I'm just starting to learn about electronics by working through Charles Platt's Make: Electronics book and I'm on experiment 14, which is a relaxation oscillator for an LED. I should note that in the schematic below, I wasn't able to figure out how to add a PUT (2n6027), so 'D4' is in fact a PUT.
When I increase R3, my intuition tells me that since I'm increasing the capacitor time constant, the capacitor should take longer to charge and thus I should see a longer time in between flashes. In practice, exactly the opposite happens. As I increase R3, the LED flashes more and more frequently. If I put in a 10K resistor for R3, the LED flashes so quickly that it appears to be staying on continuously (or the voltage across the anode of the PUT stays continuously above the gate voltage for some reason). Why does the frequency of the LED flashes increase, rather than decrease, when I increase R3?
simulate this circuit – Schematic created using CircuitLab
Best Answer
The basic operation can be seen more clearly if you first assume that C1 is shorted at time = 0. In that case you have a simple voltage divider created by R1 and R3, so increasing the value of R3 just increases the initial voltage at the anode connection. C1 then begins to charge and the gate trigger point (set by R2,R4) is reached much quicker.
Then because the charge time was very short only a small amount of charge accumulated on C1, so the discharge time (also the LED on time) is also very short.
Placing a diode in parallel with R3 may give you the result you expected (with cathode to C1). In that case R3 is nearly out of the circuit during C1 charging, but during C1 discharge R3 would extend the on time.