Saturated NPN transistor

Discussion in 'Homework Help' started by TsAmE, Apr 19, 2010.

Apr 19, 2010
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Question: Design a simple logic inverter with an NPN transitor and a 1k collecter pull-up resistor. Choose the base resistor (E12) as large as possible (but as to ensure that the transistor saturates when the input is at + 5V)

My answer: I collecter = V / R collecter = 5 / 1 = 5 mA

i base = (Vin - 0 .7) / R base

B = i collecter / i base
i collecter = B * i base
i collecter = B * (Vin - 0.7) / R base
R base = i collecter / B * (Vin - 0.7) = 5/ (50 * (5 - 0.7)) = 0.02k

Right answer: With 1k pull-up saturation current will be (5-0) / 1k = 5mA so minimum base current required is 5mA / 50 = 100 micro A. so (5-0.7)R base >= 100 micro A, so R <= 4.3 / 100 micro A = 43k. We much therefore choose the next lower E12 value i.e. 39k.

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2. SgtWookie Expert

Jul 17, 2007
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1,793
I don't know who told you that saturation is a forced beta of 50, but it's normally specified as Ic/10.

Since there is a 1k pull-up resistor on the collector, and Vcc=5v, the highest collector current possible (if Vce could be made = to 0v) is (5v-Vce)/1k Ohms, or (5v-0v)/1000 or 5/1000 = 5mA.

The standard calculation for base resistor for a saturated transistor is:
Rbase = (Vin - Vbe) / (Ic/10)
where:
Vin = the voltage applied to the left side of the base resistor
Vbe = the voltage from the base to the emitter; typically anywhere from 0.63v to 0.9v. 0.7v is a common value used.
Ic = desired collector current.
Rbase = (5v - 0.7v)/(5mA/10)
Rbase = 4.3v/0.0005 A
Rbase = 8.6k Ohms. That is not a standard E24 value of resistance. Normally you would look at a chart of standard values, and select the closest resistor.
Here is such a chart:
http://www.logwell.com/tech/components/resistor_values.html
You will see that 8.2k and 9.1k are the closest standard values. In this case, it would be OK to use either.

3. Jony130 AAC Fanatic!

Feb 17, 2009
4,496
1,266
You did mistake in this
R base = i collecter / B * (Vin - 0.7) = 5/ (50 * (5 - 0.7)) = 0.02k

Rb =[ B*(Vin-Vbe)] / Ic = 215/5mA = 43K