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How could a frequency be generated using one inductor and no capacitors? The waveform is unimportant. The active component could be transistors, logic gates (including schmitt triggers), or if necessary, op-amps. I would try to hack something together but I wonder if there are any elegant solutions that already exist. I have trawled the net and my books, and not found a thing. Thanks

Edit 1: Circuit 1 below is along the lines of the circuit suggested by Andy aka and Wouter van Ooijen. (Since the 555 timer is basically a schmitt trigger).

Circuit 1 seems like it could work. Assume ideal 5v logic. On power up, assume the schmitt output is 0v, and current flowing through R and L will be zero. The inverter's input will receive 0v. so the output will go high straight away. Then, as current starts flowing through L1 and R1, out of the schmitt's (high) output, the inverter's input will slowly rise. So far so good. When the schmitt's input rises enough, its output drops to 0v. At this point, its input is held at 5v, and as the current through R and L starts to fall, the voltage at the schmitt's input starts to fall. The problem: at this point, although the output of the schmitt is zero volts, current is still being drawn from it. This is because the inductor is acting a bit like a battery; it assumes a voltage that will maintain the current that was previously flowing through it (and R1), which was 5v. So to the schmitt trigger, this is equivalent to connecting the output to a -5V power rail via R1. Would this blow the schmitt trigger? (TTL? CMOS? 555?)

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ how about using a PUT, programmable unijunction transistor? \$\endgroup\$ – Bobbi Bennett Aug 27 '13 at 2:43
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Your circuit can work, but to protect the chip (in case it cannot source current while the output is zero), you would need a diode to conduct the current when the chip output is zero.

schematic

simulate this circuit – Schematic created using CircuitLab

Reducing the inductance or increasing the resistance gets you higher frequency.

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  • \$\begingroup\$ You mean decreasing the resistance, right? \$\endgroup\$ – biggvsdiccvs Feb 27 '16 at 8:36
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    \$\begingroup\$ Increasing. I do not know the formula, but I simulated this in multisim and 0.1H/1k gives 5.8kHz and 0.1H/10k gives 55.3kHz. Inductors work on current, so higher resistance = lower current = less time for it to decrease. An ideal inductor with zero resistance would keep the current constant forever. \$\endgroup\$ – Pentium100 Feb 27 '16 at 10:27
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    \$\begingroup\$ It was a little counterintuitive, but you're right, of course. The time constant is L/R. \$\endgroup\$ – biggvsdiccvs Feb 27 '16 at 13:25
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Using a schmitt trigger inverter with a resistor feeding back to the input and a capacitor from the input to 0V makes a conventional RC oscillator.

I suspect, with a little more care you can use an inductor feeding back to the input with a resistor to 0V too.

You might need diode clamps on the output to both supply rails to prevent back-emfs damaging the schmitt trigger.

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You could try making a H-Bridge driver with transistors to make a square wave, then use the inductor to "round" the current of the output using the inductor.

Through the power of more beer I got another Idea! You could put the inductor in series with you load and have a circuit periodically charge it and then allow it to discharge through your load.

schematic

simulate this circuit – Schematic created using CircuitLab

Please note the switch is a place-holder, you would need something capable of switching very quickly.

If you had the "switch" toggle around when the inductor is fully charged (There is no impedance) and then toggle again when it is discharged (there is no current) you could get a interesting wave moving through your load.

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  • \$\begingroup\$ If you have to manually flip the switch back and forth, how is this an oscillator? I think you missed the point of the question. \$\endgroup\$ – Dave Tweed Aug 27 '13 at 2:00
  • \$\begingroup\$ The switch is meant as a place holder for ease of concept. I'll add that in. \$\endgroup\$ – gr0undsk33per Aug 27 '13 at 2:06
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Check the famous 555 chip for an RC oscillator: the capacitor is charged via a resistor. When the capacitor voltage reaches a fixed threshold (2/3 of power supply) a flipflop is set that activates a discharge transistor. This transistor discharges the capacitor, until its voltage is below 1/3 of the supply voltage. Then the flipflop resets and the capacitor starts to charge again.

You want an inductor to take the role of the capacitor. The rule is that when you want that, you must swap voltage and current in the description. So it becomes:

A voltage is applied to a coil. When the current through the coil reaches a set maximum a flipflop activates a discharge path for the coil. When the current reaches a set minimum the flipflop resets, the discharge path is deactivated and the cycle repeats.

You won't find much LR oscillator circuits that are advertised as such, because inductors are more troublesome components than capacitors (more expensive, more bulky, higher tolerances, etc). One area where you might find such designs is very compact switched-mode power supplies, because they must include an inductor anyway. But even those are often driven by an RC oscillator.

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I have built a few LR Oscillators .They use an inductor and a current comparitor with hysteresis instead of a capacitor and a voltage comparitor with hysteresis .This is not normally done because caps are generally more ideal ,easier to buy ,have closer tolerances ,and are smaller than coils. The LR osc can be looked at as a dual of the more familiar RC osc .Colleges tend not to teach them despite thier validity .However if the inductor is already present perhaps in the form of a relay coil ,solenoid or motor or even a buck converter choke then ...Why not .I have used this scheme as a basis for relay coil power saving where the initial duty cycle is 100% for relay pull in and 50% for hold .

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