Help guide me through the valley of the UJT ???

[jalbers@bsu.edu]

I have been doing some reading on the UJT. Much has been written
about what it takes to turn them on and also about the negative
resistance region but not much discussion on what exactly happens in
the saturation region and aslo what it takes to turn them off. I also
have some other related questions.

Suppose that B2 is connected to a fixed voltage source (10V for
example). Suppose that E is connected to a variable voltage source
(0-10V for example) and B1 is connected to ground. Assume that Etta
is around 0.6 .

Starting out with Ve at zero and raising Ve to just over 6.6V should
cause a small amount of current to flow from E to B1 . I assume that
we are now in the negative resistance region. Beginning to lower Ve
should cause more and more current to flow from E to B1. I assume
that I should be able to stop at any voltage between 6.6 and Vvalley
and have the current hold at some value between Ip and Ivalley. It
appears from the snake shaped graph that after lowering Ve to Vvalley,
raising Ve will even produce more current from E to B1. What would
happen if Ve was brought down below Vvalley? What happens if Ve keeps
increasing after initially being brought down to Vvalley? Also what
has to happen to reset the UJT in other words turn off the current
from E to B1? Also when Ve is between Ve(saturation) and Vvalley, how
does the UJT know which current to use? Acording to the graph the
current could be in the negative resistance region or in the
saturation region?

Suppose that B2 is connected to a fixed voltage source (10V for
example). Suppose that E is connected through a resistor and switch
to the same 10V source. Also B1 is connected to ground. Assume that
Etta is around 0.6 . When the switch is turned on Ve is definitely
above Vp. I assume that in a split second the UJT will travel
completely through the negative resistance region and settle
somewhere? Will Ve settle at Vvalley or Ve(saturation) or somewhere
else? Also what has to happen to reset the UJT in other words turn
off the current from E to B1?

Any help would be greatly appreciated. Thanks

 

[Bob Monsen]

<jalbers@bsu.edu> wrote in message
news:ae4ee837-4a60-496b-9743-5a886e0a2b45@a17g2000prm.googlegroups.com...

I have been doing some reading on the UJT. Much has been written
about what it takes to turn them on and also about the negative
resistance region but not much discussion on what exactly happens in
the saturation region and aslo what it takes to turn them off. I also
have some other related questions.

Suppose that B2 is connected to a fixed voltage source (10V for
example). Suppose that E is connected to a variable voltage source
(0-10V for example) and B1 is connected to ground. Assume that Etta
is around 0.6 .

Starting out with Ve at zero and raising Ve to just over 6.6V should
cause a small amount of current to flow from E to B1 . I assume that
we are now in the negative resistance region. Beginning to lower Ve
should cause more and more current to flow from E to B1. I assume
that I should be able to stop at any voltage between 6.6 and Vvalley
and have the current hold at some value between Ip and Ivalley. It
appears from the snake shaped graph that after lowering Ve to Vvalley,
raising Ve will even produce more current from E to B1. What would
happen if Ve was brought down below Vvalley? What happens if Ve keeps
increasing after initially being brought down to Vvalley? Also what
has to happen to reset the UJT in other words turn off the current
from E to B1? Also when Ve is between Ve(saturation) and Vvalley, how
does the UJT know which current to use? Acording to the graph the
current could be in the negative resistance region or in the
saturation region?

Suppose that B2 is connected to a fixed voltage source (10V for
example). Suppose that E is connected through a resistor and switch
to the same 10V source. Also B1 is connected to ground. Assume that
Etta is around 0.6 . When the switch is turned on Ve is definitely
above Vp. I assume that in a split second the UJT will travel
completely through the negative resistance region and settle
somewhere? Will Ve settle at Vvalley or Ve(saturation) or somewhere
else? Also what has to happen to reset the UJT in other words turn
off the current from E to B1?

Any help would be greatly appreciated. Thanks

A UJT is a diode with two cathodes. The 'B1' and 'B2' leads are the
cathodes, and the 'emitter' is the anode.

Without any voltage on the emitter, there is a typical high resistance
between b1 and b2 that is defined by the doping levels of the N silicon. The
emitter won't accept current until it is above _both_ of the 'base' leads,
since the higher base lead will prevent the diode effect from occurring.
However, when the emitter gets above the higher base lead, and starts to
pass current, that current overcomes the effect of the higher base lead,
causing the thing to continue conducting while the emitter is higher than
the _lower_ voltage base lead. This looks alot like a negative resistance,
since a small increase in current causes an immediate decrease in voltage
required to keep it forward biased.

The typical circuit is a resistor to each of Vcc and Gnd from the bases, and
a cap/resistor on the emitter. When the emitter gets up to .6 above the
higher base, it'll immediately start conducting. That will cause it to
conduct until it discharges the cap/resistor to near ground, at which point
it'll go back into the 'high resistance between the bases' mode, and block
current between the bases again. At this point, the cap/resistor can charge
back up and start another cycle.

Regards,
Bob Monsen