Voltage Regulator Thermal Dissipation

[NivagSwerdna]

Given a regulator such as NCP1117 providing 100mA @ 5V

If I provide the regulator with say 6.2V, 9V or 12V or 20V etc. how does that affect the thermal dissipation?  i.e. does it get hotter the higher the input voltage?

I looked at the datasheet but can't puzzle that out. I would expect that I should supply nearer to 6.2V than 20V?  (1.2V max dropout)

Thanks in advance

[rdl]

Dissipation in watts for a linear regulator is Vin - Vout x I

The higher the input voltage the more power must be dissipated as heat.

[cdev]

I have a related question about voltage regulation and heat in a GPSDO controller context which I think I'm going to post in its own thread, maybe in the metrology section.

[Zero999]

Dissipation in watts for a linear regulator is (Vin - Vout) x I

The higher the input voltage the more power must be dissipated as heat.

Corrected: you forgot the brackets.

Yes, the higher the input voltage, the higher the power dissipation.

Although it's usually negligible, the current flowing out of the regulator's 0V pin can be significant too. Especialy in low drop-out regulators, when the output voltage is fairly close to the input voltage. This normally isn't a matter for thermal design. It's more important when dealing with power consumption.

[Damianos]

I think that it is better to answer by yourself. It needs just a little thinking.

A linear voltage regulator is a box that somehow observes the output voltage and tries to keep it stable, by adjusting the conductance from input to output. Lets say that someone inside measures the output voltage and adjusts a resistor that the input with the output. This creates a power loss that converted to heat.

The input power is P_in = V_in * I_in
Respectively the output power is P_out = V_out * I_out
Also according to the above description the I_in is about equal to the I_out

The power loss is the difference of the above two. Calculate it.

[NivagSwerdna]

Thanks for the responses.

Looking at a related datasheet I found...

PD = (VIN – VOUT) IOUT

TJ = TA + PD (omegaJA)

and a no heatsink 65 °C/W

I think that probably tells me what I need to know.

Thanks