# Voltage Follower with Resistors and Voltage Source

Discussion in 'Homework Help' started by bluetooth tamer, Mar 24, 2015.

1. ### bluetooth tamer Thread Starter New Member

Jan 16, 2015
23
0
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?

File size:
10 KB
Views:
8
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,963
1,098
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.

3. ### bluetooth tamer Thread Starter New Member

Jan 16, 2015
23
0
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?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,963
1,098
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).

Last edited: Mar 24, 2015
bluetooth tamer likes this.
5. ### WBahn Moderator

Mar 31, 2012
17,788
4,808
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.

bluetooth tamer likes this.

Jan 16, 2015
23
0
7. ### WBahn Moderator

Mar 31, 2012
17,788
4,808
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.

bluetooth tamer likes this.
8. ### bluetooth tamer Thread Starter New Member

Jan 16, 2015
23
0
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.

9. ### #12 Expert

Nov 30, 2010
16,355
6,852
No, according to jony's circuit, a voltage follower is an active amplifier, not a resistor in a KVL calculation.

10. ### WBahn Moderator

Mar 31, 2012
17,788
4,808
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.