boylesg
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Be careful with this drive circuit. Q1 may not switch off with all types of 555 timers. The older TTL type will only go high to within about (1.7-1.5V) from Vcc. In fact it can be as high as 3V with manufacturing spread. This will cause problems with turning off Q1 because the base will still be more than 0.5V lower than the emitter and thus keeping the transistor on when it should be off.
First you calculate the base resistor for the BU941P based on your desried maximum emitter current.
... doesn't the base resistor for the BU941P effect the emitter current you get from the BC327?
So presumably I have to raise the output voltage of the 555 or lower the emitter voltage of the BC327.......a voltage divider on the emitter some how?
Sorry to harp on about this but I am just trying to ensure that I have the right idea. I did medical science at uni and have been trying to teach myself this stuff from behind the eight ball a bit. Although I have done C++ programming at Latrobe University which came with a bit of digital electronics....... which sort of prepares you a tiny bit for analogue electronics.Base-emitter current.
It affects the BC327's collector-emitter current. (Check your dictionary and compare affect and effect.)
The simplest thing to do is add a couple of diodes, or a zener, in series with the base and the 555's output pin. With your 12V supply, adding a 4V7 zener, for example, would mean that the 555's output would have to drop below 5-and-a-bit to switch the BC327 on.
You may want to change the series R to suit, so that your desired BC327 base-emitter current figure is maintained.
Really!
Well if I attach the collector of BC327 to an LED + 820R and add a 1000u timing capacitor then I can see the LED flash. So the 555 timer is working with BC327 in that context.
It must be something to do with combining the BC327 with the darlington and its base resistor.
So presumably I have to raise the output voltage of the 555 or lower the emitter voltage of the BC327.......a voltage divider on the emitter some how?
I did medical science at uni and have been trying to teach myself this stuff from behind the eight ball a bit.
What I meant is this:
You choose the CE current that is within the limits of your transistor then plug this, along with input voltages and BE voltage drops and hFE etc, into the appropriate equations and you get the required BE current and the base resistor value required to achieve it given the input voltage.
Then you use the calculated base current of the Darlington as the CE current for the BC327.........
I have the correct idea don't I?
So how come the base resistor for the darlington transistor does not figure in the calculations for the BC327 (in switching mode)?
That resistor is also part of the BC327's CE circuit.
Let's assume the worst case scenario that the 555 'high' voltage is 9V compared to the BC327 emitter voltage of 12V from the rail.
So when the 555 is low the BC327 is conducting from the emitter to the base and into the 555 and it is conducting a larger current from its emitter to its collector......what I call an on state but you seem to be calling an off state. Have I got this right so far?
So you are saying that when the 555 goes high, the voltage level may not be high enough to block the EB current of the BC327 and hence it continues to conduct across its EC. Current flows from high potential to low potential. Am I still OK here?
So if we put a forward biased 4.7V zener on the base of the BC327 then the 555 will only sink the BC327's EB current when the voltage differential exceeds the nominal zener voltage. And that can only occur when the 555 goes low.
I have used normal diodes as a means of clipping a AC / pulsed voltage signal, i.e. in series from the signal line to ground, and have my head around how they work in that situation. But I am have trouble getting my head around how normal diodes would work in this situation. Would you mind explaining it to me?
You would be able to get away 1N4007 or similar at 75Hz. But if you were to use this circuit in the kHz range you would have to use Schottky diodes and have to string more of them together than 1N007s.......which would be far less practical.
Ohhhh. So the Darlington base resistor is includedimplicitly in the calculations rather than explicitly as I was expecting.The BC327's base-emitter current is chosen, with hFEmin in mind, so that the stage will saturate.
Syncopator, can you explain to me how the signal diodes work in place of the zener diode. Because we would not want them to break down in the same way as a zener diode.
So how would these acheive the same end when connected in the same way between the 555 and BC327 base
No we certainly don't.
If we were to connect them in the same way they wouldn't achieve the same end!
This first diagram shows the skeleton output circuit of the 555. Ideally, one or other of the two transistors will be on, the other off, at any one time. And ideally, the output will be either at 0V or, in this case, 12.
But we know that the output won't necessarily be quite at 12V, and may be inconveniently lower, as in the next diagram.
The two easy remedies are shown in the next diagram.
I think they are self explanatory.
I don't know why the design needs Q1 really. The peak current into the base of Q2 is only 60mA. This is well within the capability of a TTL type 555. It would be easier to just alter the duty cycle of the 555 and drive Q2 directly.
Base-emitter current.
It affects the BC327's collector-emitter current. (Check your dictionary and compare affect and effect.)
The simplest thing to do is add a couple of diodes, or a zener, in series with the base and the 555's output pin. With your 12V supply, adding a 4V7 zener, for example, would mean that the 555's output would have to drop below 5-and-a-bit to switch the BC327 on.
You may want to change the series R to suit, so that your desired BC327 base-emitter current figure is maintained.
Check your circuit. You are doing something wrong.
Are you sure you have the zener connected the right way around? The only way Q1 won't switch off is when the base is not pulled high enough.
What happens if you leave the base disconnected? Is the led off then?
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