Current Sources

hello,

I have an interesting problem for you guys , I hope that you will
advice me on it.
I used the circuit diagram of the current sources given by texas
instrument. The link is given below
http://focus.ti.com/lit/ds/symlink/ina2133.pdf. If this link does not
work then go to www.ti.com and type "INA2133" in the search box and go
to its data sheet.
I used figure# 16 on page#13.

Now, the problem is that If you have two of these "voltage to current
converters" having same gorund and power supplies but input signals are
coming form different channels of the Digital to analog conveter. Now,
If one current source is outputing some charge and other one is
outputing zero charge then the current source who is outputing zero
current will act as a return path for the current source who is
outputing some current. Am I right or wrong?

Regards
john

 

Without consulting the datasheet you mentioned... two quick comments:

1) I assume that you're putting both current-source outputs into the
same load, right?
2) Current sources by 'definition' are high impedance output
devices/circuits. Thus, they should not present a significant load (or
provide a current return path) to a current source in parallel with it.

 

You are wrong. Zero current means zero current (eg an open circuit). Its as
if that output were disconnected from the load.

Lets assume the current sources are called Ia and Ib and they are connected
together to the load. Then the current through the load Il is:

Il = Ia + Ib

If Ib = 0 then

Il = Ia - 0 = Ia

It's as if Ib didn't exist and was disconnected.

Try another example... If Ia were positive and Ib were negative then you
could say that Ib is sinking some of the current provided by Ia.

Colin

 

Hello,

Thanks very much! Would you please advice me that If "Ia" is 1A then
what should be the "Ib" to sink the total "Ia". How can I calculate it?
I also would like to mention that the two current sources is providing
charge to RAT's different tissues. so, the tissue is providing some
kind of impedance between the outputs of the current sources.

Thanks
John

 

In any circuit if you look at a _node_ the sum of all the current going
into/outof the node always equals zero.

So your node _as_described_previously_ had three branches so...

Ia + Ib +Il = 0

Now substitute

Ia = 1A
Ib = ?
Il = zero

so substitute and re arrange and if Ia = 1A then Ib -1A.

Simple really all the current going into the node comes out again and none
goes through the load.

But then you said...

BOING... that changes everything. So the load is actually between Ia and Ib
and the above analysis is junk.

The circuit as you describe "does not compute" because you can't have two
different current sources in series. (If you connect a garden hose in series
with another one the water rate in each has to be the same.) In practice you
need to model the load with THREE elements.....

1) an unknown impedance connecting it to ground on the output Ia and
2) an unknown impedance connecting it to ground on the output Ib and
3) an unknown impedance between Ia and Ib.

Any ideas how these compare?

 

Hello,

The impedance between the current source#1 and ground will be around
50Kohm and similar impedance between current source#2 and ground.
Ground is common between the sources. It means two different loads
having same grounds with two different current sources.

Now, what would you suggest,inorder to get one current source absorbing
and the other delivering charge.

Thanks
Regards
John

 

Ok here's how I would do the analysis...

1) Since the load to ground is the same at each end (a and b) I would keep
the whole thing symetrical and make Ib = -Ia.

2) Now if you were to measure the _voltage_ at either end of the load (eg
"a" and "b") wrt ground the value would be the same but one would be +ve and
the other -ve.

3) Now imagine the load as a resistor with the +ve voltage at one end and
a -ve at the other. The voltage at the _middle_ of the resistor/ load would
be roughly 0V. Lets call it exactly 0V or Ground.

4) This allows you (For the purpose of doing the analysis only) to draw two
equivalent circuit diagrams each representing half the original. In this
equivalent circuit - at point "a" you have 50K to ground in parallel with
_half_ the value of the load.

5) Now look at the current flowing into point "a". The current going in is
Ia. The sum of the currents is zero but how much goes through the load and
how much through the 50K? The answer is a simple ratio.. Iload =
Rload/(Rload+50K.). Now you know the load current in terms of Ia. Since
Ib=-Ia that's the problem solved.

As an exercise when you get bored...

Imagine a cube with a 1R resistor fitted along every _edge_ and soldered
together at all 8 corners. What would be the resistance between measured
between any two diagonally opposite corners of the cube? The method used to
solve this problem is the same as used above.

 

Hello,
_____________Va Vb _____________
| | | |
| | | |
| | | |
| 50kOhm 50KOhm v - 2A
^2A
| | | |
|____________|___________|_____________|
|
|
_____
___ Gnd
-
I have following circuit in my mind and I did not understand the steps
4 and 5 you adviced me in the last email. Two, current sources
connected in parallel supplying current to two different loads having
same ground. Va and Vb are the voltages across the loads ( 50KOhm).
Now, would the -2A current source become the return path for the 2A
current source. I hope ypu will forgive my drawing.
Thanks
Regards
John

 

Hello,
_____________Va Vb _____________
| | | |
| | | |
| | | |
| 50kOhm 50KOhm v - 2A
^2A
| | | |
|____________|___________|_____________|
|
|
_____
___ Gnd
-
I have following circuit in my mind and I did not understand the steps
4 and 5 you adviced me in the last email. Two, current sources
connected in parallel supplying current to two different loads having
same ground. Va and Vb are the voltages across the loads ( 50KOhm).
Now, would the -2A current source become the return path for the 2A
current source. I hope ypu will forgive my drawing.
Thanks
Regards
John

 

Hello,
_____________Va Vb _____________
| | | |
| | | |
| | | |
| 50kOhm 50KOhm v - 2A
^2A
| | | |
|____________|___________|_____________|
|
|
_____
___ Gnd
-
I have following circuit in my mind and I did not understand the steps
4 and 5 you adviced me in the last email. Two, current sources
connected in parallel supplying current to two different loads having
same ground. Va and Vb are the voltages across the loads ( 50KOhm).
Now, would the -2A current source become the return path for the 2A
current source. I hope ypu will forgive my drawing.
Thanks
Regards
John

 

According to your drawing above, the current sources _are not_ in
parallel.

What you show is two independent circuits, each one a current source
and load, that just happen to have their ground nodes connected
together. If your drawing is correct, there will be no current in the
segment of the ground node between the two loads, and neither current
source will have any effect on the other.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
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Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

He missed out the load between Va and Vb and the fact that he appears to be
trying to get Va = -Vb (or at least that's my understanding)

I'll try and explain it with diagrams....


Note if Ia = -Ib then Vb = -Va so...


_____________Va___Load___-Va_____________
| | | |
| | | |
| | | |
2A 50kOhm 50KOhm -2A
| | | |
|____________|____________|_____________|
|
|
_____
___ Gnd
-

and this is the same as...


_____________Va___0.5 _x__ 0.5___-Va_____________
| | Load Load | |
| | | |
| | | |
2A 50kOhm 50KOhm -2A
| | | |
|____________|_____________________|_____________|
|
|
_____
___ Gnd
-

Note that the voltage at x = Ov

So you can re draw the circuit like this

_____________Va_________
| | |
| | |
| | |
2A 50kOhm 0.5 Load
| | |
|____________|__________|
|
|
_____
___ Gnd
-

Now it's easy to calculate the current through the load.

 

In effect yes. If you need to know the current through the load (Rat!) then
here is how you do the sums...

I'll try and explain it with diagrams (change to fixed font if necessary)


Note if Ia = -Ib then Vb = -Va so...


_____________Va___Load___-Va_____________
| | | |
| | | |
| | | |
2A 50kOhm 50KOhm -2A
| | | |
|____________|____________|_____________|
|
|
_____
___ Gnd
-

and this is the same as...


_____________Va___0.5 _x__ 0.5___-Va_____________
| | Load Load | |
| | | |
| | | |
2A 50kOhm 50KOhm -2A
| | | |
|____________|_____________________|_____________|
|
|
_____
___ Gnd
-

Note that the voltage at x = Ov

So you can re draw the circuit like this

_____________Va_________
| | |
| | |
| | |
2A 50kOhm 0.5 Load
| | |
|____________|__________|
|
|
_____
___ Gnd
-

Now it's easy to calculate the current through the load.