How is (1/2)V_id expressed in this circuit?

In summary, the conversation discussed the use of two VCVS's in a circuit and the need for a specific voltage level for VID. LTSpice was suggested as a tool for configuring the VCVS's and some tips were given on how to properly simulate the circuit. The conversation also touched on the effect of VCM on both input op-amps and the advantage of using VCVS's to control the differential inputs to the circuit.
  • #1
song youngill
6
0
Homework Statement
instrumentation amplifier in ltspice
Relevant Equations
vin=1/2(vid)+1/2(vcm)
df.jpg

As you can see in the picture, 1/2vid is entered in each of Vin1 and Vin2 on the left. Assuming Vcm is 0, I need to configure this circuit with ltspice, but I am not sure how to configure this vcvs. I'd appreciate it if you could tell me how.
 
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  • #2
Your diagram does not show any Vin - neither Vin1 nor Vin2.
That said, I would guess that Vin1 and Vin2 are the "+" nodes of the top and bottom op amps, respectively.
If that is the case, then neither Vin is ##V_{CM}/2 + V_{ID}/2##.

One other thing. Those diamonds are most commonly a symbol for a current source. But they are labelled with a voltage value (##(1/2) V_{ID}##). So I take them as controlled voltage sources.
 
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  • #3
.Scott said:
Your diagram does not show any Vin - neither Vin1 nor Vin2.
That said, I would guess thaVin1 and Vin2 are the "+" nodes of the top and bottom op amps, respectively.
If that is the case, then neither Vin is ##V_{CM}/2 + V_{ID}/2##.
sorry. vin1 is upper amp's + input and vin2 is lower amp's + input. The does 1/2vid can't be configured as vcvs in ltspice?
 
  • #4
song youngill said:
sorry. vin1 is upper amp's + input and vin2 is lower amp's + input. The does 1/2vid can't be configured as vcvs in ltspice?
Oh, I see. Those diamonds are supposed to be Voltage Controlled Voltage Sources (VCVS's).
I am not familiar with ltspice. Perhaps you would need to actually draw out the voltage divider.
But do you realize that the value you specified before, ##V_{CM}/2 + V_{ID}/2##, is not a voltage level used in this circuit?
 
  • #5
OK. I just checked out LTSpice.
It looks like you need to create two more voltage sources - each ##V_{ID}##. Then use those with "e" components (VCVSs), each with a gain value of 0.5.
That should do the trick.
 
  • #6
.Scott said:
OK. I just checked out LTSpice.
It looks like you need to create two more voltage sources - each ##V_{ID}##. Then use those with "e" components (VCVSs), each with a gain value of 0.5.
That should do the trick.
You can probably replace it with an equivalent independent voltage source. I need to observe the vo value according to the change of the vid, so I hope I can solve the vid.
 
  • #7
Here is a bit more detail.

VID is Voltage, Input, Differential (or differential input voltage)

The simulation circuit shows that the VCM (Common Mode Voltage) affects both input OpAmps.

Tha advantage of using the two VCVS's is you need to vary only one input voltage (independent voltage source) which then controls the two VCVS's and both differential inputs to the circuit.

Have Fun!

Cheers,
Tom
 

FAQ: How is (1/2)V_id expressed in this circuit?

What is (1/2)V_id in this circuit?

(1/2)V_id is the expression for half of the input voltage, V_id, in the circuit. It is used to calculate the voltage drop across certain components in the circuit.

How is (1/2)V_id calculated in this circuit?

(1/2)V_id is calculated by taking half of the input voltage, V_id. This can be done by dividing the input voltage by 2 or multiplying it by 0.5.

Why is (1/2)V_id important in this circuit?

(1/2)V_id is important because it helps to determine the voltage drop across certain components in the circuit. This information is crucial for analyzing the behavior and performance of the circuit.

Is (1/2)V_id the same as V_id/2?

Yes, (1/2)V_id and V_id/2 are equivalent expressions and represent the same value.

Can (1/2)V_id be negative in this circuit?

Yes, (1/2)V_id can be negative in this circuit if the input voltage, V_id, is negative. This means that the voltage drop across certain components will also be negative, indicating a reverse polarity.

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