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?
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.
- 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).
So as you mentioned it says that the transistor is essentially two diodes.
You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows:
As the resistance across a diode is very low when this voltage is applied we can work out that the current would be extremely high: I = V/R, simple to see that the lower R is the higher the current, and this can be very damaging to the base terminal, I believe a datasheet of the particular transistor will give you more information on what size current it can take.
What this is saying is you need to have a current limiting resistor in front of the base terminal on the transistor which does exactly what its name describes, limits the current. As the voltage drop across the transistor will remain at 0.6-0.8V we can work out the size resistor we would need quite easily. R = (Vin - Vdrop)/I, 'I' being the base current that it can take, Vdrop being the voltage drop from the base to the emitter and Vin being the supply that is going into base, you also need to look at the hfe of the transistor so see if it will be able to give you the amount of current you need, which coincidentally can be limited, or 'tailored' with a resistor at the emitter pin so the transistor is less reliant on the hfe, but I am sure you will get on to that in the future!
Best Answer
Figure 1.
Figure 2. Slayer Exciter simulation done by LTSPICE. Source: ElectroBoom which explains the operation in further detail. (Click image for full resolution view.)