I've been making my way through "MAKE: Electronics: Learning Through Discovery", but have gotten stuck on Experiment 11, where I am making an oscillating circuit.

The book calls for a 2.2uF capacitor, but I only have a 1000uF capacitor. I decided it would be fun to try to create a circuit that functions similarly with the parts I have (or at least to understand why doing so would be impossible)

The circuit specified by the book is this: Figure 2.98

R1: 470K Resistor, R2: 15K Resistor, R3: 27K Resistor, C1: 2.2uF electrolytic capacitor, D1: LED, Q1: 2N6027 PUT

The first thing I did was replace R1 with a 6.7K resistor so it wouldn't take so damn long to charge the capacitor. Next I replaced R2 with a 26K resistor and R3 with a 96K resistor so that the PUT would only let charge through when the capacitor was near the peak of its voltage.

I was expecting the LED to turn on once the capacitor charged to ~5v, and turn off once the capacitor discharged to less than ~5v. Instead, the capacitor charges for a few seconds, and the LED remains dimly lit while the capacitor's voltage remains steady at ~2.7v.

With my very limited knowledge of electronics, I am stumped by this behavior. Am I misunderstanding how a capacitor works? Thanks in advance for your expertise!

UPDATE: I still don't exactly understand the relationship between the resistor values and the LED/capacitor getting "stuck" (where getting stuck means the LED will stay lit and the capacitor voltage will stay constant around 2.5v). After some more testing it appears that:

  1. The larger R2 and R3 are (keeping the R2:R3 ratio approximately constant), the more likely it is the LED/cap will get stuck
  2. The smaller R1 is the more likely the LED cap will get stuck.

For example, with R2 at 15K, R3 at 21K, and R1 at 66K, the LED/cap will oscillate properly (although slowly). If I change R1 to 46K the LED/cap gets "stuck"

Does anyone know of an explanation for this behavior?

I believe Mark has the right answer (based on some testing) so I've accepted it. If R1 has much less resistance than R2 and R3, the cap charges much faster than it discharges so that it oscillates quickly while it appears to the multimeter that it is "stuck" at one voltage.

However, I would appreciate if Mark (or anyone else) can explain how to come up with that insight about Rg from the datasheet

Load Line Concept

  • \$\begingroup\$ 6k7, 26k, 21k, 66k,... weird resistor values! \$\endgroup\$ Commented Jun 11, 2011 at 17:13

5 Answers 5


never really played with a PUT before (actually never heard of em) but i was interested and read the datasheet.

It looks like the current through the PUT is dependent on the resistance between gate and ground, which explains why when the cap is feeding the LED it doesn't get really mad about the LED not having a current limiting resistor. In this case the Rg gate resistance is your R3. My guess is that when you moved R3 up to 96k your limiting the current so much that your LED isn't getting to full brightness.

Additionally the low limit of this current combined with a really big cap means your capacitor discharges much slower. Combine this with the very small R1, which charges the cap quickly, and i'm betting you are getting some oscillation, but its happening very, very fast.

Try a larger R1, smaller R3 and whatever sized R2 you need to keep the divider ratio the same. Ideally track down a smaller cap, it would make finding the resistor sizes needed easier.

  • \$\begingroup\$ Very interesting! I will test your hypothesis. Out of curiosity, how did you determine that "the current through the PUT is dependent on the resistance between gate and ground"? Was "Figure 4. Effect of Supply Voltage" in the data sheet? \$\endgroup\$
    – Andrew L
    Commented Oct 10, 2010 at 21:31
  • \$\begingroup\$ Did you mean smaller R3? If so I believe I confirmed your hypothesis (see my update to the question). An R1 of 6K at 6v actually takes a few seconds to charge a cap, but I'm wondering if it is taking very little time to charge it back above the gate voltage immediately after the voltage drops below the gate voltage. \$\endgroup\$
    – Andrew L
    Commented Oct 11, 2010 at 3:50
  • \$\begingroup\$ oops, yea i did, after further reading of the datasheet, the value of Rg = R2*R3 / (R2+R3) so lowering that resistance will increase the peak current, that is the current at the peak of the decay, which lowering R3 and maintaining the same ratio will result in. \$\endgroup\$
    – Mark
    Commented Oct 11, 2010 at 4:34

You're probably not misunderstanding how a capacitor works. It's probably the programmable unijunction transistor that's acting weird.

My understanding is that a PUT stays on as long as the current through it is greater than a certain threshold. Since you decreased R1, I bet the current when the cap is discharged is higher than that threshold, so the PUT never quite turns off.

Try changing R1 back to 470k and see if it works. (It will be a little tedious to test.) Then you can decrease R1 and see how far you can go while still having the PUT turn off.

  • \$\begingroup\$ According to the book, the PUT allows current to flow when the anode (top pin of Q1) has a higher voltage than the gate (middle pin of Q1). In my example, the voltage across the gate should be ~4.7v (I think). So as far as I can understand, the PUT shouldn't allow current through when the voltage across the capacitor is only 2.7v. \$\endgroup\$
    – Andrew L
    Commented Oct 10, 2010 at 19:27
  • 1
    \$\begingroup\$ That's true-- the PUT stays off until the voltage rises above the gate threshold. However, if I understand it correctly, once it turns on, it stays on until the current drops below another threshold. I bet it's turning on right, but turning off wrong. \$\endgroup\$
    – pingswept
    Commented Oct 10, 2010 at 20:35
  • \$\begingroup\$ According to the book - "If anode voltage increases above the threshold point, current bursts through and flows from anode to cathode. If anode voltage drops back down below the threshold, the transistor stops the flow." So although I agree it's not turning off correctly, I'm not convinced it's because of a different threshold for closing the gate. \$\endgroup\$
    – Andrew L
    Commented Oct 11, 2010 at 1:42

Why don't you try a spice simulator like ltspice?


You can change values and quickly see the differences.


  • \$\begingroup\$ Does it have models for PUTs?? \$\endgroup\$ Commented Jun 11, 2011 at 17:11

I was stuck with this same problem and did some research. I'm a beginner but looking at the 2N6027 PUT datasheet, and from personal experimentation I suspected like @pingswept user stated, that the problem was in the R1 resistor value and it's relation with Valley current when the capacitor discharges.

Look at the http://www.allaboutcircuits.com/vol_3/chpt_7/8.html examples and you'll find how to achieve the proper resistor values for UJT and PUT oscillator circuits.


PUT (Unijunction) is a "strange" thing, like "Gunn diode" (used in radar circuitry) which has a characteristic curve with a "negative" region of working.

Some "PUT" are "programmable" and in fact are composed of 2 transistors (1 NPN and 1 PNP) connected correctly (B to c, b to C), these are also called thyristors.

The resistor connected to electrode E must be greater then the "negative value" of this PUT in the "working" area (use charge line concept to chose correct value.)

  • If too low, the circuit is a "monostable".
  • If greater than the negative resistance, it is an "unstable" oscillator.
  • If too much greater, it is a "stable circuit", but non-linear.

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