The circuit schematics

I am trying to tune this circuit, dimensioning the values of resistors Ra,Rb and the capacitor C... The circuit work in this way: a voltage Vin regulates the current flowing in the inductor, via the VCO block. The operational give a Vctrl to NE555 configured as astable oscillator that (adjusting the duty cycle) will impose the right current. This current could be red from the OP Amp through the feedback 2 ohm resistor. I ask you what is the quickest and fastest method to get this circuit work. I followed this procedure: First i calculated the time constant of the inductor on the drain, with L/R formula. Then i chose a frequency far away the frequency of the pole, to get a "low pass effect" of the current in the inductor, about 100kHz. So, I supposed that the op amp is not rail to rail, and then I did a calculation to impose that with Vin = 1V the charge of the capacitor (from Vdd/3 to about 11V, the rail of the OP Amp) would last about 90% of the discharge time, then getting a current of 90% of the maximum current that can flow in the drain (a kind of 12V/22Ω). Then I tried with a different value of Vin (0.1V), and verified if the duty cycle would be reduced.

I know that it looks so confusing, so please i ask you how do you dimension this circuit.


1 Answer 1


The rule for inductors is V = L * di/dt. So, at 100 KHz, when you turn on for the first cycle, you have V of 12 volts across the inductor. With L = 0.1H, di/dt = 120 amps per second. At 100 KHz and 90% duty cycle, your pulse width is 9 usec, so your current is a ramp starting at zero and increasing to 9 usec * 120 a/sec or about 1.08 milliamps peak. At this low first-pulse current, the 2-ohm resistor has negligible effect.

Then the FET opens up, and the voltage on the drain will swing positive to 12.7 volts where the diode starts conducting. Then the inductor will discharge through the 20-ohm resistance, and the current will drop until it either stops conducting (discontinuous mode) or the FET switch turns back on and the current ramps up from that point.

For an ideal inductor, you can calculate a point where the inductor current increase during the "on" part of the cycle charged through the 2-ohm resistor will equal the decrease during the "off" cycle when the current in the inductor discharges through the diode and resistor. The current waveform will have ripple at your switching frequency for this reason.

  • \$\begingroup\$ Yes, but what is the calculation to find out values for resistors and capacitors? \$\endgroup\$
    – Simus994
    Jun 20, 2016 at 18:52
  • \$\begingroup\$ Your circuit operates exactly like a flyback converter, so I would suggest you use a flyback controller to perform your duty cycle / feedback function instead of a 555. \$\endgroup\$ Jun 20, 2016 at 19:06

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