hi to everyone..

i would like to seek your expertise once more, this time with the voltage to current converter. I have attached a circuit which resembles the voltage to current circuit i am working with now.

i cannot interpret the formula that was used in obtaining the equivalent ampere.


I = (frequency input * 0.0008mA + 4 mA)
whereas

frequency input = Vinput/0.0005

i cannot obtain the or multiplier 0.0008 or even that data 4 mA that ws added to the bove equation.

How they obtained this formula.....
The value of the zener above is 2.495 V with Ik = 10mA.

Is that a reference for the circuit??? How could i obtain the output if that is true...


Additional data:

for frequency ranging 200 - 20000 Hz, the value of current output that could be obtained is 4.16mA - 20 mA, that is for 20kHz, you got a 20 mA. please help me explain the other part of the circuit..... similarly, 200 - 20000 Hz is equivalent to 0.1 to 10 V... thanks, i would appreciate your help...

 

Once I realized that what you have is an F-to-V stage that preceeds the circuit in your schematic, everything fell neatly into place.

The constant value of 4 mA can be calculated by the following route.

Knowing the voltage present at the zener diode's cathode to be 2.496V, I proceeded to use Millman's Theorom (see the tutorial on this website for a detailed description of this powerful circuit analysis tool) to construct the equation for the voltage at the input to the positive terminal of the op-amp. I assumed that the 4 mA current is the constant value of current at the output with 0 volts applied to the Vinput terminal of the schematic you posted.

V(+) = (2.496/10000)/((1/10000)+(1/10000)+(1/976))
solve this equation and you get 0.203823 volts

Looking at the circuit I see the that U2 forms a unity gain voltage follower that contains the JFET in the feedback loop. The JFET is used to implement a unity gain current mirror. That means the current set up in the 5.1K resistor R7 is mirrored in the 5.1K resistor R6.

The current in R6 is calculated by taking the voltage present at U1's positive terminal and dividing it by 5.1K. Since R7 equals R6 then for each milliamp that flows through R7 there is a milliamp flowing through R6. R6 acts as a current to voltage converter so since R6 equals R7 then the voltage across R7 then appears across R6.

U2 is another unity gain voltage follower with Q2 in the feedback loop. Since the opamp does everything it can to make its negative terminal equal to it positive terminal then the voltage dropped across R6 will be equal to the voltage dropped across R5. The current flowing out of Q2's collector will be that voltage divided by R5.

If we start with the voltage that we computed above at the positive terminal of U1, there is no voltage gain in U1 and there is no current or voltage gain in Q1 and there is no voltage gain in U2. So if we divide 0.203823/51 we get 3.9996 mA or 4 mA.

You should be able to use Millman's Theorem together with the range of Vinputs of .1 to 10 volts and you will see that the output current will come to 4.16 mA for 0.1V and 20 mA for 10V.

hgmjr

 

Oh yeah, I almost forgot. You asked about the origin of the 0.0008 mA in the expression for the current.

Once again I resorted to Millman's Theorem for the answer.

If I assume a 1 volt signal applied to the Vinput, Millman's Theorem yields the equation:

V(+) = ((1/10000)+(2.496/10000))/((1/10000)+(1/10000)+(1/976))
Solving this equation, I get the voltage at the positive input to U1 equal to 0.285483 volts. I divide that voltage by the value of R5 to get 0.285483/51 = 5.5977 milliamps.

Next, I subtract 4 milliamps from 5.5977 mA since that is the constant current present at the output and I get 1.5977 mA.

Finally, I multiply 1.5977 mA by the F-to-V conversion factor of 0.0005 and I arrive at the magic number 0.0007988. That can be rounded up to 0.0008 mA.

Let me know if you have any questions.

hgmjr

 

you were great, i'm relly expecting you to answer my question since you give me complete detil of the f to v converter. i have known the theory of millmn, nd i', just going to crefully study all what you've sid..


thanks... you were really of great help...