This is about understanding characteristics plot of a UJT:

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I have read about UJTs but I have a problem understanding the following UJT characteristics plot:

enter image description here

This doesn't look like other plots of active elements I have seen so far. I will try to explain why and write down how I interpret it maybe it helps you to see where my logic is flawed.

I interpret that plot as follows:

I start by focusing on y-axis hence the emitter voltage Ve. My plan is to observe on the above plot how the emitter current Ie changes with respect to Ve. So I will see the relation as: Ie = f(Ve) where Ve is the variable.

So I increase the emitter voltage Ve i.e I go upwards on the y-axis. Until I reach up to Vp everything seems straightforward. Upto Vp because of the reverse bias very small current flows through E and B1.

But what is happening after this point? I cannot follow anymore. I can just say Increasing Ve further beyond Vp would forward bias the diode in equivalent circuit of the UJT. But Ve is not increasing on the plot anymore it is decreasing. I lose the feeling of Ie = f(Ve) relation. What kind of plot is this? It has two identical voltage values for a given current.

How is this plot obtained? My way of thinking was "increase Ve and see what happens to Ie" but Ve does not increase after Vp. Obviously something wrong in my way of thinking here.

How can we interpret this plot or how is this obtained? I'm really confused and feel stupid why I cannot get this.

  • \$\begingroup\$ It probably wouldn't be a good idea to hook up a variable power supply. But imagine how it is actually used. Often with an RC junction at the emitter. The resistor charges up the capacitor and, at some point, the voltage reaches high enough that the device kind of "collapses" (it's resistance from emitter to B2 rapidly declines, if that's how it's hooked up.) Because the resistance suddenly declines so much, the capacitor dumps charge (as current) and the capacitor's voltage declines rapidly. But the UJT keeps on going, even then, until the cap is almost completely discharged (valley.) \$\endgroup\$
    – jonk
    Feb 26, 2018 at 7:59
  • \$\begingroup\$ I'm more struggling how this curve is obtained rather than how it is used as an oscillator. Something is held constant or? \$\endgroup\$
    – pnatk
    Feb 26, 2018 at 8:02
  • \$\begingroup\$ Hopefully, the answer helps some. \$\endgroup\$
    – jonk
    Feb 26, 2018 at 8:29

1 Answer 1


Oh, heck. I'll write it up as an answer.

Start by imagining how it is actually used. Often with an RC junction at the emitter.


simulate this circuit – Schematic created using CircuitLab

\$R_3\$ charges up \$C_1\$ and, at some point, the voltage reaches high enough that the device kind of "collapses" (it's resistance from emitter to \$B_1\$ rapidly declines.) Because the resistance suddenly declines so much, the capacitor dumps charge (as current) and the capacitor's voltage declines rapidly. But the UJT keeps on going even then, until the cap is almost completely discharged (valley.)

Now, there are two possibilities here.

  1. You are in a situation where you can continue to ramp up the voltage again while also sustaining and even increasing the current into the emitter. This is a highly unusual situation. But of course scientists will test it. So the curve shows this fact. But you must have arranged things so that when the valley is reached that the voltage can be sustained and then increased again while also the emitter current continues to increase further.
  2. You are in a situation where you cannot continue to ramp up the voltage at this rather higher emitter current. This is the usual case and is the one intended as shown in the above schematic. In this case, the current is greatly reduced again and you are returned to the very early part of the curve. And this is the common oscillator situation you see in the above schematic.

So in the above schematic, the voltage on \$C_1\$ increases as \$R_3\$ feeds more current than the tiny trickle that the UJT's emitter accepts -- the rest going to charge up \$C_1\$. Then, when the voltage on \$C_1\$ rises up high enough, the trickle of current is then enough that suddenly the resistance drops between the emitter and the bases. When that happens, and if \$R_3\$ is too large of a value to supply all that current, then \$C_1\$ suddenly finds itself adding current to what \$R_3\$ is providing and the voltage on \$C_1\$ declines. But the UJT doesn't care, because with yet-increased current flowing now in the UJT body, the body resistance dropped still more and so it doesn't require much voltage for the newly increased currents to continue.

At some point, the voltage on \$C_1\$ reaches the valley point. At this point, the UJT is still willing to keep up the high currents at a low voltage. But now it will require still more voltage to go there. However, unfortunately in this circuit's case, \$R_3\$ is designed so that it cannot supply all that current by itself. (It's too big of a value.) And the capacitor doesn't have a magic wand to wave in order to increase its voltage either. So the voltage on \$C_1\$ can't go up and \$R_3\$ is suddenly left all by itself trying to supply this rush of emitter current without any help from the capacitor, now. And it just can't. Not even close, in fact.

So the high UJT current at the valley point cannot be sustained, at all, since \$R_3\$ is too weak. So you return yourself to the early part of the curve and \$C_1\$ starts charging all over again.

If you want to measure the curve that occurs after the valley point, then you have to have a source of current that is now sufficient to supply not only the high valley current but to also be able to supply still more so that the capacitor can start to charge back up again. So if you make \$R_3\$ a very low value and make \$C_1\$ a very large value and then place a scope on all this, you might be able to capture a picture. Start the capacitor completely discharged. The small valued \$R_3\$ will charge it up pretty rapidly so it will be very hard to see it hit the trigger/peak point before the whole thing collapses to the valley. But then, if you have made \$R_3\$ small enough it will still be able to charge up the capacitor even more while also supplying all that current. (Meanwhile, I kind of predict your UJT will be undergoing some serious stress it was not designed for.)

But for most uses the whole idea is to make sure that \$R_3\$ cannot supply all that current.

There are other creative uses for this behavior, as well as some other artifacts of its construction. (The UJT is affected by the Hall effect and responds to magnetic fields, for example.)

The body resistance is divided up into two parts (because the diode is fashioned at that point along the body.) So there is a value for \$R_{B_1}\$ and for \$R_{B_2}\$. A special parameter is given on the datasheet that will tell you what the divider ratio, \$\frac{R_{B_1}}{R_{B_1}+R_{B_2}}\$, is. This tells you what the voltage will be at the cathode of the diode in the UJT (emitter open.) With no emitter current yet, the emitter should be about that voltage if you don't load it. Your peak voltage will have to be this divider voltage plus a diode drop.

  • \$\begingroup\$ Thanks for the answer, Im in a hurry go somewhere now I'll read and try to understand it as soon as Im back. \$\endgroup\$
    – pnatk
    Feb 26, 2018 at 8:29

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