Timeline for Figuring out the principal working of constant current source
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2019 at 20:16 | comment | added | TimWescott | While we're picking this apart, note, too, that unless the IRLZ44Z has exceedingly low gate-source and gate-drain capacitances, that circuit is going to oscillate like mad. | |
| Mar 2, 2019 at 19:26 | answer | added | D.A.S. | timeline score: 1 | |
| Mar 2, 2019 at 18:28 | answer | added | AnalogKid | timeline score: 2 | |
| Mar 2, 2019 at 18:03 | comment | added | Janka | Related: electronics.stackexchange.com/questions/268462/…. | |
| Mar 2, 2019 at 17:54 | comment | added | Jack Creasey | @G36 The OP27G requires at least 4-5V negative supply so the voltage across R2 and R3 must be at least 4V. Only under these conditions would this be a CC configuration. | |
| Mar 2, 2019 at 17:42 | comment | added | G36 | Yes, it's true. But only if RL x I_Load + V_R3 is less than the supply voltage | |
| Mar 2, 2019 at 17:42 | comment | added | Neil_UK | two comments. First - It's best to break this up into 2 parts. (1) Iout depends on voltage across resistor, which depends on voltage at input of amplifier. (2) Voltage at amplifier depends on Vload+ R1 and R2. Normally you'd use something more accurate than 'Vload+' to set the voltage to the opamp. Second - OP27 won't work here, it's an old-style dual-rail not R2R output, so the output won't go down to GND, neither will the input common mode go to ground. Use dual rails, or a -ve amplifier supply of -5v or more, or a modern R2R opamp. | |
| Mar 2, 2019 at 17:37 | comment | added | adfsg | @G36 So, I can write: $$\text{I}_\text{load}=\frac{\text{V}}{\text{R}_3}\cdot\frac{\text{R}_2}{\text{R}_1+\text{R}_2}$$ Is that true? | |
| Mar 2, 2019 at 17:36 | history | edited | adfsg | CC BY-SA 4.0 |
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| Mar 2, 2019 at 17:28 | comment | added | G36 | The voltage drop across R3 will be equal to the voltage divider output voltage. | |
| Mar 2, 2019 at 17:26 | comment | added | adfsg | @G36 So, how does the loadcurrent depend on the resistor values? | |
| Mar 2, 2019 at 17:25 | comment | added | G36 | This circuit work thanks to the negative feedback magic. The opamp simply compares the voltage divider output voltage with the voltage produced by the current flowing through R3 resistor and the MOSFET V_R3 = I_load x R3. And if this voltage is lower than the voltage divider voltage the opamp will turn on the MOSFET harder to make sure that the voltage drop across R3 is equal to the voltage given by the voltage divider. | |
| Mar 2, 2019 at 17:24 | comment | added | adfsg | @Huisman I've have realy no idea (Dutch: en goed om een Nederlander te zien hier :) & English: good to sea another Dutchman over here)! | |
| Mar 2, 2019 at 17:21 | comment | added | Huisman | Assume a fixed 3.3V on the positive input of the OpAmp. Next, assume there is a current of 0A running through R3. Use Ohms law for R3 and consider what happens to the OpAmp and next to the mosfet. Then, assume 2A is running through R3 and consider again what happens to the OpAmp and to the mosfet. Last, for 1A. | |
| Mar 2, 2019 at 17:10 | review | First posts | |||
| Mar 3, 2019 at 3:49 | |||||
| Mar 2, 2019 at 17:07 | history | asked | adfsg | CC BY-SA 4.0 |