Re: Question about reporting of the value of a current

On Wednesday 20 May 2015, [email protected] wrote:
> VDS D 0 3.3
> VSUB SUB_V 0 0
> VG G 0 3.3
> .PRINT OP V(G) V(D) I(IDS.MN) I(VDS)
> .OP
> .END
> 
> The result of running this is as follows: 
> #           V(G)       V(D)       I(Ids.MN)  I(VDS)
>   27.        3.3        3.3        0.0023209 -0.0023209
> 
> Why is the result of I(VDS) negative?

Current flowing INTO the first node.
But in this circuit, current flows out.
So it is negative.

For another interesting experiment, try P(VDS).
That too is negative.
Negative power dissipated means it is sourcing power.