Hi,
I have a problem that is DC steady state involving inductors. The problem asks to find Vo and Io.






I simplify the schematic where I find Va with the KCL (10-Va)/10k = Va/10k because in an ideal op amp there is no current entering it. But can current exit an op amp? I take it because Vo = 5V the current through the 5k resistor is 1mA (which equals Io) but how can this be if Vb has 1/2mA exiting into the left branch and 1/2mA entering from the middle branch? Shouldn't Io be zero?

 

Yes, op amp can deliver current to the load. And this current is deliver by an external voltage source. Because every real world op amp need external power supply source. In theory op amp is nothing more then a voltage controlled voltage source. And this is why no any external voltage source is shown in the diagram.
http://www.allaboutcircuits.com/vol_3/chpt_8/2.html

 

Thank you, now I understand that current can exit an op amp. What I still don't get is that if I do KCL at node Vb then there is 1mA entering from the right branch, 1/2mA entering from the middle branch, and 1/2mA exiting toward the left branch. This doesn't add up to be zero. Can you please explain what I might be misunderstanding about this problem?

 

This Io current is not coming from this 10V input voltage source. All this load current (Io) is deliver by op amp internal voltage source (build into op amp).

 

The EXTERNAL voltage source that is powering the opamp is connected to the common of the circuit and the current that the opamp supplies at it's output comes from that supply and returns to that supply. Your diagram doesn't show that other voltage source, and so it is easy to overlook it.

 

According to this link http://www.gadgetgangster.com/tutorials/428 the two hidden terminals of the op amp are ground and a voltage source which goes in line with what you guys drew and wrote.

 

In general, the two hidden nodes are the positive and negative voltage supplies. You usually have both a positive and a negative supply with reference to the circuit's reference node, i.e., "ground". If one of the supply pins (almost always the negative supply) is directly referred to the circuits ground, this you are running the opamp in "single-supply" mode. Some opamps are designed for this, some will tolerate it, and may will misbehave pretty badly without due care being given to its needs.

 

So this means according to jony130's diagram that a voltage follower is an op amp in single supply mode.
Cool, I didn't know about these intricacies of op amps before.
Thank you, WBahn.

 

No, according to jony's circuit, a voltage follower is an active amplifier, not a resistor in a KVL calculation.

 

So this means according to jony130's diagram that a voltage follower is an op amp in single supply mode.
Cool, I didn't know about these intricacies of op amps before.
Thank you, WBahn.

You can make voltage followers without using opamps. Also, there is nothing magical about single supply mode with regard to a particular opamp use. Running an opamp from a single supply allows you the obvious advantage of just having a single supply. But it imposes constraints on your circuit that are tighter than if you run from a dual (or split) supply. Those are practical engineering details that have very little to do with the conceptual operation of the circuit.

When all is said and done an opamp is nothing more than a differential amplifier with bitching high gain (100,000 to 10 million, give or take) -- and a host of practical limitations that have to be respected.