EDIT: Previously sort of deleted my post, but it's back now! I got most of the answer, and I thought I understood it, but then... I was missing something. Here's the circuit, again.


The output from the capacitor connected to the BJT is a sawtooth from 11V to 6V like so:


The general premise is that the current source sinks 100uA. The capacitor is initially uncharged so it's at 12V relative to ground and discharges with the current source until we hit 6V. At 6V, the emitter of Q1 is 0.6V less than the base, the transistor switches on, and the capacitor (somehow) charges back up from being shorted to +11V. The thing I'm missing is the switch. The circuit operates well when the capacitor is not there, and it is steady - there are no shorts to be found. That's what confuses me - I've probed all the points in SPICE and can't seem to understand why the circuit switches to let the capacitor instantly charge to 11V again.

Here are some clues: When the switch happens, it seems like the base-collector junction of Q1 is forward biased. As well, the current spikes up, and then is clamped by the JFET. The current spike appears to leak out to ground through the JFET's gate. But this makes no sense, does it? How can the JFET pass current from its gate? So one might chalk that up to SPICE.
Anywho, I'm stumped on this one. Has anyone seen something like this before? Any help is much appreciated.
Thanks in advance.
Sam Gallagher

 

What was the answer?

 

What was the question?

 

NOTE: Question has been updated. See initial post.
Thanks again,
Sam Gallagher

 

Not trying to answer the whole scenario but, a j-fet has a high impedance gate when operated in its normal fashion because the gate-source junction is reverse biased. If you try to raise the gate to a voltage more positive than the source, it will conduct just like a diode.

 

First, let's simplify. See below.

I have removed the unneeded parts. Now what you have is a version of a programmable unijunction transistor.

Note that the J-FET has been removed. This is because it does not do anything. Why? Because it is a current source in series with another current source. The J-FET has a much higher current when biased with zero volts gate to source than 100 ua.

You asked about the missing switch. The switch is the combination of the PNP and NPN transistors. They form a very high gain amplifier with positive feedback.
As soon as Q3 starts to conduct, its current is amplified by Q1 which increases the current into Q3 which increases current into Q1... Snap, the cap is discharged.

Once the cap is discharged, there is no longer enough current available to keep Q1 and Q3 conducting. Now the current source discharges the cap. When the voltage gets down to voltage set by the resistor divider, Q3 starts to conduct again. Snap.

Alternately, note that Q1 and Q3 act like an SCR which also turns on very quickly with a small input current compared to the load current.

I said that I removed unneeded parts. Actually the diode connected transistor has a purpose. It biases Q3. The bias sets the threshold closer to the voltage set by the resistors and compensates this voltage for temperature changes. I am not quite sure what the diode does. I have never used either of these parts in my circuits.

 

Richard, I couldn't have asked for a better answer! That was great, thanks. You put it all out, and removed the fluff to make it easier. The one part that confused me is the dynamic between the transistors, which I thought would be stable. But I looked harder - it's when it is brought below the base voltage, past the 0.6V point, that the transistor starts to really turn on hard, thus allowing a ton of current, free of charge (pun intended). The transistor saturates; its base starts to short! I get it now. Or otherwise I seem a fool, and need an even better explanation...
Thanks again,
Sam Gallagher

 

You might be interested in this thread about a _very_ fast circuit similar to yours.

http://forum.allaboutcircuits.com/threads/fast-risetime-falltime-pulsers.124518/#post-816114