Current amplifier with OpAmp

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
The circuit I have is this:
2.2.png

I' not sure if I'm supposed to add that current source there or if that symbol is just for information purposes!

I also want to ask if the 2 thumbs-up rules also applies in this case! The rules are:
1 - no current flows into any of the input terminals
2 - the OpAmp tries by any means to keep the voltage drop as 0V between the 2 input terminals!
 

MikeML

Joined Oct 2, 2009
5,444
...

I also want to ask if the 2 thumbs-up rules also applies in this case! The rules are:
1 - no current flows into any of the input terminals
...
Add the word "almost" in front of #1

If in doubt, look up the data sheet for your opamp, and check the parameter "input bias current".

Similarly, check the parameter "input offset voltage".

in.gif
 
Last edited:

Jony130

Joined Feb 17, 2009
4,969
I' not sure if I'm supposed to add that current source there or if that symbol is just for information purposes!
Yes you need to add a current source

I also want to ask if the 2 thumbs-up rules also applies in this case! The rules are:
1 - no current flows into any of the input terminals
2 - the OpAmp tries by any means to keep the voltage drop as 0V between the 2 input terminals!
Yes, this two rules hold for this circuit as long as opamp is not in saturation.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
And what about the voltage source Vcc in series with the current source? Do I have to add it too? If so, what is the purpose of having a voltage source and a current source on the same branch?
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Ok, I think the Vcc is not needed as LTSpice current source is able to work without a voltage source to "activate it".

So I have the attached circuit and I had made sure that I_Load = I_R1 + I_R2

Now I need to come up with an expression that gives me I_load in function of I_in. The only thing I can remember of by now is that I_Load = I_R1 + I_R2. But I think this is not what I'm expected to find!

Any tips?
 

Attachments

MikeML

Joined Oct 2, 2009
5,444
Ok, I think the Vcc is not needed as LTSpice current source is able to work without a voltage source to "activate it".
Not just in LTSpice... By definition, a current source will have whatever voltage between its terminals that is required to force the specified current. It makes no difference to the current-source what external voltages are around....

So I have the attached circuit and I had made sure that I_Load = I_R1 + I_R2

Now I need to come up with an expression that gives me I_load in function of I_in. The only thing I can remember of by now is that I_Load = I_R1 + I_R2. But I think this is not what I'm expected to find!

Any tips?
I(Rload) =I(R2) + I(R1); KCL
V(noninv pin) = V(invert pin) = V(output pin); per ideal opamp.
should be enough for you to get started to write an expression for I(Rload) in terms of I(in), R1 and R2...
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Not just in LTSpice... By definition, a current source will have whatever voltage between its terminals that is required to force the specified current. It makes no difference to the current-source what external voltages are around....


I(Rload) =I(R2) + I(R1); KCL
V(noninv pin) = V(invert pin) = V(output pin); per ideal opamp.
should be enough for you to get started to write an expression for I(Rload) in terms of I(in), R1 and R2...

Ok, I'm aware of that about voltages and input terminals and Vout!

Is it possible to come up with an equivalent circuit without the OpAmp?

I've been working with some equations to see if I'm going anywhere with it, but I'm not going anywhere!
I've tried to convert the OpAmp circuit into another circuit taking into account those 2 rules but probably I can't do it!
 

Attachments

MikeML

Joined Oct 2, 2009
5,444
The first two purple equations in aac.png are a good start. You shouldn't care what Vx is, so manipulate the second equation to get rid of Vx. What is left?
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
The first two purple equations in aac.png are a good start. You shouldn't care what Vx is, so manipulate the second equation to get rid of Vx. What is left?
You mean to get rid of vx by simplification or by substitution for something else? I'm not getting anywhere by trying to simplify the equation!
 

MikeML

Joined Oct 2, 2009
5,444
1. Iin*R2 = (Vo-Vx)

2. Ir1*R1 = (Vo-Vx)

3. therefore Iin*R2 = Ir1*R1

4. solve 3 for Ir1:
Ir1= Iin*R2/R1

5. Iload = Iin + Ir1 = Iin + Iin*R2/R1 = Iin*(1+R2/R1)

6. Iload/Iin = (1+R2/R1)

Try your sim for R2= 9*R1

2.2.gif

As load goes to 10KΩ, what power supply would the opamp require?
 

Attachments

Last edited:

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hum, ok... I wasn't close to that solution to the current relationship...

About the OpAmp requirement it would require a 55V supply or it would saturate and those 2 rules wouldn't hold anymore! That's what I think...
 

Jony130

Joined Feb 17, 2009
4,969
Hum, ok... I wasn't close to that solution to the current relationship...
For type of a circuit (if we interest in current) you should write your equation in terms of a current.
For example
IL = Iin + IR1

IR1 = (Vo - Vx)/R1 = [ (Iin*R2 + IL*RL) - IL*RL]/R1 = Iin*R2/R1

IL = Iin + Iin*R2/R1 = Iin * (1 + R2/R1)

About the OpAmp requirement it would require a 55V supply or it would saturate and those 2 rules wouldn't hold anymore! That's what I think...
Yes you right, but do not forget that the real op amp will have some saturation voltage. And this is why Vsup must be larger than this 55V.
http://e2e.ti.com/blogs_/archives/b/thesignal/archive/2012/05/08/op-amp-voltage-ranges-input-and-output-clearing-some-confusion
 
Top