# Homework Help: Adding op-amp to a speed gauge

1. Jun 1, 2013

### Femme_physics

1. The problem statement, all variables and given/known data

In a system to regulate speed, a speed gauge is connected to the following circuit

[Broken]

What's the purpose of this circuit? What's the difference between Uout and U*out?

3. The attempt at a solution

I assume that the purpose of this circuit is to help prevent overload, since it's a voltage follower.

The difference between U*out and Uout...well, one just goes to earth and the other one to an unknown source. Is that what they mean?

Last edited by a moderator: May 6, 2017
2. Jun 1, 2013

### rude man

Op amp outut is low impedance, your source is probably high impedance.

If you put a resistive load on your source it will change the voltage. If you do the same to the op amp output it will not.

3. Jun 1, 2013

### Staff: Mentor

Yes, it isolates the load (not shown) from the sensor (so there can be practically no interaction of the load with the sensor).

They are both referenced to earth, so there's no difference in that respect.

4. Jun 1, 2013

### Femme_physics

Then how come Uout has the ground to it at its other end, whereas U*out does not? I assume U*out is where the rest of the system is... Like a controller and such

5. Jun 2, 2013

### collinsmark

I have to agree with NascentOxygen. In circuits like this it is assumed that the potential of the output (traditionally to the right) is in reference to Earth (ground). (Which in this case is U*out).

So why is Uout explicitly shown in reference to Earth (ground)? Because Uout is not an open ended output to the right as is the convention for most outputs. Without the Earth symbol, Uout would just be a node or dot on a wire and the reader student would wonder, "what the heck is that supposed to mean?"

I agree that the author could have removed any possible ambiguity by also placing a similar Earth (ground) symbol under U*out. But even in the form it is now its mostly understood that U*out is in reference to Earth (ground).

6. Jun 2, 2013

### Staff: Mentor

The speed gauge terminals are [most likely] independent of the OP-AMP circuit, so in order for the OP-AMP to be able to make sense of the speed gauge voltage, one of the speed gauge's pair of terminals is connected to one of the OP-AMP circuit's terminals, viz., ground. This gives a common reference point, and now both the speed gauge and the OP-AMP produce a voltage relative to a common level (designated circuit ground here).

The OP-AMP is shown being powered by a +ve rail and ground, so its output is known to be relative to ground.

7. Jun 3, 2013

### Femme_physics

The logic is clear to me now...makes sense :) Thank you very much

8. Jun 3, 2013

### Staff: Mentor

A term appropriate here is "floating". Consider a battery. It is an example of a floating voltage source. If you wished to measure its terminal voltage, you could connect the red lead from a multimeter to the top terminal of the battery. But with only one lead connected the meter won't register any* voltage. You also need to connect the meter's other lead, the black one, to the other end of the battery before the meter will register a potential difference.

* on closer inspection of the arrangement, one could say the meter could register some tiny tiny voltage when you connect only the red lead to the battery, because what we really have is the black lead from the multimeter also connecting to the battery's negative terminal but via a very very high resistance path through the air!

At this point, FP's sharp mind pounces on the familiar potential divider :surprised arrangement implicit in what I described for the case of only one multimeter lead being connected to the battery.

9. Jun 3, 2013

### marcusl

EDIT: Removed due to error.

Last edited: Jun 3, 2013
10. Jun 3, 2013

### Staff: Mentor

Hmm. You must be seeing a different circuit to the one on my local web cache.

11. Jun 3, 2013

### marcusl

Oops! I must have been looking at it upside down! Sorry, my mistake