130-28176 Parallax Inc, 130-28176 Datasheet - Page 75

• Manufacturer Part Number: 130-28176
• Description: KIT PARTS PROCESS CONTROL
• Manufacturer: Parallax Inc

Specifications of 130-28176

• Accessory Type: Education Kit
• Product: Microcontroller Accessories
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• For Use With/related Products: Board of Education Full Kit
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant, Contains lead / RoHS non-compliant

amount of light that is produced (lumens) as well as the heat produced and dissipated to
the air around it. Any device that has current flowing through it and a voltage drop
across it coverts electricity into power. This power may be useful work (light, motion) or
heat to be dissipated – which could also be useful at times, such as your clothes dryer.
When the power takes the form of heat, it must be dissipated from the device either by
convection to the air or through other means. The CPU in your computer system
consumes a large amount of power, and if the heat is not removed, damage to the CPU
will quickly occur. Heat sinks provide heat conduction from the device, a greater area of
cooling and often fans are added for forced convection to remove heat more efficiently.
Transistors, such as the ones in your CPU, have current flowing through them and have a
voltage drop across them. Since a transistor operates in 3 distinct areas, when is the most
power consumed?
Maximum power is used by the transistor when it is mid-point biased where V
V
For our circuit, the highest power can be calculated:
The maximum power that the 2N3904 can dissipate (P
sinks and fans. We are well within the device’s specifications.
CUTOFF
•
•
When in cutoff, the voltage drop is at maximum (V
minimum (theoretically 0).
When in saturation, the current flow is at maximum (I
minimum (0.3 V).
P
Vdd = 5 V, I
P
and current is ½ I
Q1 Max
Q1 Max
DC Power Dissipation: The power dissipated by an element in a DC circuit is given by the
voltage across the element multiplied by the current flowing through it:
Power = Voltage x Current
P = VI
= ½ (Vdd) x ½ (I
= (0.5 )(5 V) x (0.5)(5 mA) = 6.25 mW
SAT
= 5 mA
SAT
(these occur at the same time).
SAT
)
D
) is 200 mW without adding heat
CUTOFF
SAT
) but the voltage drop is
), but current flow is
CE
is ½