How is this oscillator circuit supposed to work?

Question

This answer shows a circuit including a LM311N voltage comparator. Its stated purpose is to be a oscillator, allowing identification of an unknown inductor L1. It has similarities to the Free-Running Multivibrator shown in the datasheet. I expected that the resonance frequency of the tank circuit L1 and C7 would introduce a square wave due to the comparator, thus allowing to calculate L1 due to a known C7.

Trying to understand it, I simulated it in SPICE with a LM311 model from a forum. Furthermore I replaced the LM311 with the LT1011, which claims pin compatibility to the LM111 (~LM311 with extendet ratings). The tank circuit oscillates with a starting peak of 12..20mV until it dies out within less than 200 µs in both simulations. I also tried to change resistor values and introduced series resistances for L and C to start the oscillation, but to no avail.

Building the physiscal circuit on a stripboard provided no useable results when I measured the available potentials with a scope.

How can the circuit be used? What behaviour should it show?

I had problems understanding the circuit, as it connects the collector output straight to Vcc, while the emitter is connected to ground over its internal 4 \$\Omega\$ Resistor.

Answer 1

This is a slightly modified circuit used in many copies of an original circuit from AADE. The designer (the late Neil Heckt) was generous (or naïve) enough to publish all the details including the equations used internally so the design was widely copied, usually without credit, and never with any licensing fees.

Anyway, as others have pointed out, the short to Vcc should be replaced by a resistor, maybe 1K. The LC oscillator operates at the resonant frequency of L1/C7, so:

\$L_1 = \frac{1}{C_7(2\pi f)^2}\$

The original oscillator (from here):

The Oscillator

The key to L/C Meter IIB's operation is the oscillator circuit of FIGURE 1. The LM311 is a voltage comparator. When power is applied, the voltage at pin 2 is 2.5 volts causing the output to be at a level of 5 volts. This charges capacitor C4 through resistor R4 until the voltage at pin 3 equals 2.5 volts. As it reaches 2.5 volts the output switches to a low level inducing a transient into the tank circuit composed of L1 and C1. The transient causes the turned circuit to ring at it's resonant frequency. The ringing causes a square wave at the resonant frequency to appear at the output of the voltage comparitor. The square wave is coupled back to the tuned circuit through R3 and C3 sustaining oscillation.

For the nominal values of L1 (68 uH) and C1 (680 pF) an increase in L of 1 nH (.001 uH) or an increase in C of .01 pF produces a frequency change of slightly more than 5 Hz. A 0.2 second measuring period can resolve 5 Hz and therefore .001 uH or .01 pF.

I don't think the order of magnitude increase in hysteresis from using a 10K resistor for R5 in your circuit is an improvement. It may well cause other problems. Nor is reducing the RC product of R4/C4 an improvement.

Answer 2

You are quite right, the two output pins cannot both be connected to DC supply. With the output emitter pin (1) connected to ground, the collector output pin (7) should not be attached to the supply. But pin 7 should be pulled up to the supply voltage with a small resistor like 1000 ohms perhaps.

Answer 3

It is a relaxation oscillator - formed by R4/C10.

The LC side has no meaningful function.