1
\$\begingroup\$

There is one more question about Experiment 11 from "Make: Electronics" by Charles Platt.
The trouble is that my LED blinks only once, when I connect circuit to power source. (I tried 6V, 9V and finally 12V DC)
As I understand, capacitor should discharge, but it holds the voltage and there is no "charge/discharge" process described in the book that forces LED to blink.

There is the original circuit from the book:

Original circuit

Photo of my circuit is below (yes, it is connected to power source):

Circuit photo

Also I created Fritzing breadboard diagram and marked some points:

Fritzing breadboard diagram

I checked voltage between points displayed on the fritzing diagram (I use 12V DC power source):

  • Vab = 0.04V (after multimeter touch voltage reduces to 0.01V)
  • Vcd = 11.4V (after multimeter touch voltage reduces to 9.47V)
  • Vef = 4.9V
  • Vgh = 8.83V
  • Vij = 0V

Any ideas why LED blinks only once but not constantly?
Finally, one more noob question. As I understand, 'gate voltage' is Vgh. What is then 'anode voltage'? Is it Vcd?

\$\endgroup\$
9
  • \$\begingroup\$ Many breadboards require jumpers to make the strip along the edge continuous. I.e. D and H may not currently be connected to J depending on your breadboard \$\endgroup\$
    – crasic
    Jul 15, 2015 at 2:25
  • \$\begingroup\$ Adding to what crasic said: you also need to connect the two red (Vcc) rails with a jumper, and the two blue (ground) rails with another jumper. I don't know if you've done this or not... \$\endgroup\$
    – bitsmack
    Jul 15, 2015 at 4:20
  • \$\begingroup\$ @crasic This type of breadboard is such that D and H are connected; blue and red rail holes are connected from start to end \$\endgroup\$
    – hard-code
    Jul 15, 2015 at 5:39
  • \$\begingroup\$ @bitsmack Could you, please, explain why do I need to connect Vcc rails (and ground rails) together? I tried to do that, but it didn't help... \$\endgroup\$
    – hard-code
    Jul 15, 2015 at 5:40
  • \$\begingroup\$ @hard-code its not necessary for this simple experiment, but if you are drawing power from both rails you need to make sure both sides are connected together. I did a quick visual check of all the resistor values and the transistor and capacitor orientation and it seems to be in order. As a first trouble shooting step verify with a multimeter on "beep" mode that you have continuity everywhere you expect to, also verify LED orientation and capacitor value (can't see the value code) and transistor (verify all digits/letters in code against datasheet, extra letters can mean different packages) \$\endgroup\$
    – crasic
    Jul 15, 2015 at 5:52

2 Answers 2

Reset to default
1
\$\begingroup\$

You are correct in saying that the anode voltage is Vcd (Anode pin with reference to ground) and Vgh is the gate voltage (Gate pin to ground). That is the simple part.

Now, I cannot say immediately what is wrong. But a few things strike me as odd, so maybe we can work this through together.

If Vab is zero, Vcd must be roughly 12V, this stands true. But, Vef is 5V and Vgh is roughly 9V while there is no current through the LED (I am assuming it was off when you took these measurements). If no current is flowing through the LED, I would think that Vef+Vgh would also amount to 12V.

Now, since Vcd is 12V, this means that the capacitor is fully charged. Which makes sense. Now, we would like this to discharge, but it's not. So I would expect that there is no path to ground in order to discharge. The original schematic doesn't say where the gate or anode should go. Maybe the 2N6027 isn't properly setup?

I would recommend going over the data sheet for the 2N6027 if you haven't already. Looks like it should behaving something like a switch, i.e. Once Vcd reaches a certain value "turn on" and allow discharge, which would illuminate the LED.

Hope this helps. Let me know if you have more questions, I will do my best in being prompt with helping in any way.

Good luck! - Josh

\$\endgroup\$
4
  • \$\begingroup\$ I would also like to mention a couple things you can try as well as switching transistor pins. Try a new transistor, you may have burnt that one. And it looks like the capacitor is polarized. Make sure that the "+" is on the right side. That is all, good luck! \$\endgroup\$
    – Josh Jobin
    Jul 15, 2015 at 2:01
  • \$\begingroup\$ Thanks for answer! The thing I forgot to mention: I use 12 DC adapter plugged into 220V/50Hz public electrical network, it’s output is approximately 14V, so Vef + Vgh = 14V. According to 2N6027 datasheet, anode is on line 17 of breadboard (see Fritzing diagram), gate is on 18, cathode is on 19. I tried several 2N6027 transistors and the result was always the same. I replaced capacitor with 15K resistor and… LED is working. But anode voltage Vcd=4.84V is less than gate voltage Vgh=5.5V! According to definition of PUT, anode voltage should be greater than gate voltage... \$\endgroup\$
    – hard-code
    Jul 15, 2015 at 6:31
  • \$\begingroup\$ Hmmm interesting. Are you in North America (60Hz) or somewhere in Europe (50Hz)? If your adapter says 50/60Hz, it shouldn't matter. But maybe the adapter isn't what you think it is. Also, what voltage does the experiment suggest you use? \$\endgroup\$
    – Josh Jobin
    Jul 15, 2015 at 9:49
  • \$\begingroup\$ I'm from Europe :) Author says that it doesn't matter 6V or 12V, the main idea is difference between anode voltage and gate voltage. Maybe my 2N6027 doesn't correspond to datasheet. \$\endgroup\$
    – hard-code
    Jul 15, 2015 at 10:59
1
\$\begingroup\$

I found the answer here: http://www.electronicspoint.com/threads/problem-with-put-2n6027.262368/. I have the same 2N6027 901 PUT transistor, so the answer is just replace it by 2N6027 610.

\$\endgroup\$

Your Answer

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

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