As the title explains, I have a schematic of the elecronics in a model-train:
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It's split in three sections like the original (on paper).
At my level of education, I'm not supposed to understand all of it, but because it is included as preparations for a test, it would be nice to roughly understand what happens in the different sections.
My best guess is that the "voltage regulator" is some sort of on/off for the sound generator, when the train has enough voltage from the tracks to start. * Is the sound generator two oscillators? Then what does D2 do? * I recognize the last part as a class A-amp connected to an AB, with feedback(?).
because it is included as preparations for a test, it would be nice to roughly understand what happens in the different sections.
the sound generator, when the train has enough voltage from the tracks to start.
feedback(?).
Hi, Andreas. You're on track in understanding the circuit action.
The bridge rectifier is there to prevent things from being hooked up bass-ackwards and smoking everything.
The first segment (voltage regulator) is a rather poorly conceived method of trying to get the transistor Q1 to turn on (making the input to IC1, pin 1 low) whenever the input voltage from the bridge rectifier exceeds about 6.9V. Since this is a classroom, you should understand that zener diodes don't have a perfect V-I curve -- there is some current if the voltage across the zener is below I(zk). That will result in fairly quick turn-on of Q1 in the real world, far below 6.9V.
The signal from "voltage regulator gates on/off what you correctly perceived to be an oscillator, and its output gates on/off another oscillator. D2 is there to bypass the second 330K resistor, making the node charge-up of the 220nF cap through only R4, where charge-down is through R4 and R5. This should give IC1A ouput (pin 3) a 2/3 duty cycle. It's not a very complicated trick, but it does work.
You're also pretty much right about the last segment of the circuit.
6.9V? is that because of the 6.2V over the zener (D1), and the other 0.7V as Ube / Vbe voltage drop in Q1? What is the purpose of C1 and R2? C1 should get charged very quickly, right? If not, I would have thought it was there to allow quick changes/spikes in the track voltage to pass.. or something.
** Yep - plus allow some drop across the 4.7 kohms.
** The voltage on the rails could raw, rectified AC rather than smooth DC - analysing the schematic is impossible without knowing important details like this.
C2 likely stores charge for long enough to keep the voltage supply to the zener approximately constant.
R2 discharges the C2 down to zero volts when the rail voltage is zero.
D2 affects the waveform of the first oscillator stage - makes it asymmetrical in time.
Thanks for the catch, Phil. You're right, of course -- many controllers just provide raw, rectified AC, which would make this circuit a non-functioning mess for other reasons, too.
I hope the OP got enough out of the circuit explanation to help him with his test.
I don't see the first section as being a voltage regulator. It's a voltage-dependant switch. The threshold would be about 8.3V (two BR1diodes, one Vbe, plus the zener).
The the first oscillator (should we call it a modulator?) will be gated off whenever Q1 turns on. The second oscillator section will always run, but will be gated (modulated) by the first oscillator as long as Q1 is off.
The sound should change when the rails hit about 8.3V. Have you built one of these? Is this what happens?
It is standard practice to quote the AVERAGE value of rectified AC as a DC voltage.
Model trains normally run from a raw, rectified AC or a pulse width modulated DC supply - both of which work better than pure DC for controlling the speed.
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