100nF

LDO dropout and headroom calculator

Dropout voltage is the minimum input-to-output difference at which an LDO still regulates, and it is a resistance times the load: the pass transistor in dropout "is simply a resistor" (TI SLVA079). A part specified at 175 mV dropout at 200 mA is 0.875 Ω, so at 150 mA it drops 131 mV and a 3.3 V output needs 3.431 V in. Enter the output, the load, the datasheet dropout with the current it is specified at, and the input range to get the dropout at the real load, the headroom at the lowest input, and the dissipation and efficiency at both ends of the range.

output against input3.30 V0needs 3.43 VV_IN 3.60 V–5.50 Vdissipation (V_IN − V_OUT)·I_O371 mW0330 mW at 5.50 V45.0 mW at 3.60 Vinput voltage
Fig 1 — 3.30 V out at 150 mA: the LDO regulates above 3.43 V and tracks V_IN − 131 mV below it. Every volt of input above the knee is dissipated in the pass element.
Pass-element resistance, V_DO / I_test
875 mΩ
Dropout at 150 mA
131 mV
Input needed to regulate
3.43 V
Headroom at 3.60 V in
169 mV
Dissipation at V_IN min · max
45.0 mW · 330 mW
Efficiency at V_IN min · max
91.7 % · 60.0 %
Most load 3.60 V in can supply
200 mA (the rated maximum)

How this is calculated

Standard: TI SLVA079, SLVA207, SSZTAC2

VDO=IO⋅RonV_{DO} = I_O \cdot R_{on}
SLVA079 §1: in dropout the pass element is a resistor.
Ron=VDO(spec)Itest,VDO=Ron⋅IloadR_{on} = \frac{V_{DO(spec)}}{I_{test}}, \qquad V_{DO} = R_{on} \cdot I_{load}
SLVA207 eq 1 and 2: the datasheet figure scaled to the operating current.
VIN≥VOUT(nom)+VDO,VOUT(dropout)=VIN−VDOV_{IN} \geq V_{OUT(nom)} + V_{DO}, \qquad V_{OUT(dropout)} = V_{IN} - V_{DO}
SSZTAC2 eq 1 and 2.
PD=(VIN−VOUT)⋅IOP_D = (V_{IN} - V_{OUT}) \cdot I_O
SLVA079 §12; the package limit (T_J,max − T_A)/R_θJA is the thermal calculator's.
η=IOVO(IO+IQ) VI×100 %\eta = \frac{I_O V_O}{(I_O + I_Q)\, V_I} \times 100\,\%
SLVA079 §4 as written. Its printed example omits I_Q in the denominator.
IO,max=VIN,min−VOUTRonI_{O,max} = \frac{V_{IN,min} - V_{OUT}}{R_{on}}
The most load a given input floor can supply before dropout.

Assumptions

What sets an LDO's dropout voltage

Dropout voltage is the smallest difference between input and output at which the regulator still regulates. TI's terms-and-definitions note puts it as "the input-to-output differential voltage at which the circuit ceases to regulate against further reductions in input voltage", and its LDO Basics article gives the rule in one line: the input must stay at least VDO above the nominal output. Below that the loop has nothing left to give — "the output voltage begins to track the input voltage", less the dropout.

What sets the number is a resistance. In dropout the pass transistor is driven as hard as its gate drive allows and, in SLVA079's words, "is simply a resistor", so VDO = IO × Ron. That is why the datasheet gives dropout at a current, and why the figure at the rated maximum is the wrong one to use at a lighter load. SLVA207 makes the correction explicit: divide the specified dropout by its test current to get the resistance, then multiply by the operating current. Its TPS79901 example — 160 mV at 200 mA — is 0.8 Ω, so at 100 mA the specified dropout is 80 mV, not 160.

The resistance itself depends on how hard the gate can be driven, which is why architecture matters. In a PMOS LDO the error amplifier pulls the gate toward ground, so a higher input gives a more negative VGS and a lower Ron: the TPS799's dropout falls as its input rises. An NMOS pass element needs its gate above the output, and as the input approaches the output the amplifier runs out of swing — which is why NMOS LDOs with very low dropout carry a bias rail or an internal charge pump to drive the gate from a voltage higher than the input.

Worked example: TPS799 at 3.3 V, 150 mA, from 3.6–5.5 V

The calculator's defaults. TI's LDO Basics article specifies the TPS799 at 175 mV maximum dropout at 200 mA when regulating 3.3 V, and shows it losing regulation at 3.375 V in with the full 200 mA. At 150 mA from a rail that can sag to 3.6 V:

R_on          = 175 mV / 200 mA                = 0.875 Ω
V_DO(150 mA)  = 0.875 Ω × 150 mA               = 131 mV
V_IN needed   = 3.3 + 0.131                    = 3.431 V
headroom      = 3.6 − 3.431                    = 169 mV      → regulates
P_D           = (3.6 − 3.3) × 0.15             = 45 mW   at 3.6 V
              = (5.5 − 3.3) × 0.15             = 330 mW  at 5.5 V
efficiency    = 3.3 / 3.6                      = 91.7 %  at 3.6 V (I_Q = 0)
              = 3.3 / 5.5                      = 60.0 %  at 5.5 V
most load at 3.6 V in:  (3.6 − 3.3) / 0.875 Ω  = 343 mA  → more than the 200 mA rating

The two ends of the input range ask different questions. The low end is the dropout question — 169 mV to spare here — and the high end is the thermal one, 330 mW that the package has to shed, which is what theLDO thermal calculator checks against RθJA. SLVA079's efficiency formula includes the quiescent current in the input current; its printed example (100 mA, 3.3 V out, 17 mA quiescent) gives 73.3 % at 4.5 V and 82.5 % at 4 V, which are the figures without the 17 mA — with it they are 62.7 % and 70.5 %. The calculator uses the formula as written; enter IQ = 0 to reproduce the printed numbers.

Where the dropout model stops being valid

Common dropout mistakes

Further reading