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Similar to this question, I need a replacement for a programmable unijunction transistor [PUT]. Specifically, I'm trying to emulate the behavior described in the 2N6028 datasheet. However, I would like to build my own from normal transistors rather than buying one.

I found this page on the Makezine forums where someone asked the same question. This links to this page on edaboard which includes a schematic, but I'm not sure what the OP means by "two base B1 B2" or whether the following schematic:

enter image description here

indicates that I need exactly those particular transistors or if I could substitute in other bipolar transistors (e.g. BC548B).

This is a curiosity, I am not saying that it is a functional project, but I am instead interested in attempting to build a component out of others. I may learn something, I may not. I look forward to finding out. I'm working through the Make: Electronics book, and several of the critical beginning experiments call for a PUT. I am aware that PUTs are old, but I'm interested in this as a curiosity.

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I have added a bounty to just get an answer to what seems like a relatively simple problem. I feel like I almost feel comfortable answering it, but transistors have never been my specialty and I might incorrectly answer. – Kortuk♦ Apr 13 '12 at 18:05
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I have cleared out many unconstructive comments. The OP and the author of the book are aware the PUTs are old, let's just answer the question! – Kevin Vermeer Apr 13 '12 at 18:38
Why not put it together on a breadboard and see what it does? I'd do it, I have the components lying around, but I don't have access to the schematic it'd go into, nor what to expect. And indeed unijuntion transistors are old. I've been into electronics for over 25 years and I've only rarely come across them. – jippie Apr 13 '12 at 21:03
@jippie, if you do that you might earn 500 rep? – Kortuk♦ Apr 18 '12 at 3:08
Sure but I need a test circuit to verify if it works. – jippie Apr 18 '12 at 7:11

4 Answers

up vote 0 down vote

With the circuit given in the question above and Question #5 on this page, the circuit acts like a mono-flop. Tried it with BC546B and BC547B at 12V. The voltage at TP1 increases from 0V at power on to approx. 11 volts, then the pseudo-PUT triggers and the voltage at TP1 drops to approximately 0.8V. It does not reset.

(My first try on KiCad, bear with me ...)

I experimented with 15E, 150E and 1k5 for R3. Attached a 1M, 560k from the 'floating' base-collector to ground and to +12V => The circuit stays a monoflop. R3 must be < 100E.

Hints, tips anyone?

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up vote -1 down vote

There are a few questions and a lot of irrelevant information with op amps given.

  1. Can you substitute BC548B for BC547A ? Yes ..... The BC548B (NPN) has much more hFE and the BC547A has much higher breakdown voltage.
  2. The B1 B2 designation is not longer used in favour of the thyristor terms Gate G and Cathode K.

Comments:

  1. the 2N6027 is not as sensitive as the 2N6028 (eg. 1uA vs 0.1 uA so higher hFE is more sensitive)
  2. The best part specs to examine may be the Philips BRY39 4 pin PUT device.

I remember 40 years ago touching a theramin glass ball that changed synthesizer frequencies connected to metal contacts all over the globe that acted as the pull up resistance via the fingers to many PUT's, caps and speakers that acted all over the globe as a network of gated Relaxation oscillators. Note the power dumped via the 20 ohm resistor on the cathode(K) to ground is spec'd as Vo. is enough to drive a speaker.dumped while discharging the cap. It almost sounded like porpoises.

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2  
I don't understand the point of this answer: where is the part in which you answer the question? – clabacchio♦ Apr 18 '12 at 5:54
Formatting is lost when CR is stripped from text. but my answers were there...."1)Can you substitute BC548B for BC547A ? Yes – Tony Stewart Apr 18 '12 at 19:13
"what the OP means by "two base B1 B2" ..... 2) The B1 B2 designation is no longer used in favour of the thyristor terms Gate G and Cathode K capiche? – Tony Stewart Apr 18 '12 at 19:15
up vote 2 down vote

Disclaimer #1: This is not a direct answer to your question, as I've never used a PUT, nor I've read the book you've taken the exercise from. BUT I've attended a course about oscillators, and these may be workarounds for your problem.

Disclaimer #2: You asked for transistor based solutions; this solution is op-amp based, but I found it clear enough.

So, what do you need is a circuit with negative differential resistance in order to make an oscillator (and eventually other stuff). There are two main types of dipoles with that characteristic, and they're called 'S' and 'N' respectively, due to their I(V) characteristic.

Below is illustrated the difference between the two dipoles.

Link to the original page (in italian)

These dipoles can be used to create an oscillator with a passive dipole constituted by a RLC network. In order for the circuit to oscillate, the resistance has to be chosen in a way that its V-I curve intersects the 'S' characteristic in three points:

enter image description here

But back to the problem

Using an Op-amp, is quite easy to build a 'N' dipole, to achieve the same effect that you have with the PUT.

enter image description here

The analysis of this circuit, to demonstrate the function, can be done separately for the three operating regions of the Op-amp. The V(I) characteristic is:

enter image description here

$$ R_{eq} = \frac{V}{I} = - \frac{R_1 R_3}{R_2} $$

in the high-gain region of the Op-Amp.

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I'll probably edit this answer – clabacchio♦ Apr 15 '12 at 10:04
You've got the negative differential resistance section of the V-I curve passing through the origin - but a real PUT is a passive component, and so cannot achieve this. – Nathaniel Apr 15 '12 at 12:37
@Nathaniel that's not completely true, as if you look at the V-I curve of the PUT there is a part of it in the second quadrant. Plus, I say in the first notes that I'm not directly trying to emulate the PUT but to provide an alternative solution – clabacchio♦ Apr 15 '12 at 12:43
To which PUT V-I curve are you referring? If it's the one in figure 1C of the 2N6028 data sheet I confess I don't quite understand it ($V_A$ is never defined) but surely it's referring not to the passive component alone but to the two powered circuits in figure 1A and figure 1B. (Fair enough about providing an alternate solution rather than an emulation - but the OP is asking for a drop-in replacement for use in the beginner-oriented projects in the Make: Electronics book.) – Nathaniel Apr 15 '12 at 12:56
@Nathaniel yes but he dold that the component is used because of the negative resistance characteristic, so I think that he can apply the same principles with this solution. About the curve, I know that is not the same, but the important part is to have that negative resistance; then, it remains only to tune the passive circuit in order to obtain the intersections. – clabacchio♦ Apr 16 '12 at 6:34
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up vote 2 down vote

I faced the same issue when working through the Make: Electronics book, ultimately I ended up just buying some 2N6027s from DigiKey but prior to that I was able to get by something working using a couple BJTs as shown on this site: http://encyclobeamia.solarbotics.net/articles/put.html

equivalent PUT circuit using BJTs:

enter image description here

with programming resistors:

enter image description here

If I recall, I used a 2N3904 (NPN) and a 2N3906 (PNP).

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2  
That looks much more like a SCR than a PUT. – Olin Lathrop Apr 13 '12 at 18:49
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@OlinLathrop, Answer with what it really is if you disagree. They say a PUT is very similar to a thyristor. – Kortuk♦ Apr 13 '12 at 18:53
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@OlinLathrop, Gate placement is different. – Craig Apr 13 '12 at 18:54
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@craig, I have been previously told you can purchase an SCR with a gate that is active high or active low, this is the active low version. – Kortuk♦ Apr 13 '12 at 18:55
@OlinLathrop, by the datasheet it seems it is very very similar to an SCR. I will have to look at the numbers longer. – Kortuk♦ Apr 13 '12 at 19:05

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