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
    Commented Nov 25, 2013 at 4:04

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.