I am currently learning electronic on my own. And I have a question about the circuit below.

Given: 2N6027 is a programmable unijunction transistors. (I'm not sure if that is the right schematic for it)

Question: The objective here is to make the LED oscillates on and off. I have tried all of the three circuits below, but only Circuit 1 works. Why circuit 2 does not work? I believe the resistors R2 and R3 in circuit 1 are connected in series anyway right?

Why does the PUT's gate have to be in between R2 and R3?


simulate this circuit – Schematic created using CircuitLab

Below are the pictures of the Circuit 1

http://examples.oreilly.com/9780596153755-files/mkel_02/mkel_02_098.pdf http://examples.oreilly.com/9780596153755-files/mkel_02/mkel_02_103.pdf

  • 1
    \$\begingroup\$ The 2N6027 has three leads. Your schematics make it look like it has four. Indeed, your symbol isn't right. The favored symbol these days for a unijunction is an encircled diode with a third line entering the anode at an angle, such as shown about halfway down the page allaboutcircuits.com/vol_3/chpt_7/8.html This article starts off with another symbol, like a JFET with the gate at an angle, for the non-programmable UJT. Good article overall, by the way - go read it and then refine your question. \$\endgroup\$ – DarenW Nov 25 '13 at 4:04

I refer to my previous answers on this exercise:

Make electronics - charles pratt - experiment 11

PUT Base Resistors Question

The resistors aren't exactly in series; they're constructing a voltage divider, putting the PUT's base voltage at $$ 27/(15+27) * 6V $$ This is the programmable part of a PUT which sets the on voltage for it. In circuits 2 and 3 the gate is connected to the negative rail and is therefore at 0V, preventing it from ever turning on.

As in my other answers, PUTs are an unusual near-obsolete part and I wouldn't bother learning about them.


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