LM2907-N: LM2907

Hi Vinod,

The 1200 mW Power Dissipation rating the Absolute Maximum Ratings table is an indicator of the package maximum power dissipation capability, and not the expected internal power dissipation of the LM2907. In actuality, it will be far less than that and a function of the supply voltage, and current running through the IC and current it delivers to the load.

For a first-order approximation you can use the information in the LM2907 datasheet Typical Characteristics curve, Figure 3, as seen below. It appears that for a 24 V supply voltage the supply current current is about 4.5 to 5 mA. Then, you apply the basic power formula that you mentioned: PDISS = Vcc x Icc. So for the simple quiescent case PDISS = (24 V) (0.005 A) = 0.12 W.

The LM2907 in F-to-V mode has the emitter follower output transistor driving an external load. The current that flows through the transistor to the load will increase the power dissipation over that of the quiescent condition calculated above. The actual power dissipation in that transistor will depend on the output voltage level, and the current it is supplying to the load at the output voltage.

The additional power dissipation incurred by the LM2907 when driving a load will be the difference between Vcc and Vo, divided by the load current. It probably won't amount to more than a few tens, or hundreds-of-milliwatts so the power dissipation won't even come close to the 1.2 W package rating.

Regards, Thomas

Precision Amplifiers Applications Engineering

Hello Vinod,

I would not expect the current to be as high as 50 mA, unless the output is driving a low resistance load such as relay or solenoid. You show a 10 k load so the LM2907 current will not be anywhere near that high.

The LM2907 when operating as a F-to-V as you have shown is in a dynamic operating condition, and not a quiescent state so the current and power dissipation will be varying with time. The total operating current will be close to the quiescent current, plus whatever the output load current is at a particular moment.

We saw from the Figure 3, Total Supply Current graph, that the current will be about 4.5 mA with a 24 V supply voltage applied to the LM2907. The worst-case dissipation in the NPN output transistor stage will be when the voltage across the output transistor and 10 k load are equal at 12 V each. The total power dissipation will be approximately:

P_DISS = (V_CC ∙ I_Q) + [(V_CC / 2)² / R_L]

P_DISS = (24 V ∙ 4.5 mA) + [(24 V / 2)²  / 10 k]

P_DISS = 96 mW + 14.4 mW = 110.4 mW

That should be pretty close to actual.

Regards, Thomas

Precision Amplifiers Applications Engineering