transistor as a switch

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ahmedragia21 said:
Roff , how did you calculate the base resistor (3.9K) ?

i think it should be 2.25/1.6mA =1.4Kohm
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The 89S52 can sink 1.6mA. You are not required to run it at that current. The 2N3906 collector current is about 18mA. It is a high beta transistor, and will easily saturate with the ≈1mA of base drive provided by the 3.9k. It will also work with 1.4k, but you will just be wasting power.
 
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thanks , i got this question in my , how the 89S52 sinks current like 1.6mA at 0.4VCC and it should output VOL is zero volt then at the base resistor there should be zero volt so that the PNP turns on , if this is true then how did you estimate your base resistor , you assumed that the base voltage is 0.4VCC should it be 0 volt ?

thanks
 
The exact amount of base drive current to the PNP isn't critical, so it is not important whether VOL is zero, or 0.4V, or somewhere in between. I just assumed the resistor has about 4V across it, and that about 1mA would be sufficient base drive. You could use any value between 2.4k and 4.7k and it should still work. Electronics is almost never an exact science.
 
I just assumed the resistor has about 4V across it

how did you assume that and there's a zero volt in the base ?
im so confused
 
ahmedragia21 said:
I just assumed the resistor has about 4V across it

how did you assume that and there's a zero volt in the base ?
im so confused
Click to expand...
The emitter of the PNP is connected to +5V. The base will be ≈0.7V below that, or about +4.3V. When the 89S52 output is low, it will be at ≈0.3V. Therefore, the resistor will have ≈4V across it.
 
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