What calculations do I need to use to get the resistance value required to protect a transistor from burning out which is in full saturation mode if the load is basically a short in both sinking and sourcing a load ?


Is it recommended to use a forward biased diode between a pic output to a transistor base or mosfet gate to protect the pic in case of breakdown of the semiconductor when switching high voltages that would destroy the pic layout would like this picoutpin->baseresistor->diode(<|)->base or gate ?


Why do pnp mosfets or transistors source a load current more easily than a npn one does to sink?

 

Here is the answer Sgtwookie provided me. Since you are shorting something, the Transistor will burn up. BJT's use current for saturation, If you limit the collector current to 800ma(for a 2n2222) , then the following answer will apply.

"When you are using a transistor as a saturated switch, the basic rule of thumb is that the base current needs to be 1/10 of the desired collector current.

A generalize formula that you can use is:
Rbase = (Vsupply - Vbe)/(Ic/10)
where:
Rbase is the resistance in Ohms needed to limit current flowing through the transistor's base.
Vsupply is the voltage available to source current to the base.
Vbe is the voltage difference from the base to the emitter when the transistor is saturated. For most purposes, you can use 0.7v to 0.8v, unless you are operating the transistor near it's design limits. In the latter case, you are better off to change to a more capable transistor.
Ic is the desired collector current.
In your case, you have a 12v supply, and you want 800mA collector current.
So:
Rbase = (Vsupply - Vbe)/(Ic/10)
Rbase = (12v - 0.9V)/(0.8a/10)
Rbase = 11.1/0.08
Rbase = 138.75 Ohms
140 Ohms is the closest standard value.

For Q1, you need 80mA collector current to drive Q2.
So that means you need around 8mA base current. LM339's have a pretty wimpy open-collector output stage; their Vce gets pretty high when you're trying to sink more than around 6mA, but let's try it anyway.
Rbase = (Vsupply - Vbe)/(Ic/10)
Rbase = (12v - 0.8v)/(.08a/10)
Rbase = 11.2/0.008
Rbase = 1.4k Ohms.
That's pushing it a bit much for the LM339, so I bumped it up to 1.5k.
11.2v/1.5k = 7.46mA, which should be enough to saturate the 2N3906. It's not that far off. Had it been 1/20 the desired collector current, you might run into difficulties. "
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Okay i understand better now.
Just to clear things up how would i solve the following for calculating limiting collector resistor value required to protect the transistor if the the load is being sinked on the collector and is a basically consists of a short ? and that goes with the other configuration where the load is sourced shown in the attachment picture.

 

This looks suspiciously like a homework question, but here goes. I have a loose grasp on the subject, and any better qualified posters here, please feel free to chime in. Vcc = 9v, Transistor current = 250mA(max)

9v/.25A = 36Ω

it appears your 1K base resistor at 8.1v limits the base current to 8mA.

As the load drops to 500Ω the CE current drops further, and this is good. I would think the limiting resistor protects the transistor form the danger of a short between the transistor and ground. (good design) The second circuit I think is referred to as a voltage follower, but I never looked beyond Wookie's explanation as to the difference.