Yes, I know that. The coil I was using in the simulation last night has 2 ohms of resistance. That really helped to reduce the current through the diode(s). Keep in mind, I did not design that circuit, I am only trying to analyze it.
That is probably true. I will leave it to MrAl to confirm it.
Yes, I already said previously that transistor saturation is what turns the diode(s) off.
Even at that high value, the transistor still burns up. But it's naive to think the coil would protect the transistor anyway. Trying to get a high enough value works against other circuit goals.
Absolutely true. No confirmation required.
EDIT: The other "diode" is an arrow indicating the dirction of current. (Yeah, yeah I know the direction of current is already implied, but that's what it is nontheless)
Then there should be no reason to ask why I don't consider the diode. That is the answer.
MrAl,
Let us both quit postulating about what happens, and try to determine what is going on from a circuit analysis program. It has been years since I used one, but yesterday I downloaded LTSpice free for nothing from Linear Technology's Website. I spent a good part of the evening installing it, and getting up to speed using it. I was mostly using generic models that came with the program, and the result were not too good. I showed a startup oscillation of 4 or 6 cycles before Q2 stabilized in the cutoff state. This causes the diode(s) to fry, unless I add some resistance to the coil.
So, what I need from you is some information.
1) What resistance are you using for the coil?
2) How many diodes are there? Your schematic shows a LED and a iLED, which I infer is a infrared LED. Would you clear that up?
3) The program came with a PSpice model of a 2N4403, but not a 2N4401. So could you send me the PSpice models of the components you are using, or tell me where I can download them.
I want to duplicate as close as possible what you are doing. Then when I get to run satisfactorily, we can compare notes.
Until I hear from you.
Ratch
.model 2N4401 NPN(Is=26.03f Xti=3 Eg=1.11 Vaf=90.7 Bf=4.292K Ne=1.244 Ise=26.03f Ikf=.2061 Xtb=1.5 Br=1.01 Nc=2 Isc=0 Ikr=0 Rc=.5 Cjc=11.01p
+ Mjc=.3763 Vjc=.75 Fc=.5 Cje=24.07p Mje=.3641 Vje=.75 Tr=233.7n Tf=466.5p Itf=0 Vtf=0 Xtf=0 Rb=10 Vceo=40 Icrating=600m mfg=Fairchild)
.model 2N4403 PNP(Is=650.6E-18 Xti=3 Eg=1.11 Vaf=115.7 Bf=216.2 Ne=1.829 Ise=58.72f Ikf=1.079 Xtb=1.5 Br=3.578 Nc=2 Isc=0 Ikr=0 Rc=.715 Cjc=14.76p
+ Mjc=.5383 Vjc=.75 Fc=.5 Cje=19.82p Mje=.3357 Vje=.75 Tr=111.6n Tf=603.7p Itf=.65 Vtf=5 Xtf=1.7 Rb=10 Vceo=40 Icrating=600m mfg=Fairchild)
.model NICHIAWHITE20MA D(Is=5.12809n Rs=2.25 N=8.1397 Cjo=42p Iave=20m Vpk=5 mfg=Nichia type=LED)
Ratch....do us all a huge favour......keep your interpretations of junctions to yourself...........
Thirty four pages of stuff that is basically a waste of time.
All the knowledgeable guys here have posted here and there. They have been kind to you. They have had the patience to discuss stuff with you.
They have explained stuff to you. They have proven stuff to you.
Note the word YOU.
How do You feel about it???
If I were you I would never mention the subject again.
A lot depends on how long the transistor is in saturation and what its current is. Like I said before, the diode(s) are just as vulnerable when the transistor is in cutoff.
There are two listed names, LED and iLED, which might be a superfluous current symbol, or a incomplete diode symbol. Only MrAl knows for sure.
I don't understand, the diode is a integral part of the circuit. When I get the simulation going, I will remove the diode and see what happens.
Again, the sequence of events is as follows:
1. Q2 turns on.
2. Current rises in inductor
3. Current exceeds that which Q2 can hold with it's current bias
4. Vce voltage starts to rise
5. Vcb gets higher, transistor comes out of sat.
6. Voltage rises Vce
7. Rising voltage couples through cap to Q1 base, turning it off quick
8. Q1 turning off turns off Q2
9. Q2 turning off gives rise to inductive kick back and high boosted voltage Vce
If you dont agree with that sequence of events then list your own idea of how you think it happens. You really should do a circuit simulation first though because it looks like you havent done that yet.
But how it is possible that ideal coil connect to the voltage source can not longer sustain a rising current?MrAl,
When the 3 volts can not longer sustain a rising current in the coil, the back voltage drops to zero.
Ratch
But how it is possible that ideal coil connect to the voltage source can not longer sustain a rising current?
transistor has resistance too.
For Ib=340uA--->Ic stop rise at 34mA
And in the oscillator Ic stop rise at 33mA , very strange coincidence?
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