saturation vs active mode energy dissipation question

circuitashes

Joined May 13, 2008
37
I read somewhere on this site that a transistor in saturation mode dissipates less energy than a transistor in active mode. I would like to know why. Please help as soon as possible.

AlexK

Joined May 23, 2007
34
The dissipated power for BJT (assuming DC operation) is approximately

P = Vce * Ic

And for a saturated transistor Vce is always lower compared to one in the active mode.

Papabravo

Joined Feb 24, 2006
13,972
There are actually two things to consider. As the previous poster has indicated, power dissipation is the product of current and voltage. In the linear region with current and voltage in the middle of their range on the load line, power dissipation is greater than in either saturation or cutoff.

To prove this to yourself, draw a straight line with any negative slope from the vertical axis to the horizontal axis. Now as you move along the line multiply the x-coordinate by the y-coordinate and see where the maximum occurs along the line.

If you know any calculus, then ask yourself where the maximum of the function
Rich (BB code):
 y = x*( m*x + b) = m*x^2 + b*x

for  m < 0, b > 0

where m*x + b is the equation of the load line in a CE configuration.
will occur

Last edited:

hgmjr

Joined Jan 28, 2005
9,029
To elaborate on AlexK's comments, Vce for a BJT is generally approximated at around 0.2 volts. A BJT biased into the active region might have at least an order of magnitude larger Vce. Between the active mode and the saturated mode Vc can change dramatically but the current will often only change by a factor of 2 for a BJT biased to place its Vc halfway between the collector supply and ground for example.

Using the formula that AlexK has already provided, that means that in the active mode the BJT will dissipate say 2*i Watts and at saturation it will dissipate 0.2*2*i or 0.4*i Watts.

hgmjr

circuitashes

Joined May 13, 2008
37
Thanks guys - Alexk,Papabravo, and hgmjr. Comments were really helpful. Out of curiousity, i would like to know: if Vbe = 0.7 and Vce = 0.2, would it be right to say
Vcb = Vce - Vbe = -0.5 volts?

Caveman

Joined Apr 15, 2008
471
That's right.

Papabravo

Joined Feb 24, 2006
13,972
Thanks guys - Alexk,Papabravo, and hgmjr. Comments were really helpful. Out of curiousity, i would like to know: if Vbe = 0.7 and Vce = 0.2, would it be right to say
Vcb = Vce - Vbe = -0.5 volts?
You must also remember that if voltage drops are positive then a votage rise is negative. So the base is 0.5 volts above the collector and that by symmetry Vbc is the additive inverse or 0.5V