For 5V/33 Ohm case(worst case in three load conditions), power dissipation will be around 758mW.(P=VI)

Why didn't you answer the entire question?

 

Why didn't you answer the entire question?

Vce sat @ 150mA for 2N2222 is 0.4V

Transisor dissipation will be (Vce * 152mA)= 60mW

The dissipation across collector resistor is [(15-(5+Vce))V * 152mA]= 1.459W [2.22W during 0V at load]

Load dissipation is 5V * 152mA=0.760W

So I will prefer 2W ratings for 22 Ohm(or 1W/47 Ohm may be replaced with) and 1W ratings Loads with 2N2222 as driver. 1/8W for 1K as less portion of 152mA flows through it.

 

Transisor dissipation will be (Vce * 152mA)

You asked about the collector resistor, also.
You need to calculate numbers. Equations are impressive, but they don't tell you how hot a part will get until you put numbers in place of variables. I think you know that.

I am not your calculator.

 

You asked about the collector resistor, also.
You need to calculate numbers. Equations are impressive, but they don't tell you how hot a part will get until you put numbers in place of variables. I think you know that.

I am not your calculator.

Cool guru
I have edited my previous post...

 

Vce sat @ 150mA for 2N2222 is 0.4V

Transisor dissipation will be (Vce * 152mA)= 60mW

The dissipation across collector resistor is [(15-(5+Vce))V * 152mA]= 1.459W [2.22W during 0V at load]

Load dissipation is 5V * 152mA=0.760W

So I will prefer 2W ratings for 22 Ohm(or 47 Ohm can be replaced with) and 1W ratings Loads with 2N2222 as driver. 1/8W for 1K as less portion of 152mA flows through it.

If the transistor is saturated, your circuit won't work. You need to select the collector resistor so that, at maximum emitter voltage, Vce will be at least 1 volt or more. If you make Vce too high, the transistor will get too hot. You need to calculate the dissipation over the entire range of emitter resistance and emitter voltage. It can be higher with less current and more voltage.