The rotor of the motor distributor has small ferrite rods imbedded. Those have to be detected when they pass the very small (11mm high E ferrite) transformator. One of the windings on it is the L3 from the diagram, generating 600mH 12V sinus. Another winding detects the inducted signal. Detection rate can be up to 10.000 times X 6 (for 6 cylinders)/minute. Due to the winding of the transformator the signal at the pick up coil's ampitude becomes lower (1Volt) when the ferrite passes. This is detected by another circuit. So it is important that the peak amplitudes don't change a lot due to temperature since this will interfere with the passing of the ferrite rod.But you're pulling out the voltage, not the current!
In this case, it is necessary to stabilize using feedback on the current. Also it will reduce the maximum power! If you want to stabilize the magnitude of the magnetic field, then it is possible, but this other generator circuit will be different. And I do not understand, your attachment to the transformer.
1. You mean L3?; how small?I do not know of a simple way to stabilize the current.Only complex ways come to mind.I present the circuit as follows:
1. a small resistor connected in series with the inductance (this is a current-voltage converter);
2. Amplifier on the operational amplifier;
3. Peak detector;
4. The filter;
5. Scheme of adjustment of the oscillator current;
6. Buffer amplifier.
Yes the inductance changes when the ferrite rods comes nearIs it possible to understand the mechanism as follows: The ferrite rods are periodically approaching the inductor, which causes a change in inductance?
Do you want to get the modulation frequency and control the rotation speed?
Until I understand what you want, I can not help you.But the current stabilization circuit will not be easy!
PS is with LDO regulator at 8V.Generally, the voltage instability may have at least three reasons.
First is instability of PS itself. Remedy - 78XX.
Second is thermal point of set instability. Bordoyunov is solved that brilliantly.
Third is - what the generation regime You are looking for? If mild, it is principially instable yet has good (low harmonics) form factor. Choosing a harder loop factor You get a hard generation regime what gives a stable voltage but rather bad form factor. Thus, its always the compromis between both mentioned, quality and quantity.
Sounds interesting. Resistance of the coil is 1.2 ohms. I possibly have to adjust the inductance, so it would be interesting to have the formulas used. Many thanks for yr help!I need to know the inductance resistance of 60 μH, or even better the Q-factor at some frequency.This is required to determine an additional, series resistor with this inductance, and the value of the current through the inductance, which must be stabilized.A small resistor is needed to isolate a voltage proportional to the current through the inductance.Further, this voltage will be amplified, straightened and filtered.The resulting voltage is used to change the mode (and transconductance of the transistor) of the oscillator.
Looks very nice, very stable current I(L1) in #36 ... but since I am not an expert as you apparently are.... , can you explain yr voltage source B1 and V(I)?… and the functioning of yr current stabilization. Thanks fr yr time and effort!
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by Jake Hertz