cascode mystery

How did the base current during the switch on time using the new methods compare to the base current with just the resistor?

"kell" wrote in news:1143263007.820537.134600 @v46g2000cwv.googlegroups.com:

It may just be me, but I don't see any advantage over just the points in this circuit....

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Doesn't look like a cascode to me! Shouldn't it have two transistors?

Leon

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Saves the points from getting guff from the flyback (in the regionof

300V).

Cascode doesn't imply the use of two transistors. (BTW, The name actually comes from valve/tube circuits.)

In the context of this thread, the bottom switching element is labeled "points" in the diagram.

Regards, Allan

For the linear regulator I did this:

darlington in--,------ -----,----out | \\c \\_/e | | \\__/ | 1K | | | | Radj '-----+---- | \\c | |--+ e / | | 2k2 | | '----+ | ground

With the 10 ohm power resistor I had half an amp running into the base at about nine tenths of a volt. I tried adjusting this regulator for higher voltages and currents than that. It always comes around to the same thing. The way I physically break contact at the emitter matters with the active circuit, but not with the resistor circuit. The power resistor circuit works no matter how carelessly I break contact with the emitter. With the active circuit, I have to take great care to break contact as quick and snappy as I can, or I get no spark. If I press two stiff wires against each other and snap them apart, I can get a spark from the coil. If I just touch them and pull them away, I don't get a spark. With the passive drive, any old way of dragging wires around or brushing them against each other will get a good spark.

Perhaps the 10-ohm resistor provides a way to help the BJT turn off quickly, as its emitter soars upward after the points open, and you removed that base-discharge path from the circuit. Try adding a diode,

. ,-----| | darlington | Rb

--
 Thanks,
    - Win

Could be a lot of things:

(1) You don't have anything drawing charge out of the base. Since you're pumping a good .6 amps in, it could take several microseconds for the charge to dissipate, meanwhile the transistor is still conducting and drawing away spark energy. I'd put a 100 ohm resistor from emitter to base to help draw out the stored charge.

(2) Depending on the size and health of that capacitor across the coil, there might be enough kick to avalanche the transistor. Try using a higher voltage transistor, such as those ubiquitous horizontal-output transistors, good for 1200 or more volts.

First of all, I don't understand what the advantage of such a circuit would be, above a single resistor, as far as power consumption goes.

If you built a switching regulator with a 3 volt output, the base resistor power could be cut in half.

But assuming that you are more interested, right now, in why you are not getting a spark, remember that the spark occurs because the cascode transistor switches off very quickly when the points open, because the collector current detours out through the base (and base resistor). This removes the stored base charge very quickly, so the collector current can be chopped off very quickly. In any other version, you still need a low resistance path for the coil current that appears at the base the moment the points open, in order to have most of the energy stored in the coil come out the spark.

The clue here is the observation of spark on "snap" break of the points simulation versus a sliding break. With the 10R drive, the capacitor discharges through the coil in parallel with the loop consisting of the

10R in series with the forward biased BC junction. There is no parallel path for the Darlington drive because it cuts off at ~0.9V above GND, its control voltage reference. So the 10R drive rapidly discharges C to 0V at which point the coil continues to drive C in reverse as well as reverse current through the saturated CB junction. The transistor recovers quickly allowing the collector node to charge to an indefinitely large voltage clamped by the spark. The Darlington will either zener to GND through the 2x Vbe or a single Vbe to its BC to V+ depending on rating until the power transistor CB junction recovers. However, if there's any conduction Re to GND during this time, due to sliding the wires apart causing bounce, the power transistor BE junction may avalanche into the base circuit causing some kind of sustained conduction condition which depletes the coil energy at low voltage.

Well, I finally got the circuit to work. I built another buck converter and drove the base with about two and a half volts, and this time it works good. Funny thing happens: sever the converter's ground, the damn thing still works nearly as well as it does wired correctly. Weird. Anyway, thanks for all the responses!

Kell