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# Transistor Switch Application Circuit

Basically I am reading a transistor switch example from Horowitz Hill book - The art of electronics.

He starts by explaining that there are 4 golden rules for transistors:

1) Collector must be more positive than the emitter
2)base-emitter and base-collector act like diodes
3)transistors have max values for Ic, IB and VCE that cannot be exceeded
4) when rules 1-3 are applied Ic is approximately IB that is (Ic=βIb)

Going deeper into the example he explains when the switch is closed that.

1) base rises to .6V
2) voltage drop across base resistor is 9.4V
3) base current is 9.4mA
4) application of rule 4 leads to Ic = 940mA (for a typical β=100) which is wrong, wrong, wrong! Because rule 4 only holds when rule 1 is obeyed

So here is where things get confusing for me.

If Ic is 940mA how are you breaking the rule that collector voltage is not greater than emitter voltage?

He says to have a current of 940mA can only occur if you would pull the collector below ground. I totally dont understand this statement!

What exactly does this mean?

How would you pull a collector below ground?

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### Replies to This Discussion

I have to take issue with golden rule #1

1) Collector must be more positive than the emitter

This applies for an npn transistor but for a pnp transistor the collector would have to be more negative than the emitter.  So for now let’s just apply these rules for the npn transistor.

If Ic is 940mA how are you breaking the rule that collector voltage is not greater than emitter voltage?

The answer is Ic cannot be 940mA.  Your circuit plainly shows that Ic must be the current of the lamp which is 0.1A or 100mA.  No matter what, that is the limit of your current.  We call this saturation.  When the transistor is in saturation then the formula for Collector Current using Beta times the base current does not apply.  It only allows you to confirm that your transistor is in saturation.