Author Topic: trying to understand voltage reference circuit  (Read 318 times)

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Offline hsn93

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trying to understand voltage reference circuit
« on: February 25, 2019, 09:31:00 am »
hello, im tring to understand the circuit so i will try to explain what i understand. if i get somthing wrong and you can correct me i appreciate that.

the circuit is for RTD 3 wire configuration from TI (http://www.ti.com/lit/an/snoa481b/snoa481b.pdf)


as much as i understand:
U1D / U1B / U1C are for voltage reference which is concerning this topic.

i don't know about op amps and their types, drift, noise, common noise rejection, ...etc.
its like black box for me so I'm trying to understand with idea of (these are ideal amplifiers).

1- we have 1.25V from LH4140ACH-1.25.
2- I'm going to ignore that this 1.25V is connected to R19 for now.
3- U1D positive input has 1.25V.
4- U1D will try to follow the input so its output will be 2.5V. because R21 = R22
5- we have (the output going to the RTD (R6) with the resistors shown:
   (1636 / (1636 + 634)) * 2.5V = 1.802V is going to positive of U1B

6- U1B is just buffer so its output is 1.802V.
7- i assume that (R15, R16, C1, C2) is just filter and what is going to U1C is just pure DC 1.802V
8- U1C is just buffer

9- at top of R19 we have output of the U1C which is 1.802, and at bottom of R20 we have 1.25V (difference = 0.55V)
10- across R19, R20 is 0.55V so that will be /2.

11- input of U1D = (1.25 + (0.55/2))  = 1.525V

12- go to step 4. except we have bigger voltage

this means it will keep increasing the voltage?  :scared:

 

Online mikerj

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Re: trying to understand voltage reference circuit
« Reply #1 on: February 25, 2019, 10:17:00 am »
U1D / U1B / U1C maintain a constant voltage drop over R18, i.e. it's a constant current source for the RTD. 
 

Offline Damianos

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Re: trying to understand voltage reference circuit
« Reply #2 on: February 25, 2019, 11:41:02 am »
There is a reference on the first circuit presented in the mentioned application note.

Because U1B and U1C are buffers (for DC), the output of the last is equal to the input of the first. Lets call it Vfb (feedback).
At the input of U1D we have VinU1D=(Vref+Vfb)/2.
As the gain of U1D is equal to 2, the output is VoutU1D=Vref+Vfb.
The voltage across R18 is VR18=VoutU1D-Vfb=Vref+Vfb-Vfb=Vref.
So the output current is IOUT=Vref/R18, independent of the output voltage (if the circuit works inside its limits).
 


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