1
\$\begingroup\$

Hybrid pi high frequency equivalent model

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.

\$\endgroup\$
  • \$\begingroup\$ Imagine a high voltage source and a high value resistance in series in a "black box. Bring out one battery terminal directly to a contact and the other terminal via the resistor to another terminal. Connect the two terminals together with a 0.000000000000000 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. What happens INSIDE the 'black box' is what allows the zero Ohm link to simply carry the current generated with the internal potential. \$\endgroup\$ – Russell McMahon Jul 15 '19 at 12:26
  • \$\begingroup\$ the point about a current source is it can have any terminal voltage, including zero, whatever's needed to get the external currnet to flow and stay constant. Just like a voltage source can support any current, whatever's needed to hold the voltage constant. \$\endgroup\$ – Neil_UK Jul 15 '19 at 13:54
  • \$\begingroup\$ In an ideal circuit, if a current source is shorted out, the voltage will be zero. This does not create any difficulty for the current source. From an analysis perspective, you can remove the current source from the circuit (keeping the short). \$\endgroup\$ – mkeith Jul 16 '19 at 1:34
1
\$\begingroup\$

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.

schematic

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Ok I get it now, the source can provide current even when there is no potential difference. The relationship between voltage and current is true in cases where there is relation, like a resistor. But I didn't get your example though, is the 0 resistance link in parallel with the high resistance. Can you please explain that ? Thank you \$\endgroup\$ – Allen Jul 15 '19 at 13:02
  • \$\begingroup\$ Can you maybe provide a picture of the current source that you suggest in the example ? I don't understand what you refer to as the battery ? \$\endgroup\$ – Allen Jul 15 '19 at 13:26
  • \$\begingroup\$ @Allen See amended and extended answer. \$\endgroup\$ – Russell McMahon Jul 16 '19 at 2:47
  • \$\begingroup\$ Thank you @Russell McMahon I see it now. \$\endgroup\$ – Allen Jul 16 '19 at 5:24
2
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Why would the voltage have to be 0 ? What if there is resistance in parallel with the current source, then there would be a potential difference but still current would be constant. \$\endgroup\$ – Allen Jul 15 '19 at 13:04
  • \$\begingroup\$ The voltage across the current source is whatever it needs to be to drive the stated current through the load, i.e. V=IR. If R=0, then V=0. If the total load is \$R_{tot}= r_o ||R_L\$ then the voltage across the current source would be \$g_mV_{\pi}R_{tot}\$ \$\endgroup\$ – Chu Jul 15 '19 at 13:25
  • \$\begingroup\$ Oh ok you are talking about the case in the question and not in every case \$\endgroup\$ – Allen Jul 15 '19 at 13:26
  • \$\begingroup\$ No, every case. If a current source needs to drive a current through a short circuit across its terminals, the voltage across the source is zero. Ohm's law! V=IR \$\endgroup\$ – Chu Jul 15 '19 at 13:29
  • \$\begingroup\$ ... definition of a current source is that it develops the voltage necessary to supply its nominated current. \$\endgroup\$ – Chu Jul 15 '19 at 13:32
1
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I get it now. I thought there's always a relation between current and voltage. \$\endgroup\$ – Allen Jul 15 '19 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.