# Current source with grounded Load and floating power supply

The following circuit represents a basic op-amp current source and can be found in the "Art of electronics 3rd edition" in page 228.

$I_{load}$ is easy to calculate in this situation.

The problem with this circuit is that the load is floating (neither side is grounded).

This problem is fixed by connecting the load to GND and with a floating power supply as showed in the following circuit.

I can't understand how $I_{load}$ is calculated here.

I know that any node in the circuit can be used a reference node (ground node) which, i think, implies that the potential of the rest of the nodes are calculated in function of the new ground node.

When i try to analyse the circuit something goes terrible wrong. By the expression of $I_{load}$ (present in the last figure)

$V_{in}= (V_+R_1)/(R_1+R_2)$

But looking at the circuit $V_{in}=v_+=v_-= V_{GND}$

Shouldn't $V_{GND}$ be seen as 0V to the other nodes?

What i am missing?

Thanks

• Vgnd != com. You're free to choose your own reference (as long as you're consistent), so choose com as 0V to make your calculations easier. It doesn't matter that Vgnd != com. – brhans Sep 4 '18 at 16:37

## 2 Answers

A caveat to this circuit is the current does not flow to ground, which is hard to wrap your head around. Because the circuit is floating and power is not sourced from ground (like it normally is) the current returns back to the source through R3. All of the current through the load goes back through R to return to the source. The vgnd or com node is negative relative to the ground.

The ground in this circuit merely is a point at which to analyze the circuit. If you wanted to you could also put this node at 1000V and analyze it, and the current would be the same through the load.

It's much easier to wrap your head around the circuit if you put the 0V node here, as you can easily spot the voltage divider for the load, and the current sense resistor R (or R3 in the pic below):

• I'm analyzing the circuit with the new information. The second circuit e much easier to analyse but its the circuit of the first image i posted. – AmssmA Sep 4 '18 at 17:50
• What I drew is an example of the current, it does not flow into ground, if flows through R, because it flows through R com goes negative. The current flowing away from the ground (and none flowing into ground) is what I was trying to represent. – laptop2d Sep 4 '18 at 17:52
• ok. But as it has a topology similar to my first picture initially i didn't understood your purpose. – AmssmA Sep 4 '18 at 19:35

Here's your 2nd circuit, redrawn:

simulate this circuit – Schematic created using CircuitLab

Note that it is identical in every meaningful way to your 1st circuit. The only difference is that the "ground reference" has been moved and the value of $V_\text{IN}$ is referenced against the floating supply rail rather than this new "load ground reference" that the load itself "sees." It's otherwise the same, exact thing.

In short, all that's happened is that the "load ground reference" itself floats above the supply's common.

• A comment to my own post. I used to be able to shrink down the image by adding an "m" to its URL. While adding an "s" still works to make it really small (too small) and "b" still works to make it a little bit bigger than the "s" does, the "m" no longer appears to work in selecting a better, more intermediate size. This has been occurring only recently and my older posts using "m" still look correct to me. It's just new posts that seem to fail here, now. Anyone else experiencing the same problem? – jonk Sep 4 '18 at 19:28
• Discussion of that problem: meta.stackexchange.com/q/314028/159388 meta.stackexchange.com/q/315004/159388 – Kevin Reid Sep 4 '18 at 21:50