Current Sources

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

CWatters wrote:

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 go> I also would like to mention that the two current sources is providing

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

CWatters wrote:

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.

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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.