Why is this current source with both sides grounded still supplying current

Question

Now if we write a node equation at \$C\$ we get

\$I_C = g_m − sC_μ V_π \$

But my question is, is it correct, how do we define the current through the branch with \$g_mV_π \$ since both of its terminals are connected to ground now, which means that there is no potential difference between both terminals ?

My initial thought was that since there is a constant current source then the current would be held constant by it. But then that would mean there is a potential difference between those ends. Now that means potential difference across \$C_μ\$ is not \$V_π \$ since one end is connected to the current source which if working would create potential difference (since current is held constant).

But here the equation suggests that the current source is working and as well as the capacitor \$C_μ\$ is grounded too.

Where are my wrong ? Please point it out. Thank you.

Answer 1

What happens INSIDE a current sink is what allows the zero Ohm link to simply carry the current generated with the internal potential.

In any circuit that you deal with you would have no problem with the concept of current flowing in an ideal zero Ohm wire - having the current provided by a current source should not make any difference.

Imagine a high voltage source and a high value resistance in series in a "black box.
Bring out one high voltage source terminal directly to a contact and the other terminal via the resistor to another terminal.
Connect the two terminals together with a 0.00000000... Ohm link.
Current in the link will be I = V/T = Vhigh/Rhigh.
Potential drop across the link = 0.0000 ... V.
No "laws" have been violated.

For example, here is a hypothetical 1 amp near ideal current source. I chose an internal voltage of 100 million volts and an internal resistance of 100 million ohms. In practice somewhat smaller values are used [ :-) ! ].

Short circuit this CC source with a zero ohm resistor and 1 amp flows in the resistor. This is the equivalent of you example situation. Use a 1 ohm resistor for R2 and current in R2 will be I = V/R
= 100,000,000 / 100,000,001 = 0.99999999 A.
Use a 1000 Ohm resistor for R2 and I will be 0.99999 A
ie the difference between bridging the output with 1000 or 1 or zero Ohms is minimal.
For an ideal current source the difference is zero.

^{simulate this circuit – Schematic created using CircuitLab}

Because, what happens INSIDE the 'black box' is what allows the zero Ohm link to simply carry the current generated by the internal potential.

Note also the difference between a current source and a voltage source.
We are much more familiar with the latter.

A V Volt voltage source

• Develops voltage V across its terminals as Voc when it is open circuited.

• No power is dissipated in this state.

• Shorting the output of an ideal voltage source produces infinite current.

An I ampere current source

• Develops current I at its output as Isc when its output terminals are short circuited.

• No power is dissipated in this state.

• Open circuiting the output of an ideal current source produces infinite voltage at the output.

Our AC mains supply approximates an ideal voltage source.
We open circuit it when it is not in use.
If our AC mains was configured as an ~= ideal current source
we would need to short circuit it when not in use.

Answer 2

You're interpreting the operation of the current source wrongly. For the current source to provide a current of \$g_mV_{\pi}\$ (or any current, for that matter) into a \$0\small \:\Omega\$ load it must have \$0\$ V across it. If it were supplying a load other than \$0\small \:\Omega\$, it would have a voltage other than \$0\$ V across it.

Your equation is wrong, btw, there are brackets missing.

Answer 3

You are making an incorrect assumption. Specifically, you are assuming that if there is current flowing through the current source then there must be a voltage across that source. This is not correct. The voltage across a current source is not determined by the current flowing through it; the voltage across a current source is determined by the rest of the circuit.

In this case the voltage across the current source is 0V even though the current may be nonzero.