LDO Efficiency calculation parameter

LDO efficiency calculation output current do we need to consider load current value or LDO maximum current rating?

Ex: I have selected TI# LP38841-ADJ

• Input voltage: 0.95V
• LDO maximum output current: 800mA (datasheet value)
• LDO quiescent current: 32 mA (datasheet value)
• Designed output voltage: 0.85V
• Maximum load current: 4 mA When applying the above efficiency formula, is it necessary consider LDO maximum current (800mA) or maximum load current (4mA) for output current Io?

• Why did you pick this LDO? 30 mA is a really high value for Iq, and it suggests this regulator is not designed for such a light load. Sep 2 '20 at 15:17
• I accept your comment. Now selected with low Iq current ~ 15-25uA to use very low Load current application. I too facing very challenge to select the LDO's due to Min voltage (0.67V). Sep 3 '20 at 11:02

The efficiency is a function of output current, it is not a constant.

So the efficiency of an LDO will, in general, be zero with no load current and maximum at the maximum output current.

So it really depends on at what output load current you want to evaluate the efficiency.

It's not straightforward to come up with a general closed-form solution for efficiency given Iout(t), but if input and output voltages are fixed and your output current is piecewise constant you can simply take a time-weighted average of output current.

For example, Vin = 5.0V, Vout = 3.3V , Iq = 1mA and Iout is 10mA for 10% of the time and 1mA for 90% of the time.

Efficiency with Iout = 10mA is 60% Efficiency with Iout = 1mA is 33%

Average efficiency is thus 35.7%.

• Thank you very much.. Sep 2 '20 at 8:05

Think about what "Efficiency" means and look at the formula. In general Efficiency is defined as: "what comes out" / "what goes in" x 100%

So a circuit that outputs 0.4 W, while 0.5 W is taken at the input has a 0.4 / 0.5 x 100% = 80 % efficiency.

The formula represents this, the output power: $$\I_O * V_O\$$

is divided by the Input power: $$\(I_O + I_q) * V_I\$$ and multiplied by 100%.

Note that this efficiency would always be 100% if $$\I_q\$$ = 0

This quiescent current $$\I_q\$$ is the current what the LDO itself consumes.

LDO efficiency calculation output current is we need consider Load current Value or LDO max current rating ?

Would it make sense to use the maximum current rating?

Now think, suppose that $$\I_q\$$ of the LDO is 0.1 mA.

What would the efficiency be if the load current was also 0.1 mA: $$\I_O\$$ = 0.1 mA?

What would the efficiency be if the load current was higher, 100 mA: $$\I_O\$$ = 100 mA?

Hmm, the efficiency is increased when the load current is higher. To me that makes sense as the impact of $$\I_q\$$ (which is only 0.1 mA) becomes smaller as $$\I_O\$$ gets larger.

Now do those calculations again but use the maximum current rating $$\I_O\$$ = 800 mA

Now what happens to these (wrong!) efficiency numbers?

In general: Maximum values are values that you should not exceed, if you do it can damage the device. So your actual values would usually be smaller than the maximum ratings, you have to do your calculations with the actual values, not the maximum ratings.

• Thank You so much... Sep 2 '20 at 8:05
• Input power should have Vi instead of Vo. And is not always 100% if there’s 0 Iq. Sep 2 '20 at 15:16
• In the datasheet Iq (Quiescent Current) not given directly. How i can calculate or derive the Iq current. Ex part number BD3541NUV-E2 Sep 3 '20 at 11:05
• In the datasheet Iq (Quiescent Current) not given directly Oh yes it is, for the LP38841 (datasheet: ti.com/lit/ds/snvs289c/snvs289c.pdf ) it is on the frontpage: Quiescent Current: 30 mA (typ) at full load. Where did "part number BD3541NUV-E2" come from, what is it? That does not look like an IC typenumber. Instead of mentioning "obsure" typenumbers, include a link to the datasheet. Part numbers are often pointless as they mean only something to specific companies. Sep 3 '20 at 11:13
• This is another part from ROHM, link for Datasheet rohmfs.rohm.com/en/products/databook/datasheet/ic/power/… Sep 3 '20 at 11:41

It may be better to know the absolute worst case power dissipated by the LDO.

When the LDO is has either a higher voltage drop or a larger current the power dissipated increases and therefore the temperature rise of the LDO increases.

If you operate outside the absolute max voltage drop, output current or Trise of the component you are likely to degrade its life and damage the component.

• Welcome to EE.SE, Jason. "Jumping off of the previous post" doesn't make sense on SE sites. StackExchange sites don't work like forums. Answers float up and down by votes received or user sorting preferences. As a result any attempt at using the Your Answer box for discussion turns into a mess. Please take the Tour to learn more about how the site works. Sep 2 '20 at 17:28