ADA4897: power dissipation calculation

Document created by analog-archivist Employee on Feb 23, 2016Last modified by analog-archivist Employee on Feb 23, 2016
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I have a question about the power dissipation calculation in the ADA4897-1
In the “MAXIMUM POWER DISSIPATION” section, there’s the following formula on
power dissipation. Would you like to explain the definition of “Total Drive
Power” and why the voltage is Vs/2 in the total drive power calculation.
And if the Vout is nearly equal to Vs because of R-R output, “Vs/2*Vout/RL- 
Vout^2/RL” will be negative value. I thinks this is impossible. Would you like
to give explanation on this? (Figure 1)


Please refer to the scanned diagram for the enlarged output stage. The total
drive power is the power driving through the pnp transistor in the output
stage. At the midpoint of the two output transistors (where I marked Vout), 
the DC bias of the transistor is always Vs/2. Therefore, there will be
(Vs-Vs/2) = Vs/2 Volts across the transistor. To calculate the current across
the pnp transistor, we know that iT (current of transistor) = iR(current
through resistor) and that’s why we use Vout/R to calculate the transistor

I believe at the point you are mentioning…that when Vout is very close to Vs,
the first term Vs/2*Vout/RL will still be bigger than Vout^2/RL. This is
because Vs/2 will always be bigger than Vout. (See the example in the attached

Also, this equation is only valid if Pd does not exceed the maximum power at a
particular temperature. If Pd is greater than the maximum power, it will damage
the part. For maximum power vs. temperature, you can refer to the graph on the
Maximum Ratings page.