Can anyone explain to me how you find the appropriate resistor values for R1 and R2? I'm not sure how to do that.
Thank You
Thank You
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If the LED is going to use 10ma then
So (5V  0.6V (BE drop)) / 0.001A = 4400Ω ≈ 4.7KΩ
Can anyone explain to me how you find the appropriate resistor values for R1 and R2? I'm not sure how to do that.
Thank You
Oh ok, makes sense, sorry, mistook what you were trying to say there. Yeah, getting soaked. Just got home yesterday from 2 weeks in Sunny Florida so kinda a bummer to come home to rain and winter weather in the forecast...Base current is 1/10 the collector current. I was figuring the LED took 10ma, so the base took 1ma, hence 0.001A.
The weather dumping on you yet?
Where in FL?Just got home yesterday from 2 weeks in Sunny Florida so kinda a bummer to come home to rain and winter weather in the forecast...
I realized that could be confusing, so I added a clarification say that Vcc in the formula was your base current supply voltage, and not the collector of the transistor in question.To Sarge, thanks for the further explanation. When you say Vcc (supply voltage), is that the 12V main voltage?
No, sorry for creating the confusion. The load voltage doesn't come into consideration for calculating Rb.If I have 1 LED at 20mA 2Vf, I would have 122 = 10V/.020A = 500Ω current limiting resistor for the LED. Would the Vcc then be 10V?
Assuming 10V, Rb = (Vcc  Vbe(sat)) / (Ic/10)
Rb = (10  0.8) / (0.020/10) Vbe = .8V at 0.020A and 25*C
Rb = (9.2) / (0.002)
Rb = 4600Ω
That right?
To answer your Q about where in FL, 2 weeks with my mother in Port Charlotte. Had a good time, went fishing several times, and caught some very nice grouper. Tasty!Where in FL?
I realized that could be confusing, so I added a clarification say that Vcc in the formula was your base current supply voltage, and not the collector of the transistor in question.
No, sorry for creating the confusion. The load voltage doesn't come into consideration for calculating Rb.
In your case:
Rb = (5v0.8v) / (20mA/10)
Rb = 4.2v/0.002
Rb = 2,100 Ohms
Standard resistance table: http://www.logwell.com/tech/components/resistor_values.html
2k and 2.2k are standard E24 values. You could use either.
4.2v / 2k = 2.1mA
4.2v / 2.2k = 1.91mA