see now.
When the source is zero volts and is connected "backwards", the amp
output starts at 2.5V (my Vref) and drops towards 0V as the source
voltage drops towards -50V.

I just have to make sure that my 10V supply and Vref work with the
amp's specs. Finally making sense of Figure 31 on page 15 of the
AD628 spec sheet, it looks like with a Vs of 10V and a Vref of 5V, the
amp can deal with a +46V to -38V source.

So, I can use that .1% res. voltage divider across my +/-50V source to
bring it down to +/-25V. With a Vs of 10V and a Vref of 5V, the amp
outputs 2.5V to 7.5V for a +/-50V source. I'll just divide the amp's
output again to bring it with range of the PIC's ADC.

Sounds kind of kludgy though....any higher rated (common mode voltage)
diff amps out there? Heck, is using a diff amp the right thing to do
here.

Thanks!
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[Fred Bloggs]

I'd like to monitor a slowly changing 1V to 50V source via the ADC on
a PIC16F876A. I was originally going to use a 10:1 voltage divider
(0.1% resistors) to bring the voltage down to 0.1V-5V and apply that
to the ADC. Then I realized I would eventually accidently hook up the
source "backwards" and blow the PIC.

I was considering using a difference amp (AD628) for the job. Not
only would I not have to worry about GND problems, it would drop the
comon mode voltage that might be riding on the long leads from the
diff. amp. to the voltage source.

But, while using a 10V supply and a 2.5 Vref gives me an output of
2.6V-7.5V for a "correctly" connected source (SRC POS to +IN of amp
and SRC NEG to -IN of amp), which I would either divide-by-2 before
going to the PIC or change the amp's default 0.1 gain, what happens if
the source is connected "backwards"?

Does the amp just output zero volts (or whatever its min. voltage
swing is, IIRC, 30mV) because I'm using a unipolar supply for the amp?
Or do I end up zapping the amp?

I'd like to avoid using a bipolar supply for the amp as I don't need
the neg. voltage for anything else.

And while it would be great to be able to monitor the source no matter
how it was connected, I can't have the the PIC's ADC cover a +/-50V
source with high enough resolution. I need 10-bits available at the
DAC for a 50V range, minimum.

Adding a 12-bit ADC would do the job but I'd like to keep this as
simple as possible and I think I can if the diff. amp. setup described
above just drops to zero volts (almost) when the source is connected
"backwards". I could then use a amp output of less than 2.5V as an
indicator that the battery pack was connected backwards and sound a
klaxon, flash a strobe, send out a flare, whatever, to let me know.

See http://www.maxim-ic.com/appnotes.cfm/appnote_number/1132

and

http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1963

and use resistive voltage divider to scale the input down.

 

[John]

See http://www.maxim-ic.com/appnotes.cfm/appnote_number/1132

Hmm...cool.
But, wouldn't I exceed their voltage ratings (+/-44V) when the
connections were "reversed"?

"Correct polarity" voltages from voltage divider:

50V -----+
|
.-.
| |100K
| |
'-'
+-------- 5V
|
.-.
| |11.1K
| |
'-'
|
0V -----+-------- 0V


"Reversed polarity" voltages from voltage divider:

0V -----+
|
.-.
| |100K
| |
'-'
+-------- 45V
|
.-.
| |11.1K
| |
'-'
|
50V -----+-------- 50V

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[vasile]

I'd like to monitor a slowly changing 1V to 50V source via the ADC on
a PIC16F876A. I was originally going to use a 10:1 voltage divider
(0.1% resistors) to bring the voltage down to 0.1V-5V and apply that
to the ADC. Then I realized I would eventually accidently hook up the
source "backwards" and blow the PIC.

If you're using a simple diode in series with your divider and the
voltage across the diode will be around 0.4V (schottky one) the error
measurement you'll get will be
0.6V instead of 1V and 49.6V instead of 50V.
If this is ok, then a simple diode could solve your backward connection
(if you didn't found the appropiate connector)

greetings,
Vasile

 

[Fred Bloggs]

Hmm...cool.
But, wouldn't I exceed their voltage ratings (+/-44V) when the
connections were "reversed"?

"Correct polarity" voltages from voltage divider:

50V -----+
|
.-.
| |100K
| |
'-'
+-------- 5V
|
.-.
| |11.1K
| |
'-'
|
0V -----+-------- 0V


"Reversed polarity" voltages from voltage divider:

0V -----+
|
.-.
| |100K
| |
'-'
+-------- 45V
|
.-.
| |11.1K
| |
'-'
|
50V -----+-------- 50V

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Hey- thanks for catching that!- and here after all this time, I still
can't even get simple resistor networks right. You're the man- quite
quick and clever, very impressed here. I can see you're going to go
places. Please tell us how you eventually solve this state of the art
problem, We all will be in your debt! Anxiously awaiting your teachings...

 

[Donald]

Hey- thanks for catching that!- and here after all this time, I still
can't even get simple resistor networks right. You're the man- quite
quick and clever, very impressed here. I can see you're going to go
places. Please tell us how you eventually solve this state of the art
problem, We all will be in your debt! Anxiously awaiting your teachings...

Where does the 0V in the first diagram and the 50V in the second diagram
get connected to.

The ground of the PIC processor.


I think this will be a problem.

 

[John Woodgate]

Hey- thanks for catching that!- and here after all this time, I still
can't even get simple resistor networks right. You're the man- quite
quick and clever, very impressed here. I can see you're going to go
places. Please tell us how you eventually solve this state of the art
problem, We all will be in your debt! Anxiously awaiting your teachings...

Fred - his second diagram is just upside down. Don't be too hard on him.

To the OP: The best solution is to make it impossible to connect the
supply the wrong way, by using a non-reversible connector. If you can't
do that, you can either connect a diode in series (if the 0.6 V drop
doesn't matter), or put a fuse in series and a diode across the input
after the fuse, if you can tolerate an interruption if the polarity is
reversed.

 

[Homer J Simpson]

To the OP: The best solution is to make it impossible to connect the
supply the wrong way, by using a non-reversible connector. If you can't do
that, you can either connect a diode in series (if the 0.6 V drop doesn't
matter), or put a fuse in series and a diode across the input after the
fuse, if you can tolerate an interruption if the polarity is reversed.

Or use a barrier (back to back zeners across the inputs) to protect the
device.

 

[Fred Bloggs]

Fred - his second diagram is just upside down. Don't be too hard on him.

To the OP: The best solution is to make it impossible to connect the
supply the wrong way, by using a non-reversible connector. If you can't
do that, you can either connect a diode in series (if the 0.6 V drop
doesn't matter), or put a fuse in series and a diode across the input
after the fuse, if you can tolerate an interruption if the polarity is
reversed.

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.

 

[Fred Bloggs]

Or use a barrier (back to back zeners across the inputs) to protect the
device.

Back to Electricity I review for you...there are no negative voltages in
the circuit, only negative differentials.

 

[John Woodgate]

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

I quite agree about that, especially when they contradict you! Maybe we
should just not attempt to answer their questions, but send them all
directly to sci.electonics.basics.

 

[Homer J Simpson]

Back to Electricity I review for you...there are no negative voltages in
the circuit, only negative differentials.

Back to Safety I review for you - ever worked in a plant where they make
explosive/flammable products?

If Boeing had put those on the fuel tank wiring TWA Flight 800 probably
wouldn't have exploded.

 

[Fred Bloggs]

Back to Safety I review for you - ever worked in a plant where they make
explosive/flammable products?

There is no source in the circuit that is negative wrt GND. This has
nothing to do with "safety."

If Boeing had put those on the fuel tank wiring TWA Flight 800 probably
wouldn't have exploded.

That damned thing exploded because of four factors: 1) the polymide
insulation of the exterior tank was deteriorated and shorted to tank
fuel gauge circuit, 2) no provisions for preventing O2 entry into the
tank (even to this day) despite extant technology and products for just
this purpose, 3) corrosion product buildup on the tank interior fuel
gauge terminals lowering arc breakdown immunity to several hundred
volts, terminals were exposed, and 4) lengthy runway delay on hot day
required pilot to run A/C for passengers, condenser heat rejection
significantly heated tank exacerbating fuel vapor O2 mixture situation.
Your diodes, unless grossly over-sized, would have sizzled off an
explosion in any case. To this day, the FAA has not required an O2 block
on the center tank ventilation channel.

 

[Homer J Simpson]

Your diodes, unless grossly over-sized, would have sizzled off an
explosion in any case. To this day, the FAA has not required an O2 block
on the center tank ventilation channel.

A proper barrier can sink any excessive voltages:

http://www.manufacturing.net/ple/article/CA431825.html

Intrinsically safe barrier types
Intrinsically safe barriers are used to interface between electrical devices
in a hazardous location, and electrical devices located in the safe area
(associated apparatus). The two types of barriers are passive barriers and
galvanically isolated barriers.

Passive barriers
In the passive intrinsically safe barrier, zener diodes are connected in
parallel with the input, while a resistor and a fuse are connected in series
with it (Fig. 2). Under normal operating conditions, the barrier passes
electrical signals in both directions. It is designed to withstand a fault
voltage of up to 250-Vac.

 

[Fred Bloggs]

A proper barrier can sink any excessive voltages:

http://www.manufacturing.net/ple/article/CA431825.html

Intrinsically safe barrier types
Intrinsically safe barriers are used to interface between electrical devices
in a hazardous location, and electrical devices located in the safe area
(associated apparatus). The two types of barriers are passive barriers and
galvanically isolated barriers.

Passive barriers
In the passive intrinsically safe barrier, zener diodes are connected in
parallel with the input, while a resistor and a fuse are connected in series
with it (Fig. 2). Under normal operating conditions, the barrier passes
electrical signals in both directions. It is designed to withstand a fault
voltage of up to 250-Vac.

Hmmm- that's to protect the electronics from faulting in some hazardous
way. If you located that in the fuel gauge, then when the fuse opens,
the terminals arc. If you located it upstream of the harness short, it
again does no good. Then there was always the possibility of arcing
through deteriorated internal tank harness insulation. The explosion was
an "unanticipated" event in a complex system designed with little to
none fault analysis of this type performed. And the FAA will set no
requirement because the expected cost of the estimated disaster
frequency is less than the cost of retrofit.

 

[Homer J Simpson]

Hmmm- that's to protect the electronics from faulting in some hazardous
way. If you located that in the fuel gauge, then when the fuse opens, the
terminals arc.

Outside of the fuel tank.

If you located it upstream of the harness short, it again does no good.
Then there was always the possibility of arcing through deteriorated
internal tank harness insulation. The explosion was an "unanticipated"
event in a complex system designed with little to none fault analysis of
this type performed. And the FAA will set no requirement because the
expected cost of the estimated disaster frequency is less than the cost of
retrofit.

Unless it's your ass in one of those seats. Hell, they could dump 20 lbs of
crushed dry ice in each unused tank for pennies per flight and enhance
safety, but then they always were a tombstone agency.

 

[John]

I'm just fed-the-@!#*()-up with these people jumping right into
Never contradicted Fred, just had a question because it seemed that
the voltages would be too high for the device he recommended. I did
put a question mark after my statement, not sure if I was correct.


I could have certainly posted my question to sci.electronics.basics
but thought the topic of getting a +/-50V signal down to 0-5V, perhaps
using a diff amp, was a bit beyond the basics....sounds like I was
wrong. Getting voltage dividers right though is certainly EE101.

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[John]

To the OP: The best solution is to make it impossible to connect the
Thanks.
I can't use a polarized connector for this but I think you're right, a
diode in series or use the fuse/diode rev pol protection would be the
simplest, easiest thing to do. Or use Fred's divider circuit.

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[John]

If you're using a simple diode in series with your divider and the
Can't use a polarized connector so I'm leaning towards the inline
diode (I measured a .208V-.212V drop with the 11DQ05's I have) or a
fuse/diode protector circuit.

Thanks!
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