I am reading/doing the make electronics book and I am having a hard time in understanding how is it possible to have the positive and negative poles connected (even with a resistor) to the base of a transistor.

1) Wouldn't it be a short circuit?

2) Why not just the positive?

Below is the example circuit from the book:

circuit image

Edit: Adding a schematic with a more conventional layout that aids understanding. Current flow generally top-to-bottom, and signals (such as they are in this circuit) generally left-to-right.

enter image description here

  • \$\begingroup\$ No, because resistors. Because that's how the circuit works. \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 10 '14 at 2:29
  • \$\begingroup\$ the 15K and 27K resistors form a voltage divider, to hold the gate of the unijunction transistor at about 3.8 volts. There will be about 0.15 mA flowing through the resistors - far from a short circuit. \$\endgroup\$ – Peter Bennett Apr 10 '14 at 2:48

It's not a short circuit because the resistors restrict the current flow.

You can't use just one pole of the battery, because there would not be a "circuit" if the current couldn't flow around.

It isn't a transistor, that is a programmable unijunction transistor, and it has an anode, a cathode, and a gate (no base).

By the way, contrary to what one might think, the 2N6027/6028 does not have just one junction, it has three junctions (four layers). The name "programmable unijunction" comes from the fact it can be used in similar applications to an earlier device called a unijunction transistor.

This is a simple LED blinker circuit ~ 1Hz.

Edit: Even if you don't yet understand how the circuit works, you can see that the current is limited. Whatever G (gate) does, it can't be worse than shorted to ground (it actually does something like that when the PUT fires). Under those conditions the current through the 15K resistor is 6V/15K or about 0.4mA (not much current). The current through R3, no matter what happens with that A (anode) can't be any higher than 6V/470K or 13uA, which is quite tiny.

Edit': The 15K and 27K resistors form a voltage divider. Before it fires, the PUT gate is pretty high impedance, we can ignore it for analysis. The voltage from the two resistors is 6V * 27K/(15K+27K) = 3.9V. The PUT will fire at about 0.35V more than 3.9V, or about 4.2V. RG (as shown on the datasheet) can be calculated as a bit under 10K.

So each flash, the capacitor starts off at a small voltage and charges to 4.2V through the 470K resistor, then discharges through the LED, and the cycle starts again.

Why were those values chosen, instead of different values or leaving one resistor out entirely?

  1. You don't want to make the voltage divider voltage too high or the capacitor charging will not be consistent, or may not fire at all. That means you need R2 and it has to be not too different from R1.

    The voltage has to be high enough that the residual voltage on the capacitor doesn't matter too much and you don't need a bigger capacitor than necessary. For sure you need R1.

    There is a natural voltage to use that is about 2/3 the supply voltage (1-1/e), which gives you a time that is about R multiplied by C.

  2. The parallel combination of R1 and R2, 10K in your example, must not be too high or the PUT will latch on. It must not be too low or a lot of power will be wasted and it will be harder to trigger, perhaps stopping the oscillation. A good value to use is one that's on the datasheet because you don't have to guess about what it will do. There are two options- 1M (too high) and 10K (okay).
  • \$\begingroup\$ I meant the gate when I said base. Thanks for correcting me. What I really need to understand is this basic concept of circuit. I really can't undertand with all the material I read so far why a single pole would not create the voltage by the gate of the put. Can you recommend some material that would make the "circuit" concept more clear? I build this circuit and it works. I just want to understand better. \$\endgroup\$ – Maralc Apr 10 '14 at 4:54
  • \$\begingroup\$ @Maralc I hope I made it a bit more clear. If you get some parts and play with them it would probably help. They're all cheap and worst case a few bits get smoked. Do you have a voltmeter? You can get a fairly decent one for $20 or so, good enough for low voltage stuff. Poke around and see what's going on. A used oscilloscope (the worst working scope is better than nothing) ... a bit later. \$\endgroup\$ – Spehro Pefhany Apr 10 '14 at 5:01
  • \$\begingroup\$ I have a multimeter. Still have to understand where the oscilloscope will fit in my setup (I understand it will show me the waves...). \$\endgroup\$ – Maralc Apr 10 '14 at 5:05
  • \$\begingroup\$ But in regards to circuits and transistors I saw this other question that looks like shows a traditional transistor, and in this case it just needs one polarity: electronics.stackexchange.com/questions/33902/… \$\endgroup\$ – Maralc Apr 10 '14 at 5:06
  • \$\begingroup\$ Why is it different in regards to circuit behaviour and polarities? I understand that a NPN/PNP has a different behaviour than a PUT. \$\endgroup\$ – Maralc Apr 10 '14 at 5:07

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