# transistor bias

Discussion in 'The Projects Forum' started by samuel.whiskers, Jun 9, 2014.

1. ### samuel.whiskers Thread Starter Member

Mar 17, 2014
95
2
The following schematic snippet shows a 555 output driving a pair of transistors (are they still a Darlington pair even if different??) - the load connected to the collector of the TIP3055 is a rocket ignitor. This and the leads attached to it have a combined resistance minimum about 1 Ohm.

I do have a working version of this, it ignites rockets no problem... From memory the bias resistor on the BC547 is 50 Ohm (I can't recall) - at the time I chose this value, as a higher one seemed to not allow the TIP3055 to conduct fully (I think!)....

What I was after was an educated/calculated approach to how to calculate the best bias resistor here.

The aim is for the TIP3055 to switch fully on, and allow maximum current into the load. Of course the load disintegrates a fraction of a second after ignition....

Lee

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2. ### MrChips Moderator

Oct 2, 2009
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For a single BJT switch, a Ic/Ib current gain of 10 is the general guideline.
Hence if the collector current is 12A, we need to supply the base of the BJT with a base current of Ib = Ic/10 = 1.2A

Your circuit shows two transistors configured as a Darlington pair. If we use the same rule of thumb of a current gain of 10, then we need to drive the first transistor with 0.12A.

If the 555 timer circuit is powered from 12V, then the required current limiting resistor on the output of the 555 (pin 3) is 12V/.12A = 100Ω.

Remember, the purpose of the resistor is to protect the output of the 555 and the driven transistor from too much current. Increasing the value of the resistance will lower the drive current and may or may not prevent full saturation of the driven transistors.

Going in the other direction, the 555 can provide 200mA. Hence going as low a 50Ω should not hurt it.

Last edited: Jun 9, 2014
3. ### AnalogKid Distinguished Member

Aug 1, 2013
4,290
1,173
The BC547 is not a very strong driver transistor, and a 100 ohm base resistor would mean 100 mA of base current. The part is rated for this under some conditions, but if applied continuously it would be 0.06 W of power in the base junction. That's a lot.

Is the 555 running as a monostable or free-running astable? Either way, what is the "on" time?

ak

4. ### wayneh Expert

Sep 9, 2010
11,910
2,854
Correct me if I'm wrong, but I believe this also means that the current through the igniter wire would have to approach 2A in order for the current at the 555 output to approach 200mA. If the igniter wire is reliable as a fuse - meaning it will always "blow" long before it reaches 2A - then I would argue that you don't need any resistor at all. There is almost no risk to the 555 or the transistor. A 2A fuse on the 3055 collector circuit might be appropriate to protect against shorted leads.

5. ### BobTPH Active Member

Jun 5, 2013
768
107
No, you can push too much current through the BE junction without even having the collector connected.

Bob

6. ### samuel.whiskers Thread Starter Member

Mar 17, 2014
95
2
Monostable - and really only for a fraction of a second in use, from the time the current is applied to the time the igniter burns..... It does more work on the test bench, when I change the timing a little so I can see what is happening with an ammeter.... I simulate the load with alligator clips and a 1 Ohm (10W) resistor - the resistor and the TIP3055 get a little warm after a while.

7. ### samuel.whiskers Thread Starter Member

Mar 17, 2014
95
2
The ignitor burns like a fuse - near instantaneous for all intents and purposes.... The TIP3055 is rated to 15A - I have the 555 set to deliver a 0.5s pulse - the igniter leads are fairly long and have a resistance of near 1 Ohm - so I thought even a dead short would be ok, and have done so in testing with no probs.

8. ### AnalogKid Distinguished Member

Aug 1, 2013
4,290
1,173
The 3055 collector current ends when the ingiter blows, but the BC547 base current is there for the entire length of the 555 pulse. If you want to turn it on hard but then back off for the remainder of the pulse, increase the base resistor and put a capacitor in parallel with it.

ak

9. ### wayneh Expert

Sep 9, 2010
11,910
2,854
Well now I'm confused (again). I was having trouble understanding a circuit posted on this forum a while back and, after reading up on transistors, made sense of the circuit by reasoning that the maximum base current depends on the collector current. That was news to me but that's what I read. I had previously thought that, as you just said, the base current can be too high no matter what the collector-emitter current is.

Now I need to go back and look at that other circuit, and try to find that reference again. Nuts.

10. ### MrChips Moderator

Oct 2, 2009
12,228
3,282
You're putting the carriage before the horse, or the effect before the cause.

The relationship between the base current Ib and the collector current Ic is

Ic/Ib = beta, where beta is known as the DC current gain

or Ic = beta x Ib

The base current Ib is the cause and the collector current Ic is the effect.
Thus Ic will flow if there is sufficient Ib. Ib controls Ic, not the other way around.

Note that the equation Ic = beta x Ib is not an equality.

As Bob has pointed out, you can have the collector disconnected and still have a positive base current Ib.

The base current is very similar to the current flowing in a forward biased PN junction, i.e. a rectifier diode. Hence the base current Ib is determined by the base source voltage and the current limiting resistor in series with the base.

11. ### AnalogKid Distinguished Member

Aug 1, 2013
4,290
1,173
Here is another part that might help. In a typical transistor circut with a resistor in the emitter circuit, base current causes collector current, both flow through the emitter, and the ratio s the gain of the transistor. Up to a point.

When a transistor saturates, the collector current can no longer increase in pace with the base current. This can happen with a resistance in the emitter, but is most common when the transistor is intended to act as a non-linear switch. Even though this thread involves a darlington configuration, the result is the same. With the emiter bolted to GND, the transistor still doesn't saturate until the base current times beta exceeds the available collector current. This usually happens at very low base currents, but the base still is just a diode (or a pair of diodes) bolted to gnd.

ak

12. ### samuel.whiskers Thread Starter Member

Mar 17, 2014
95
2
Thanks for everyone's input/discussion - I've certainly learned a few things about transistors I didn't previously know.....

The data I'm looking at suggests Ic of 100mA for the BC547 - perhaps I should revise this transistor.... would be annoying to have to fix if it dies!
I've got a fair few transistors these days - is there any reason I couldn't just replace it with a BD139??

13. ### BobTPH Active Member

Jun 5, 2013
768
107
An SCR might be a better device to fire an igniter. A short pulse will trigger it and it will stay on until the igniter burns out. Back when I was doing model rockets, that is what I used.

Bob