122-28176 Parallax Inc, 122-28176 Datasheet - Page 74

• Manufacturer Part Number: 122-28176
• Description: GUIDE STUDENT PROCESS CONTROL
• Manufacturer: Parallax Inc

Specifications of 122-28176

• Accessory Type: Manual
• Product: Microcontroller Accessories
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• For Use With/related Products: Propeller Education (PE) Kit
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant, Contains lead / RoHS non-compliant

Saturation Region:
As shown, there must be some limit to how much current can be developed in the
collector. The current limit is based on the supply voltage and the value of R
called saturation current (I
At saturation, when the transistor is in full conduction, V
the transistor will be conducting as much collector current as possible based on the
restriction of R
emitter junction of the transistor will always drop a small amount of voltage, typically 0.3
V. Therefore, I
Over a current range of 0 mA to 4.77 mA, the transistor will be in the active region. With
an h
Any base current above this value will cause the transistor to be in saturation and at
maximum current.
Consider an increase in the value of R
200, calculate values for I
Would the potentiometer require more or less voltage to drive the transistor into
saturation? Since 10 times less current is required, 10 times less voltage is required at the
base to drive the transistor into saturation.
Transistor Power Dissipation
Power is the work performed by a device or system per unit of time. In electronics power
is measured in watts. Light bulbs are devices we commonly purchase based on the power
output, such as 60 watt, 100 watt or even 200 watt bulbs. A light bulb's power is in the
FE
of 200, control is defined over a range of 0 mA to 23.5 µA for I
I
I
I
I
I
SAT
SAT
B
SAT
B
= I
= I
= Vdd/R
= (Vdd − 0.3 V)/R
= (Vdd−0.3 V)/R
C
C
/h
/h
C
SAT
FE
FE
(fully-on or acting as a closed switch). At saturation, the collector to
will be slightly less.
= 4.77 mA/200 = 23.5 µA
= 0.477 mA/200 = 2.35 µA
C
= 5 V/1 kΩ = 5 mA.
SAT
SAT
).
and I
C
C
= 4.7 V/10 kΩ = 0.477 mA
= 4.7 V/1 kΩ = 4.77 mA
B
at saturation.
C
to 10 kΩ. Based on a 5 V supply, with an h
CE
will be at the minimum, and
B
.
C
and is
FE
of