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I'm following instructions from Charles Platt's "Make" and trying to mount PUT oscillator. Strangely, it blinks only once when I switch on the DC current, no oscillation was observed. I have tried different combinations of resistors, changed capacitor from 1 uF to 300 uF, voltage source from 6V to 9 V, however, there is no any significant change in behaviour of my oscillator. Here is the schema:

enter image description here

And here is my plate with all data measured by multimeter:

enter image description here

Please, can anybody tell me what is my mistake?

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  • \$\begingroup\$ @Bimpelrekkie, Actually, Platt himself uses bigger resistances and lower voltage (from 9 to 6 V). But I will try. \$\endgroup\$
    – tenghiz
    Commented Mar 28, 2019 at 14:59
  • \$\begingroup\$ Just off the cuff, I think your \$20\:\text{k}\Omega\$ resistor is too stiff. Try a much larger value. (Reduce the capacitor value, as you see fit.) \$\endgroup\$
    – jonk
    Commented Mar 28, 2019 at 15:00
  • 1
    \$\begingroup\$ The resistor values shown in the picture don't match the values on the schematic, and you have an LED on the breadboard that is not shown on the schematic. Please edit the schematic to show what you actually built. \$\endgroup\$ Commented Mar 28, 2019 at 15:24
  • \$\begingroup\$ @Peter Bennett , I changed some resistors because I was trying to follow the description of this schema from other sorces. \$\endgroup\$
    – tenghiz
    Commented Mar 28, 2019 at 15:34
  • \$\begingroup\$ I added something to my answer, the LED can complicate things. You should really consider a different schematic. \$\endgroup\$ Commented Mar 28, 2019 at 15:53

4 Answers 4

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Overview

PUJTs (aka PUTs) have rather wide variations between parts. They were suddenly popular in the 1960s, but I think they had different names back then. Today, they've been renamed as the 2n6027 and 2n6028. (There may be other versions today, but I'm not aware of them.) I have bunches of both here, but I don't use them that often.

They are thyristors (SCR-like) and can be "programmed" easily with a resistor divider. There is a parameter called \$\eta\$ that was specified for the UJT (not the same device.) The resistor divider for the PUJT essentially allows you to set the value of \$\eta\$ (which you cannot do with the UJT.)

The basic canonical circuit looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

(Please forgive the weird PUJT symbol, as CircuitLab doesn't seem to include a PUJT.)

Grab up the OnSemi 2N6027/2N6028 datasheet. Note that, if your resistor divider is made up of \$R_1\$ and \$R_2\$ that they define \$R_\text{G}=\frac{R_1\cdot R_2}{R_1+R_2}\$ (the Thevenin resistance, of course) and \$V_\text{S}=V_\text{B}\cdot\frac{R_1}{R_1+R_2}\$ (the Thevenin voltage.) These will determine the valley current you need to allow the device to go below before the device can reset itself for another oscillation. Smaller values of \$R_\text{G}\$ provide for higher valley currents and larger values of \$R_\text{G}\$ provide for lower valley currents.

If resistor \$R\$, supplying both capacitor charging current and once triggered then also the anode current, is too small then \$R\$ will supply too much current after triggering and the device will have sufficient forward voltage to "sit" over the at far right of the following curve. Note that the forward current allowed by \$R\$ is higher than the valley current and so the system sits where the red dot is located. The result is that it won't oscillate.

enter image description here

On the other hand, if you arrange things so that the circuit must go below the valley current for the device, then the device will reset itself and return to the point on the curve shown below:

enter image description here

So, now the capacitor can begin to recharge and the cycle repeats. So you have an oscillator.

Design for pulsing an LED

Here's a suggestion for driving a small LED. I've arranged it so that there is a current limiter action due to using \$Q_1\$ as an emitter-follower. The values of \$R_\text{SET}\$ and \$R_\text{K}\$ and \$V_\text{S}\$ will determine what happens with the LED.

schematic

simulate this circuit

Looking at the datasheet, we see:

enter image description here

So we know that it will trigger at a range somewhat above \$V_\text{S}\$. Since \$V_\text{CC}=12\:\text{V}\$ in this case, let's set \$R_1=27\:\text{k}\Omega\$ and \$R_2=15\:\text{k}\Omega\$. (This results in \$R_\text{G}\approx 9.6\:\text{k}\Omega\$, which is close to one of the datasheet's offerings.) Therefore, \$V_\text{S}\approx 7.7\:\text{V}\$. We can expect things to trigger over a range of from \$7.9\:\text{V}\$ to \$8.3\:\text{V}\$, with a typical expectation of \$8.05\:\text{V}\$. Let's just call it about \$8\:\text{V}\$.

The valley current can be found here:

enter image description here

We will want to make sure it goes below \$70\:\mu\text{A}\$. This can be achieved by making sure that \$R\ge\frac{V_\text{CC}=12\:\text{V}}{70\:\mu\text{A}}\approx 172\:\text{k}\Omega\$. Let's go with \$R=220\:\text{k}\Omega\$.

For a blinking frequency on the order of one second, \$R\cdot C\approx 1\:\text{s}\$. So \$C=\frac{1\:\text{s}}{220\:\text{k}\Omega}\approx 4.6\:\mu\text{F}\$. We could go up or down from there, but let's go with the closest standard value of \$C=4.7\:\mu\text{F}\$.

The forward voltage during discharge is:

enter image description here

The LED current will follow an exponential decay curve (it won't be at some fixed value for the entire pulse.) Let's say we've decided on a peak LED current of \$30\:\text{mA}\$ as acceptable. We can expect that the maximum voltage across \$R_\text{K}\$ will be \$V_\text{MAX}\approx 8\:\text{V}-800\:\text{mV}=7.2\:\text{V}\$. If we assume that \$Q_1\$'s \$V_\text{BE}\$ will be, say, also \$800\:\text{mV}\$, then this means we have about \$7.2\:\text{V}-800\:\text{mV}=6.4\:\text{V}\$ across \$R_\text{SET}\$. So \$R_\text{SET}=\frac{6.4\:\text{V}}{30\:\text{mA}}\approx 213\:\Omega\$. Use the standard value of \$R_\text{SET}=220\:\Omega\$.

During discharge, a number of things happen. The gate voltage will pretty closely follow the cathode voltage on the way down. However, the anode voltage will be around a diode drop above the cathode. The equivalent circuit looks something like this for finding the minimum cathode voltage prior to reset:

schematic

simulate this circuit

This solves out to about:

$$\begin{align*} \frac{V_\text{MIN}}{R_\text{G}}+\frac{V_\text{MIN}}{R_\text{K}}+\frac{V_\text{MIN}}{\beta\cdot R_\text{SET}}&=\frac{V_\text{S}}{R_\text{G}}+I_\text{V}\\\\ &\therefore\\\\ V_\text{MIN} &=\frac{V_\text{S}+I_\text{V}\cdot R_\text{G}}{1+R_\text{G}\cdot\left(\frac{1}{R_\text{K}}+\frac{1}{\beta\cdot R_\text{SET}}\right)} \end{align*}$$

At this point, we can either choose the value of \$V_\text{MIN}\$ or else \$R_\text{K}\$. I'm going to suggest that the minimum LED current should be about \$10\:\text{mA}\$ during discharge (going from a peak of \$30\:\text{mA}\$ to a valley of \$10\:\text{mA}\$ for a very rough average of \$20\:\text{mA}\$ [discounting the exponential decay aspect of this.])

So I add \$10\:\text{mA}\cdot 220\:\Omega=2.2\:\text{V}\$ to the \$V_\text{BE}\$ of \$Q_1\$ or \$V_\text{MIN}=2.2\:\text{V}+800\:\text{mV}=3.0\:\text{V}\$. Solving for \$R_\text{K}\$ is an exercise in math (involves LambertW function, which I'll avoid here.) But if you play around, you'll find that for \$70\:\mu\text{A}\le I_\text{V}\le 150\:\mu\text{A}\$ then \$R_\text{K}\approx 5\:\text{k}\Omega\$. So I've selected \$R_\text{K}=4.7\:\text{k}\Omega\$.

Discounting the small current provided by \$R\$ during discharge and the small base current required by \$Q_1\$, this leads to \$\tau=R_\text{K}\cdot C\approx 22\:\text{ms}\$. The discharge time is complicated by the current supplied into \$R_\text{K}\$ by \$V_\text{S}\$ through \$R_\text{G}\$. In this case, using an average current through \$R_\text{G}\$ of about \$600\:\mu\text{A}\$ for the period I get an equation for the capacitor voltage of about \$V_t=2.8\:\text{V}+5.2\:\text{V}\cdot e^{-\frac{t}{R\cdot C}}\$. (The value of \$2.8\:\text{V}\$ comes from \$R_\text{K}\cdot 600\:\mu\text{A}\approx 2.8\:\text{V}\$.) So the number of \$\tau\$'s required will be about: \$\operatorname{ln}\left(\frac{8\:\text{V}-2.8\:\text{V}}{3.6\:\text{V}-2.8\:\text{V}}\right)\approx 1.87\$. So the pulse period I'd predict is about \$1.87\cdot 22\:\text{ms}\approx 40\:\text{ms}\$.

The final circuit is:

schematic

simulate this circuit


Added in Response to OP's Test Results

As I mentioned in a comment I made to your answer here on your attempts to use the PUJT as an oscillator, I will attempt to build one and see if my results are similar to yours, or different. (I'll use dead-bug construction, myself, as I pretty much don't like solderless breadboards.)

In the meantime, I went out and secured some PUJT Spice models and have set up a circuit that replicates the results of the above analysis I did to see if Spice would agree (at least, in theory, so to speak.) Hopefully, this will provide some confidence in the theoretical approach, if nothing else.

The models I found for the two PUJTs I can simulate using Spice are:

.SUBCKT 2N6028 1 2 3
*Programable Unijunction Transistor
*MODEL FORMAT: PSpice
*       anode  gate  cathode
*node:    1      2      3
Q1 2 4 3 NMOD
Q2 4 2 1 PMOD
.MODEL NMOD NPN(IS=2E-15 VAF=100 IKF=0.3 ISE=2.5E-12 NE=1.6 RE=0.15 RC=0.15 CJE=7E-10 TF=0.6E-8 CJC=2.2E-10 TR=4.76E-8 XTB=3)
.MODEL PMOD PNP(IS=22E-15 VAF=100 IKF=0.3 ISE=1E-12 NE=1.7 RE=0.15 RC=0.15 CJE=7E-10 TF=1.6E-8 CJC=2.2E-10 TR=5.1E-8 XTB=3)
.ENDS
*
.SUBCKT 2N6027 1 2 3
*Programable Unijunction Transistor
*MODEL FORMAT: PSpice
*       anode  gate  cathode
*node:    1      2      3
Q1 2 4 3 NMOD
Q2 4 2 1 PMOD
.MODEL NMOD NPN(IS=5E-15 VAF=100 IKF=0.3 ISE=1.85E-12 NE=1.45 RE=0.15 RC=0.15 CJE=7E-10 TF=0.6E-8 CJC=2.2E-10 TR=4.76E-8 XTB=3)
.MODEL PMOD PNP(IS=2E-15 VAF=100 IKF=0.3 ISE=1.90E-12 NE=1.5 RE=0.15 RC=0.15 CJE=7E-10 TF=1.6E-8 CJC=2.2E-10 TR=5.1E-8 XTB=3)
.ENDS

Given the above, I was able to construct the following schematic in LTspice:

enter image description here

As you can see, there's nothing different nor particularly special here. It's just a replication of what I had discussed, earlier. (The main thing is the models, which I took from ORCAD's set of device models.) The BJT is just a typical 2N2222 device, which is quite common and not at all unusual. The LED model is a basic "red" LED designed to operate with about \$20\:\text{mA}\$ and about \$2\:\text{V}\$.

Here are the simulation results. The first picture here is for the entire 10 second run (using UIC):

enter image description here

From above, you can see that the pulses occur approximately one second apart. Not quite, of course, because I hadn't accounted for the added load due to the BJT, itself, which will reduce the pulse duration somewhat. But this is well within reasonable expectations, I believe. And as you can see, LTspice does indicate that it will oscillate.

This second picture zooms up on one of the pulses:

enter image description here

Here, you can see the peak pulse current, the pulse duration, and also the current at which it shuts off. I think you can see that the peak pulse current is close to the \$30\:\text{mA}\$ I'd predicted from theory. You can also see that the final current is \$10\:\text{mA}\$ just before it shuts off -- also as I'd predicted from theory. And finally, the pulse duration is about \$42\:\text{ms}\$, which I consider to be extremely close to what I had predicted from theory.

In short, I'm not sure how I could have possibly gotten any closer -- entirely from pure theory and reading a datasheet -- to what LTspice gets from using a model that was extracted from a source of PUJT models that came from ORCAD models developed circa 1990 and was at the time, obviously, completely unconnected directly to the datasheet I used to make my predictions of behavior, a few days ago.

This is a confirmation of sorts that independent thinking a few days ago, working from a datasheet, can result in very similar outcomes using a Spice simulator (which takes into account many, many more details than I did.)

What's left, of course, is to build it. Because of the relatively low values of resistors, I don't expect parasitics to be much of a problem. But I'll use dead-bug construction, anyway, just in case. Perhaps I'll uncover some important issue that wasn't in the datasheet and wasn't in the Spice models.

But for now, I consider the above to be a step towards confirming the approach I took here.

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    \$\begingroup\$ \$ \text{Way to Go} \:\text{Jonk}\$ \$\endgroup\$
    – D.A.S.
    Commented Mar 29, 2019 at 5:58
  • \$\begingroup\$ @SunnyskyguyEE75 Uh. Thanks? hehe. \$\endgroup\$
    – jonk
    Commented Mar 29, 2019 at 6:05
  • \$\begingroup\$ Thank you for taking your time and explaining all the details! \$\endgroup\$
    – tenghiz
    Commented Apr 1, 2019 at 14:48
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    \$\begingroup\$ @tenghiz I'm glad it may have helped. Thanks for the kind note, too. By the way, I doubt further answers will be added at this point. If the answer I've provided is what you were looking for and if you also don't expect further, better answers you are free to select it as an answer. (Of course, leaving it open might lead to a better answer, so it's your call how you want to proceed.) \$\endgroup\$
    – jonk
    Commented Apr 1, 2019 at 17:13
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In the schematic Vbb = 12 V so it uses a 12 V supply (I'll ignore that there's no ground indicated).

Your circuit seems to be using 4 x 1.5 V batteries so that's only 6 V. That might not be enough to get the oscillation going.

Let's look in the datasheet of the 2N6027

See figure 3, more than 6 V is needed to trigger the effect that you want.

Also, there is a lot of tolerance on this parameter, I mean: it varies per transistor. You might be unlucky and have one which needs more than 11 V, see table "Electrical characteristics" row: "Peak output voltage"

Additional issues are also that you added a LED in series with the Unijunction Transistor's Cathode, that increases the required voltage even more. The LED being at that point in the schematic also changes the behavior.

You should have a look at different LED flashers using an Unijunction Transistor, in Google search for "UJT flasher" and press the images tab. Now you will see plenty of schematics. Try some of those.

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  • \$\begingroup\$ @jonk, I decreased capacitancd to 2.2 uF, changed 20 kOhm resistor to 470 kOhm, increased voltage to 15 V. The same situation as previously: it only blinks once when I plug in DC and after there is nothing. There is as previously 0.07 V across LED. Voltage between cathode and anode is 13 V. \$\endgroup\$
    – tenghiz
    Commented Mar 28, 2019 at 15:23
  • \$\begingroup\$ @tenghiz The usual problem is what I mentioned. However, it's not the only possible problem. What happens in the case I was worried about is that once the capacitor charges up to the trigger voltage (assuming other parameters are met), the PUJT fires. But a stiff resistor prevents the capacitor from discharging sufficiently to reset the PUJT because the resistor can supply too much current. That's why I suggested that idea. However, it seems it's perhaps that problem plus another problem. \$\endgroup\$
    – jonk
    Commented Mar 28, 2019 at 15:34
  • \$\begingroup\$ @ jonk , can resistors used in a voltage divider play siginificant role, for example, is combination 5 kOhm - 10 kOhm is better than 15 kOhm - 30 kOhm? \$\endgroup\$
    – tenghiz
    Commented Mar 28, 2019 at 15:38
  • \$\begingroup\$ @tenghiz In a sense, yes. The resistor divider's equivalent Thevenin resistance determines the minimum valley current that you must "go below." So yes, the resistor divider affects the design. Do you want me to write something up about how to design these beasts? If so, I may need more information about what you are trying to achieve. It's good to have a concrete example to illustrate the abstract issues. \$\endgroup\$
    – jonk
    Commented Mar 28, 2019 at 15:48
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    \$\begingroup\$ @tenghiz That circuit just won't work right. You can't just insert an LED and resistor there. Usually, you use the cathode resistor in conjunction with another BJT to get what you need for driving an LED. \$\endgroup\$
    – jonk
    Commented Mar 28, 2019 at 15:54
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Here is the outline of all my trials.

First of all I decided to follow the advice of @Bimpelrekkie and mount a scheme following the Google query "UJT flasher". My first trial was carried out according to this scheme (https://www.electroschematics.com/6904/programmable-unijunction-transistor-put-flasher-circuit/):

enter image description here

Also, the outcome: it simply doesn't work with 9 V DC source. However, I can obtain a blink of LED in case if I increase the voltage to 12 V. The duration of a blink can be extended if the capacitance is increased to 47 uF. Since this scheme gives the same result as I described in my question, I decided to check another source (https://www.youtube.com/watch?v=MJ3-wBvDowk):

enter image description here

It is not an oscillator, however, I have chosen this scheme to check whether my PUT is a working one. Outcome: nothing. I have tried both 2N6027 and 2N6028 - there is no difference.

My next step was to follow the advice of @jonk. Frankly speaking, I was deeply impressed by the theoretical part and all the calculations. This is what I like in physics! So... it doesn't work. It blinks once at the beginning - and it's all.

enter image description here

This is the point when I started to question myself whether I have some problems with my hands or whatever. Here I decided to mount an alternative oscillator scheme without PUT from the second edition of "Make Electronics":

enter image description here

As the author states, he decided to abandon PUT because "some readers complained that you can’t buy them [PUT] easily, while others said that using a PUT was just too oldschool". Let it be. Here is my outcome:

enter image description here

It blinks!

As for PUT, I decided to postpone the exploration of this device before I meet personally some experimented person who can show me on spot how to use it.

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  • \$\begingroup\$ Thanks for the update! I'll take a moment when I get a chance and test out my own work. If it also fails, I'll work out the issues further and let you know what I find out. If it works, I'll just say so. Yes, pujt devices aren't used much and I stopped using them years ago, despite having lots of them here. But being relaxation oscillators capable of using high impedance results, they can be quite efficient when they work. Thanks again for the update. \$\endgroup\$
    – jonk
    Commented Apr 2, 2019 at 10:19
  • \$\begingroup\$ I've added simulation results to my answer. It confirms the theory I applied. \$\endgroup\$
    – jonk
    Commented Apr 2, 2019 at 10:57
  • \$\begingroup\$ @jonk, thank you very much. It seems that I have to follow your example and start to learn Spice. \$\endgroup\$
    – tenghiz
    Commented Apr 10, 2019 at 14:02
  • \$\begingroup\$ Well, Spice won't fix not knowing how to interpret and apply theory. You still need to know how to read the datasheet and make sense of it after some fashion. Now simulation isn't reality and parts vary widely (especially these parts.) But at least in the simulation world you can see how closely my use of the datasheet and theory was able to predict what Spice plotted out. Pretty amazing, actually. Of course, Spice uses ideal wires and power supplies, etc. And parts vary. So while theory tells you why and the datasheet tells you approx. behavior, reality may sill need addressing, too. \$\endgroup\$
    – jonk
    Commented Apr 10, 2019 at 17:27
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...the circuit is wired wrong...check your anode, gate, and cathode connections...you'll find that the components that are supposed to be connected aren't connected correctly. LR.

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    \$\begingroup\$ Lawrence Riles - Welcome :-) Please can you Edit your answer to add more details? Are you referring to the breadboard photo in the original question, or in another answer, or to a schematic, or ... ? Also, can you be more specific about exactly which "components that are supposed to be connected aren't connected correctly"? Otherwise it's difficult for readers to actually use your answer. Thanks. \$\endgroup\$
    – SamGibson
    Commented May 26, 2021 at 20:55

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