Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] That's What it's Called, Voltage Sensor?

Status
Not open for further replies.

~analoger~

Banned
Member level 1
Joined
Jan 4, 2013
Messages
37
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Location
Mars
Visit site
Activity points
0
Hello,

In the following circuit, the open-circuit voltage at node "a" should be Vs, when connected to a voltage-follower, the open circuit voltage at Vo = Va = Vs? Hence ignoring the source resistance Rs, which could be internal resistance. This way the voltage-follower eliminates the internal resistance of the voltage source and giving the voltage of an ideal source. But the OpAmp itself has output resistance which acts as internal resistance when the circuit is closed at the OpAmp output and the voltage is measured across the connected load.

Now my questions, theoretically an ideal OpAmp can be used to convert a non-ideal voltage source to ideal, and practically if we have an OpAmp with a lower output resistance than the source internal resistance, it can make the source more ideal?

How does it sense and copy the voltage of the source as shown in the picture?

Thanks.

circ1.jpg
 

hmm you start this a different way but yes you could see the purpose of an op amp to be used as creating a more idealistic voltage source.
basically the connection from the output to its inverting input creates a feedback loop. this feedback loop is a negative feedback, meaning it will try as best it is capable at self balancing.
I am not sure how much you care to go inside the op amp but basically the innards steer current based on the input voltages, if output starts off at ground and you apply a pos voltage on the input, you it will cause the output to go up, since the output is connected to the other input, that input goes up and this feedback actively occurs until the point that again to the best of the amplifier's capability both inputs are equal and the loop has settled. so V- is now equal to V+, and since v- = vout, vout=v+.

now about the term voltage sensor. It is more commonly referred to as a voltage buffer. Because as you said with a low output impedance it behaves more like an ideal source. This means that loading the output wont affect the signal itself (again limited tot he amplifier's capabilities). Also the amplifiers input is very high impedance, so that it doesnt load the original signal. because of the high input impedance and low output impedance it is commonly used as a buffer to isolate a voltage node from the loading that you need to apply on that voltage node for application you are designing.
-Pb
 

Theoretically we can calculate the voltage at one terminal of an open circuit, like the part marked with green color in the picture. How is it possible to measure it in practice using another circuit without drawing any current from the source? That's why I called this OpAmp configuration a "voltage sensor."
 

It will draw some current, ideally opamp has no infinite resistance, but in life this isnt true, there will be pico amps or more flowing through the gates of the amplifier inputs. I think you should look into how a mosfet works and how it uses charge to create an electric field. In the professional world we do call it a buffer because you are isolating the input nodes from the effects and loading of the output circuitry. If you wish to call it a voltage sense circuit, that is your own prerogative, but confusion will follow in the discussion. There is a circuit called a voltage sense amplifier but it is used in correlation with sram elements for detecting their values.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top