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Why this curve is going down? Why the current is negative?

audioguru

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#3
The circuit is simply a 9V battery with a 1k ohms load resistor.
The sweep might be the voltage in the battery dropping and the "curve" is the resulting reduction in the current of the circuit.
 

audioguru

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Most Helpful Member
#5
The problems with the images are:
1) Too small, they should be double the size posted.
2) Negative faint grey lines and letters on a black background instead of a normal positive image of black lines and letters on a white background.
3) Saved as a very fuzzy JPG file type instead of as a very clear PNG file type.
 

ronsimpson

Well-Known Member
Most Helpful Member
#6
The circuit is simply a 9V battery with a 1k ohms load resistor.
The sweep might be the voltage in the battery dropping and the "curve" is the resulting reduction in the current of the circuit.
I agree with the AudioGuru. A sweep function takes one (or more) values and shows what happens if that value varies. In this case it looks like the battery is "swept" from 0 to 9 volts. I think this is not the function you want. Sorry I don't know this program.

Many of us use LT Spice. (free and good)
 
Thread starter #7
The problems with the images are:
1) Too small, they should be double the size posted.
2) Negative faint grey lines and letters on a black background instead of a normal positive image of black lines and letters on a white background.
3) Saved as a very fuzzy JPG file type instead of as a very clear PNG file type.
Can you help now?
 

Attachments

alec_t

Well-Known Member
Most Helpful Member
#8
The graph slopes down because the current into V1 is negative and its magnitude increases as V1 voltage increases. Spice shows current direction as into a component.
 

ronsimpson

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Most Helpful Member
#9
You have a window "DC transfer characteristic". I think you want a different option there.
Now you are plotting "V" against I. Where V changes from 0 to 9.

I think you want "DC ?????". I don't know the exact words. DC something. There should be a place for entering time. "0 to 100mS"
 

alec_t

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Most Helpful Member
#11
It isn't a fault that needs 'fixing'. It's a Spice feature.
 

ronsimpson

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Most Helpful Member
#13
How am I supposed to fix this?
I understand you can read this picture but I can't. Where did you answer the question about what program is this?
If you want help don't give out bad ratings.
upload_2018-1-2_12-31-46.jpeg
On LT Spice, and others, you can change "I(V1)" to "-I(V1).
OR
On other spices, you can change "I(V1) to I(R1). OR -I(R1) This way you are plotting the current in the resistor. OR negative current.
If I(R1) plots negative you can try removing the resistor and rotating 180 and putting it back.
 
Thread starter #14
I understand you can read this picture but I can't. Where did you answer the question about what program is this?
If you want help don't give out bad ratings.
View attachment 109920
On LT Spice, and others, you can change "I(V1)" to "-I(V1).
OR
On other spices, you can change "I(V1) to I(R1). OR -I(R1) This way you are plotting the current in the resistor. OR negative current.
If I(R1) plots negative you can try removing the resistor and rotating 180 and putting it back.
Is there any way to run this by code?(This is NGSpice engine.)
 

alec_t

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Most Helpful Member
#16

audioguru

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Most Helpful Member
#17
The red line has two kinks but it should be straight. It is correct at 3mA and 3V across the 1k resistor up to 7mA and 7V across the 1k resistor. Below 3 and above 7 the red line is wrong.
 

MikeMl

Well-Known Member
Most Helpful Member
#18
It is a static circuit. You can only do an .OP "operating point" analysis on it...

71.png

If you want to see what happens if you change V1, then use a .DC sweep. It plots the independent variable V(V1) on the x-axis, and the dependent variable(s) I(R1) on the y-axis.

71a.png

The current through R1 [I(R1)] is positive by convention. If you apply KCL to the circuit, which way does the current through the source V1 have to flow? By convention, when a source is delivering current to a load, it is a negative current. That is the way engineers (and Spice) work...
 
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