Calculating resistor values in this circuit.

Never mind the 555 and everything to the left of it - have a good understanding of that part.

But I don't understand hoiw the designers of the circuit (below) have come of with the resistor values on the base of the BC327 and the base of the Darlington.

I have had a stab based on a 555 puting out from 50 to 100mA but I have not come remotely near the resitance values they did or the base saturation voltages of the transistors.

So I need a really basic lesson on how to do this and what known values you start off with.

Have had a go at using a BD140 and 2SD2348 from a television.....actually I joined two of the latter as a darlington pair. I just used the resistor values in the below circuit but obviously it didn't work with these different resistors.

Apart from the maximum ratings there doesn't seem to be an aweful lot of difference between the saturation levels of BC327 / BD140 and the darlington / 2SD2348

I get that the idea of the resistors is to set the base current and voltage at the saturation level of each transistor. But I would dearly love to know how to do this for myself with any transistors I might like to try.

One problem is that the collector load for Q2 is a transformer. This will have a very low DC resistance ( .5 ohms?) so the output stage will have a potentially huge current before it saturates. Lets presume an output current of 5A, the darlington transistor will have a Hfe of 1000 or so, so the base current ot get this is 5/1000 = 5mA, as there are two base emitter junctions, it also need 2 X .8 = 1.6V. So if Q1 is saturated, then it will drop .3V (C->E),. Therefore the amount of current will be ( 12 - .3 - 1.6) = 10.1, the resistor is 150 so the current is 10.1/150 ~ 60 mA. Which is more then enough according to my guess of 5 mA!!! When the output of U1 is at earth , the voltage drop across R4 is 12-.8 (VB->E Q2) = 11.2, R = 2K2, so the current is 11.2/2.2 = 5mA base current which is more then enough to saturate a transistor providing its gain is > 60/5 = 12X.
Frank

With respect to the other version of this circuit I am working with that uses a complentary BJT totem pole on the 555 output.....

It seems that every where I look about driving a mosfet gate with a simple circuit, such as the 2 complemntary BJT totem pole that I have tried, inevitably one or more of the experts seem to give the advice that it just aint practical to do so in place of a specific mosfet gate driver IC.

I have pretty much exhausted all the simple possibilities I can think of. The totem pole thing works but it is driving the mosfet at a gate voltage barely above its threshold. Boosting the voltage with an op amp doesn't work because it is presumeably distorting the square wave so much that it will no longer drive the flyback transformer. And building amplfiers that minimise distortion is an entire challenge in itself for me that I am not prepared to take on just yet.

So I have bitten the bullet and purchased some of these: **broken link removed** Looks as though they will do the job nicely.

Found them on ebay in the US but then I ended finding an Australian supplier of them - brilliant.

Still want to figure out this transistor version of the circuit however so I can slot in any transistors I like.

- - - Updated - - -

http://www.allaboutcircuits.com/vol_3/chpt_4/10.html

Just found this regarding bjt base resistors

But what is the beta symbol? I am assuming it is hFE in the datasheets?

Also found this calculator: http://kaizerpowerelectronics.dk/calculators/transistor-base-resistor-calculator/#comment-14103

I have sort of gotten the picture from the previous reply that the idea is to set the base current of each transistor in the circuit to that transistor's base saturation level. Allowing a too high current does not acheive anything and probably damages or interferes with the operation of the BJT. Correct?

But one thing about this Darlington based circuit. What if the 555 voltage pulse has a peak of about 9V? That exceeds the 5V maximum base voltage of BC327. Surely you would need an appropriate voltage divider to bring this down a bit below 5V?

" What if the 555 voltage pulse has a peak of about 9V?". Because the emitter of Q1 is taken to the +12V line if the most positive level from the 555 is 9V, the base current of Q1 = 12- .8V -9V = 2.2V so the base current of Q1 = 2.2//2K2 = 1mA, which means that Q1 will not be cut off and the darlingtons will not be cut off. It would be better if some diodes are inserted in series wiith the base of Q1 artificially raiising its Vbe to 4V, so providing U1 has an output of > 8V, Q1 will be cut off.
Sorry about the confusion about Vcebo, its 40 years since I went to college and the memory fades :-(
Frank

So you are telling me that the voltage at the base of a BJT is always relative to the the voltage at the emitter regardless of what the 555 is pumping out? Presumeably due to the solid state physics of the BJT, which I am not going to try and get my head around at this stage. Would expain a lot about readings I have been getting with my multimeter. To my novice mind "Q1 = 12- .8V -9V = 2.2V " seems counter intuitive and it should be 9V - 0.8V. By the way aren't all BJTs supposed to drop a voltage of 0.7 when open?

Why a diode and not just a smaller value resistor? V=IR....actually it would be a larger value resistor rather than a smaller value one wouldn't it.....due to a voltage divider effect between the resistor and the BJT?

Might also explain why when I wired this kit up it seemed that the darlington transistor was constantly turned on, since the BC327 was not being turned off, and it wouldn't drive my flyback transformer. In the actual kit there is a 10R resistor in series with the 1n4004 rather than the voltage regulator as in the simulated circuit above. When I took a voltage reading I think it was about 9V but it was with my analog multimeter and it is not all that accurate these days - had it for about 20 years or so. Will repeat with the new DMM I now have.

Perhaps the problem of the circuit not working is due to the fact that the car battery I am using is not quite putting out 12V - its an old crappy one. This would probably lead to the problem due to "12- .8V -9V = 2.2V" actually being "11- .8V -9V = 1.2V ".

Well any way you have given me some new avenues of investigation into my problem with the darlington circuit - thanks.

Yeh!, badly expressed, in the forward direction, the base is always between +.6 and + .8 of a volt with respect to the emitter on a NPN BJT. On a PNP, such as your Q1, its always negative. So the emitter is nailed to + 12V, so the base is .8 V negative i.e. 11.2 V, but your 555 output is even more negative at 9V, so there is a voltage drop of 2.2V driving current through the 2K2 resistor, giving an Ib of 1mA.
If you just change the resistor, say from the 2K2 to 220K, the Ib would still be 2.2/220 = 1/100 mA, which is better but the gain of Q1 might be 100, giving a Ib for the darlington of 1mA, X the 1000 gain of the darlington = 1 A, quite a lot for a transistor that is mean't to be off. And when the 555 output goes to earth the voltage drive to Q1 base is 11.2V, and the current is now 11.2/220 ~ .2MA, which is a bit low for the darlington to saturate.
What I would do is to check the output of the 555, to make sure that it is working and not broken, i.e. it goes to 80% of the +12V (or whatever) and not only 5V. Suppose its 8.2V. Now at this voltage you want to make sure Q1 is off, so its Vbe must be less then .6, say .4V. So if the base is at 8.2, the emitter must be less then 8.6V. So if you put a 100 ohms between the emitter and the +12V, its got to drop 12 - 8.6V = 3.4V, so the current through it must be 3.4/100 = 34 mA. So connect another resistor between the emitter and the 100 ohms, to earth, this resistor must drop 8.6 V @ 34 mA, so its value is 8.6/.034 = ~250 ohms. Or much better use enough diode in series with the base to make sure that the 3.4 V cannot turn them on, 4 X .8 = 3.2V too low, 5 X .8 V = 4V which is OK.
Frank