Load cell and wheatstone bridge

J

James Dill

Guest
Need therory help.!!!!!

I work with load cells to weigh the weight in hoppers. These load cells are
wheatstone bridge-type strain gauges. I can find the formulas to work out
the resistence and voltage drops of the gauges. But, when you parallel the
load cells, the voltage output is addative or as the scale book says "summed
together". This is where the problem lies...... I have never run into the
situation where parallel voltages are added togetrher in DC this only
happens in series circuits like voltage dividers. Can someone tell me how
or where to find the information about summing parallel voltages together.
Thanks for any suggestions to this problem.

P. S. I have Googled this problem and come up with a blank.
 
On Thu, 28 Sep 2006 20:25:18 -0500, "James Dill"
<jamesedill@cox.net> wrote:

Need therory help.!!!!!

I work with load cells to weigh the weight in hoppers. These load cells are
wheatstone bridge-type strain gauges. I can find the formulas to work out
the resistence and voltage drops of the gauges. But, when you parallel the
load cells, the voltage output is addative or as the scale book says "summed
together". This is where the problem lies...... I have never run into the
situation where parallel voltages are added togetrher in DC this only
happens in series circuits like voltage dividers. Can someone tell me how
or where to find the information about summing parallel voltages together.
Thanks for any suggestions to this problem.

P. S. I have Googled this problem and come up with a blank.
Click to expand...
---
When you say 'parallel' are you sure the book isn't talking about
the load cells being in parallel mechanically?

--
John Fields
Professional Circuit Designer
 
These are electrically in parallel not mechanically. I built two Wheatstone
Bridge circuits out of resisters. After applying voltage to one of these, I
had 15 MV output. I then soldiered the second bridge on top of the first
bridge and I got a 31 MV output. I then soldiered 4 resisters between the
bridges to make a box like circuit. I was looking for some kind of strange
recirculating curcuit. I came up blank.

I know the two bridges in parallel double the output voltage or summed as
the scale books say but I have no ideal why.


"John Fields" <jfields@austininstruments.com> wrote in message
news:picqh25ahuhu4kh2i49nqp6kap5i4csf1g@4ax.com...
On Thu, 28 Sep 2006 20:25:18 -0500, "James Dill"
jamesedill@cox.net> wrote:

Need therory help.!!!!!

I work with load cells to weigh the weight in hoppers. These load cells
are
wheatstone bridge-type strain gauges. I can find the formulas to work out
the resistence and voltage drops of the gauges. But, when you parallel
the
load cells, the voltage output is addative or as the scale book says
"summed
together". This is where the problem lies...... I have never run into the
situation where parallel voltages are added togetrher in DC this only
happens in series circuits like voltage dividers. Can someone tell me how
or where to find the information about summing parallel voltages together.
Thanks for any suggestions to this problem.

P. S. I have Googled this problem and come up with a blank.

---
When you say 'parallel' are you sure the book isn't talking about
the load cells being in parallel mechanically?

--
John Fields
Professional Circuit Designer
Click to expand...
 
On Sat, 30 Sep 2006 20:34:32 -0500, "James Dill"
<jamesedill@cox.net> wrote:

These are electrically in parallel not mechanically. I built two Wheatstone
Bridge circuits out of resisters. After applying voltage to one of these, I
had 15 MV output. I then soldiered the second bridge on top of the first
bridge and I got a 31 MV output. I then soldiered 4 resisters between the
bridges to make a box like circuit. I was looking for some kind of strange
recirculating curcuit. I came up blank.

I know the two bridges in parallel double the output voltage or summed as
the scale books say but I have no ideal why.
Click to expand...
---
Please bottom post.

So what you're saying is that when you do this: (View in Courier)


+V>-----------+-------+
| |
[R1] [R3]
| |
+-------|------>E1
| |
| +------>E2
| |
[R2] [R4]
| |
GND>----------+-------+


E1-E2 = 15mV,


and when you do this:

+V>-----------+-------+-----------+-------+
| | | |
[R1] [R3] [R5] [R7]
| | | |
+-------|-----------+-------|----->E1
| | | |
| +-----------|-------+----->E2
| | | |
[R2] [R4] [R6] [R8]
| | | |
GND>----------+-------+-----------+-------+


E1-E2 = 31mV?


If that's true, the only way that can happen is if the resistors in
the bridges aren't proportional. That is,


R1 R5
---- =/= ----
R2 R6

or

R3 R7
---- =/= ----
R4 R8

or both.

If we set up a single bridge to give us a 15mV differential across
the outputs, one solution would be:

+1V>----------+-------+
E1 |R1 |R3
[493] [507]
| | E2
+-------|------>0.507V
| |
| +------>0.493V
| | E3
[507] [493]
|R2 |R4
GND>----------+-------+



Where:

E1 R2 1V * 507R
E2 = ------- = ------------- 0.507 volts
R1+R2 493R + 507R

Similarly,

E1 R4 1V * 493R
E3 = ------- = ------------- 0.493 volts,
R3+R4 507R + 493R


and

E2 - E3 = 0.507V - 0.493V = 0.014V


Now, if you connect two identical bridges in parallel:


+1V>----------+-------+-----------+-------+
|R1 |R3 |R5 |R7
[493] [507] [493] [507]
| | | |
+-------|-----------+-------|----->E1
| | | |
| +-----------|-------+----->E2
| | | |
[507] [493] [507] [493]
|R2 |R4 |R6 |R8
GND>----------+-------+-----------+-------+

R1 will be in parallel with R5, R2||R6, R3||R7, and R4||R8,

so the total resistance two 493 ohm resistors in parallel will be:


R1 * R5 493R * 493R
Rt = ---------- = ------------- = 246.5 ohms
R1 + R5 493R + 493R


similarly, the total resistance of two 507 ohm resistors in parallel
will be 253.5 ohms.

Since the two bridges in parallel can be simplified to a single
bridge with the corresponding resitors in the legs in parallel, we
wind up with:


+1V>----------+-------+
E1 |R1 |R3
[246.5] [253.5]
| | E2
+-------|------>0.507V
| |
| +------>0.493V
| | E3
[253.5] [246.5]
|R2 |R4
GND>----------+-------+


Solving for E2, we have:


E1 R2 1V * 253.5R
E2 = ------- = ----------------- = 0.507 volts
R1+R2 246.5R + 253.5R


and for E3:


E1 R4 1V * 246.5R
E2 = ------- = ----------------- = 0.493 volts
R3+R4 253.5R * 246.5R


Therefore, the voltage across the outputs of the bridge, as earlier,
is:


E2 - E3 = 0.507V - 0.493V = 0.014V


Consequently, the difference in the readings you're experiencing
must be due to the tolerances of the resistors you're using, or it
may have something to do with the extra resistors you _soldered_
into the circuit. There's no telling unless you post a schematic
and specify what you're using for resistors and what you're using to
measure the output voltage from the bridge.



--
John Fields
Professional Circuit Designer
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:h2fvh2l0qongh8bsedk6ennhn1ra05o0al@4ax.com...
On Sat, 30 Sep 2006 20:34:32 -0500, "James Dill"
jamesedill@cox.net> wrote:

These are electrically in parallel not mechanically. I built two
Wheatstone
Bridge circuits out of resisters. After applying voltage to one of these,
I
had 15 MV output. I then soldiered the second bridge on top of the first
bridge and I got a 31 MV output. I then soldiered 4 resisters between the
bridges to make a box like circuit. I was looking for some kind of
strange
recirculating curcuit. I came up blank.

I know the two bridges in parallel double the output voltage or summed as
the scale books say but I have no ideal why.

---
Please bottom post.

So what you're saying is that when you do this: (View in Courier)


+V>-----------+-------+
| |
[R1] [R3]
| |
+-------|------>E1
| |
| +------>E2
| |
[R2] [R4]
| |
GND>----------+-------+


E1-E2 = 15mV,


and when you do this:

+V>-----------+-------+-----------+-------+
| | | |
[R1] [R3] [R5] [R7]
| | | |
+-------|-----------+-------|----->E1
| | | |
| +-----------|-------+----->E2
| | | |
[R2] [R4] [R6] [R8]
| | | |
GND>----------+-------+-----------+-------+


E1-E2 = 31mV?


If that's true, the only way that can happen is if the resistors in
the bridges aren't proportional. That is,


R1 R5
---- =/= ----
R2 R6

or

R3 R7
---- =/= ----
R4 R8

or both.

If we set up a single bridge to give us a 15mV differential across
the outputs, one solution would be:

+1V>----------+-------+
E1 |R1 |R3
[493] [507]
| | E2
+-------|------>0.507V
| |
| +------>0.493V
| | E3
[507] [493]
|R2 |R4
GND>----------+-------+



Where:

E1 R2 1V * 507R
E2 = ------- = ------------- 0.507 volts
R1+R2 493R + 507R

Similarly,

E1 R4 1V * 493R
E3 = ------- = ------------- 0.493 volts,
R3+R4 507R + 493R


and

E2 - E3 = 0.507V - 0.493V = 0.014V


Now, if you connect two identical bridges in parallel:


+1V>----------+-------+-----------+-------+
|R1 |R3 |R5 |R7
[493] [507] [493] [507]
| | | |
+-------|-----------+-------|----->E1
| | | |
| +-----------|-------+----->E2
| | | |
[507] [493] [507] [493]
|R2 |R4 |R6 |R8
GND>----------+-------+-----------+-------+

R1 will be in parallel with R5, R2||R6, R3||R7, and R4||R8,

so the total resistance two 493 ohm resistors in parallel will be:


R1 * R5 493R * 493R
Rt = ---------- = ------------- = 246.5 ohms
R1 + R5 493R + 493R


similarly, the total resistance of two 507 ohm resistors in parallel
will be 253.5 ohms.

Since the two bridges in parallel can be simplified to a single
bridge with the corresponding resitors in the legs in parallel, we
wind up with:


+1V>----------+-------+
E1 |R1 |R3
[246.5] [253.5]
| | E2
+-------|------>0.507V
| |
| +------>0.493V
| | E3
[253.5] [246.5]
|R2 |R4
GND>----------+-------+


Solving for E2, we have:


E1 R2 1V * 253.5R
E2 = ------- = ----------------- = 0.507 volts
R1+R2 246.5R + 253.5R


and for E3:


E1 R4 1V * 246.5R
E2 = ------- = ----------------- = 0.493 volts
R3+R4 253.5R * 246.5R


Therefore, the voltage across the outputs of the bridge, as earlier,
is:


E2 - E3 = 0.507V - 0.493V = 0.014V


Consequently, the difference in the readings you're experiencing
must be due to the tolerances of the resistors you're using, or it
may have something to do with the extra resistors you _soldered_
into the circuit. There's no telling unless you post a schematic
and specify what you're using for resistors and what you're using to
measure the output voltage from the bridge.



--
John Fields
Professional Circuit Designer
Click to expand...
Yes I agree with your above equations but I think I have led you off track.
My experimental circuit was to just prove that the output voltages of the
bridges were additive (yes I used 5% resistors) a balanced bridge should
have 0 output. The two bridges had the same basic current draw and roughly
the same output. When I paralleled them, the current was additive and the
voltage was also????

Please look at the following link. This is similar to the summing box that
I used. There are no active parts. All the connections are in parallel on
the back of the circuit board. We have three load cells in parallel on our
scale. If one of them is removed the other two will continue to sum
weight??

..http://cooperinst.thomasnet.com/item/accessories/xaa-911-load-cell-summing-box/xaa-911?&forward=1





1
 
On Sun, 1 Oct 2006 13:53:49 -0500, "James Dill" <jamesedill@cox.net>
wrote:


Yes I agree with your above equations but I think I have led you off track.
My experimental circuit was to just prove that the output voltages of the
bridges were additive (yes I used 5% resistors) a balanced bridge should
have 0 output. The two bridges had the same basic current draw and roughly
the same output. When I paralleled them, the current was additive and the
voltage was also????

Please look at the following link. This is similar to the summing box that
I used. There are no active parts. All the connections are in parallel on
the back of the circuit board. We have three load cells in parallel on our
scale. If one of them is removed the other two will continue to sum
weight??

http://cooperinst.thomasnet.com/item/accessories/xaa-911-load-cell-summing-box/xaa-911?&forward=11
Click to expand...

---
Without a schematic I wouldn't even venture a guess.


--
John Fields
Professional Circuit Designer
 
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