3

Can someone help me to find the transfer function of this circuit (i want to have the tension through the capacitor I think) and the step response in order to get the voltage in the function of the time?

I have a comparator (the block to the right with " ??? ") and it compares the tension between V1 (voltage divider) and V2 (tension I try to express with respect to time) and I want to know WHEN V1 = V2.

V1 is obtained by a voltage divider then it's a constant voltage

V2 can be expressed with respect to time but the circuit is too complex for me (Simulink seems to calculate this tension with the diode D7, R12, R10 and maybe R6 and R 100k...). Could someone help me to find the transfer function and also the step function because I can't find the constant of time?

(t= 0.497s when V1=V2) I'm sorry if I made mistakes as I'm not English (and I'm not an engineer either)

Here are the pictures -

The Circuit: enter image description here

V2: enter image description here

And V1 (I can express the voltage divider and find by myself the tension given by Matlab): enter image description here

And at t=0.497 reached (with V3), the voltage is zero as you can see before the diode because of the comparators and the block "NOT" with the MOSFET in the right corner:

enter image description here

Thanks in advance.

winny
  • 13,064
  • 6
  • 46
  • 63
YuGo
  • 31
  • 2
  • Welcome to EE SE! Use the [system identification app](https://www.mathworks.com/help/ident/ug/how-to-estimate-transfer-function-models-in-the-system-identification-app.html) or [tftest](https://www.mathworks.com/help/ident/ref/tfest.html) to estimate a transfer function. – Unknown123 Apr 15 '19 at 19:42
  • Thanks for the system identification app but i can't export my workspace variable (i have a scope which logs the data to a variable in the workspace) because the format is incorrect (not an IDDATA nor IDFRD variable). – YuGo Apr 16 '19 at 09:08
  • So you've got a vector. Then use [tfestimate function](https://www.mathworks.com/help/signal/ref/tfestimate.html) instead. – Unknown123 Apr 16 '19 at 16:55
  • Ok it doesn't work well then is there any chance to get the SYMBOLICAL differential equation of my circuit solved by simulink ? I just want to know which resistance is processed to get the tension V2 – YuGo Apr 16 '19 at 20:55
  • What do you mean by doesn't work well? You can derive the differential equation easily using Inverse Laplace Transform if you have the Transfer Function. See [Transfer Function to Single Differential Equation](https://lpsa.swarthmore.edu/Representations/SysRepTransformations/TF2SDE.html). – Unknown123 Apr 20 '19 at 13:14
  • I hope this isn't a [XY Problem](https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem). You can see possible duplicate question regarding your question in other SE site. See [here](https://stackoverflow.com/questions/18684436/estimate-transfer-function-from-input-output-data) – Unknown123 Apr 20 '19 at 13:15

1 Answers1

1

You cannot find a "transfer function" of this circuit as a whole, because it's NOT linear. Even the part directly involved on the \$V_2\$ generation isn't linear because it contains a diode.

You have to reason otherwise: The circuit generating \$V_2\$ is an envelope detector which, if designed correctly, will obtain the envelope of the \$V_3\$ voltage. The value of that envelope is what it's compared by the comparator with the \$V_1\$ voltage obtained by the voltage divider.

Then it's also the 100K resistor connected between the comparator's output and the positive input, which maybe introduces some sort of hysteresis into the comparison.

If \$V_1 > V_2\$, the comparator's output will be LOW, so the inverter output will go HIGH turning the MOSFET ON. Then, presumabily (the whole circuit isn't visible) a positive voltage vill be applied at the upper input of the first "???" block (which I presume it's another comparator identical to the second one), so perhaps its output will go HIGH generating a step that will be detected with the second comparator, which will turn its output ON, which will turn the MOSFET OFF, deactivating the output of the first comparator, making \$V_1 > V_2\$ again but ONLY AFTER SOME TIME....

My guess is that's some kind of oscillator, but built in an unnecessarily complicated way.

As you see, there are many unknown things in this circuit to be able to be more precise in the answer.

Orestes Mas
  • 253
  • 1
  • 7