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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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  • \$\begingroup\$ 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. \$\endgroup\$ – Kortuk Apr 13 '12 at 18:05
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    \$\begingroup\$ 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! \$\endgroup\$ – Kevin Vermeer Apr 13 '12 at 18:38
  • \$\begingroup\$ 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. \$\endgroup\$ – jippie Apr 13 '12 at 21:03
  • \$\begingroup\$ @jippie, if you do that you might earn 500 rep? \$\endgroup\$ – Kortuk Apr 18 '12 at 3:08
  • \$\begingroup\$ Sure but I need a test circuit to verify if it works. \$\endgroup\$ – jippie Apr 18 '12 at 7:11
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I'm also working my way through the book "Make electronics" by Charles Pratt. I also stumbled on the PUT at experiment 10. The circuit simulator I'm using icircuit, based on this circuit simulator applet but it doesn't provide a PUT component, although it's a really great simulator.

I tried the first proposed alternative above (1 transistor PNP and 1 transistor NPN), but it doesn't give reliable results on my simulator. I guess ordinary transistors don't always behave as ideal/simulated transistors.

Consulting the book "practical electronics for inventors" from Paul Scherz, I think i found a good alternative to the PUT with the MOSFET N channel:

abstract from the book : "Mosfet (enhancement) n-channel : Normally off, but a small positive voltage at its gate (G)—relative to its source (S)—turns it on (permits a large drain-source current). Operates with VD > VS. Does not require a gate current. Used in switching and amplifying applications.

Please note that for the MOSFET (enhancement) n-channel, the positive voltage must be at the gate (G) and not at the source (S) as it is the case for the PUT.

I took a a printscreen of the result in my circuit simulator applet. It seems to work all right.

UPDATE 23/08 : In the end, it happened that the idea of replacing the PUT with a MOSFET (enhancement n-channel) in experiment 11 of makes electronic from Charles Pratt was a dead end. A valid alternative is a 555 timer. See following post.

MOSFET used as a switch

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  • \$\begingroup\$ If you edit your answer to include a link to the image in a publically viewable site someone will be able to include it. \$\endgroup\$ – PeterJ Aug 20 '13 at 13:18
  • \$\begingroup\$ I tried the same thing after successfully making the circuit on a breadboard and discovered that if you change the Beta/hFE to 20 for your transistors the PUT should work as expected in iCircuit. twitter.com/dtrotzjr/status/421156879926960128/photo/1 \$\endgroup\$ – dtrotzjr Jan 9 '14 at 5:50
  • \$\begingroup\$ Anonymous users said: Hi, The PUJT alternative with pnp-npn pair is OK, but simulated circuit (screen dump) is wrong - the electrodes were swapped and should be reversed as pnp's E is emitter/anode X1 and B-C is base/gate B1. Also, very often designations are different: E B2 B1 or A G K instead of X1 B1 B2. \$\endgroup\$ – Russell McMahon Sep 14 at 10:51
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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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  • \$\begingroup\$ 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. \$\endgroup\$ – Nathaniel Apr 15 '12 at 12:37
  • \$\begingroup\$ @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 \$\endgroup\$ – clabacchio Apr 15 '12 at 12:43
  • \$\begingroup\$ 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.) \$\endgroup\$ – Nathaniel Apr 15 '12 at 12:56
  • \$\begingroup\$ @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. \$\endgroup\$ – clabacchio Apr 16 '12 at 6:34
  • \$\begingroup\$ I agree absolutely - I didn't mean to attack your design, I meant more to suggest that you could improve your circuit with some kind of biasing so that the negative resistance part of the curve would be in the top right quadrant. \$\endgroup\$ – Nathaniel Apr 16 '12 at 8:10
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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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    \$\begingroup\$ That looks much more like a SCR than a PUT. \$\endgroup\$ – Olin Lathrop Apr 13 '12 at 18:49
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    \$\begingroup\$ @OlinLathrop, Answer with what it really is if you disagree. They say a PUT is very similar to a thyristor. \$\endgroup\$ – Kortuk Apr 13 '12 at 18:53
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    \$\begingroup\$ @OlinLathrop, Gate placement is different. \$\endgroup\$ – Craig Apr 13 '12 at 18:54
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    \$\begingroup\$ @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. \$\endgroup\$ – Kortuk Apr 13 '12 at 18:55
  • \$\begingroup\$ @OlinLathrop, by the datasheet it seems it is very very similar to an SCR. I will have to look at the numbers longer. \$\endgroup\$ – Kortuk Apr 13 '12 at 19:05
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For everyone else who is going through the Make Electronics book, I managed to get it working using a BC557-B (PNP) and a BC547-B (NPN) transistors and replaced R2 which was originally 15k with a 4.7k one.

Just connect the base of the PNP to the collector of the NPN and the collector of the PNP to the base of NPN. You will couple the capacitor and R1 to the emitter of the PNP, the gate with the 2 resistors is the base of the PNP and the LED goes on the emitter of the NPN. The first image is ok to follow, but scratch that resistor.

For those interested in the details the only decent link I could find that explains what this pseudo thyristor is: The Unijunction Transistor (UJT) - from allaboutcircuits.com. Scroll to the PUT section.

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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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Replace 2N6027 with 2N3904 and 2N3906 and using R2=10k and R1 = 27k. It works in LTSPICE and in breadboard. Using 10uF and 670k LED flashes 1 time after 4 sec. 0.25Hz. And use 9v supply not 6v.

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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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    \$\begingroup\$ I don't understand the point of this answer: where is the part in which you answer the question? \$\endgroup\$ – clabacchio Apr 18 '12 at 5:54
  • \$\begingroup\$ Formatting is lost when CR is stripped from text. but my answers were there...."1)Can you substitute BC548B for BC547A ? Yes \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 18 '12 at 19:13
  • \$\begingroup\$ "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? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 18 '12 at 19:15

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