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The efficiency of the LDO is a function of the input and output voltages. $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$$I_{out} \approx I_{in}$ for a series linear ...
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#5: Post edited
by
Nick Alexeev
·
2024-07-06T17:36:19Z (5 months ago)
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
In your case when *V*<sub>out</sub> is 1.2V, and *V*<sub>in</sub> is between 3V and 5V. The efficiency of the LDO will be between 40% and 24%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of significance is decided individually for each end product.- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
- In your case where *V*<sub>out</sub> is 1.2V, *V*<sub>in</sub> is between 3V and 5V, the efficiency of the LDO will be between 40% and 24%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of insignificance is decided individually for each end product.
- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
#4: Post edited
by
Nick Alexeev
·
2024-07-06T15:42:07Z (5 months ago)
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
In your case when *V*<sub>out</sub> is 1.2V, and *V*<sub>in</sub> is between 3V and 5V. The efficiency of the LDO will be between 40% and 24%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of significance is decided individually for each device.- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
- In your case when *V*<sub>out</sub> is 1.2V, and *V*<sub>in</sub> is between 3V and 5V. The efficiency of the LDO will be between 40% and 24%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of significance is decided individually for each end product.
- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
#3: Post edited
by
Nick Alexeev
·
2024-07-06T15:36:55Z (5 months ago)
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
In your case when *V*<sub>out</sub> is 1.8V, and *V*<sub>in</sub> is between 3V and 5V. The efficiency of the LDO will be between 60% and 36%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of significance is decided individually for each device.- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
- In your case when *V*<sub>out</sub> is 1.2V, and *V*<sub>in</sub> is between 3V and 5V. The efficiency of the LDO will be between 40% and 24%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of significance is decided individually for each device.
- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
#2: Post edited
by
Nick Alexeev
·
2024-07-06T15:34:53Z (5 months ago)
- The efficiency of the LDO is a function of the input and output voltages.
η = P_out / P_in = (I_out V_out) / (I_in V_in )I_out ≈ I_in for a series linear regulator (as opposed to shunt regulators, which are a minority). Input and output currents cancel.η ≈ V_out / V_inIn your case when V_out is 1.8V, and V_in is between 3V and 5V. The efficiency of the LDO will be between 60% and 36%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk.[prettier calligraphy for the equations forthcoming ]
- The efficiency of the LDO is a function of the input and output voltages.
- $$\eta = \frac {P_{out}} {P_{in}} = \frac {I_{out} V_{out}} {I_{in} V_{in}}$$
- $I_{out} \approx I_{in}$ for a series linear regulator ¹ ². Input and output currents cancel.
- $$\eta \approx \frac {V_{out}} {V_{in}}$$
- In your case when *V*<sub>out</sub> is 1.8V, and *V*<sub>in</sub> is between 3V and 5V. The efficiency of the LDO will be between 60% and 36%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. An LDO with low efficiency can still be used when current is insignificant. Threshold of significance is decided individually for each device.
- ¹ LDO's own quiescent current becomes significant when calculating standby current consumption. On the scale of 100mA current draw the LDO's quiescent current is insignificant.
- ² Series regulators, as opposed to the less common shunt regulators.
#1: Initial revision
by
Nick Alexeev
·
2024-07-06T15:12:45Z (5 months ago)
The efficiency of the LDO is a function of the input and output voltages. η = P_out / P_in = (I_out V_out) / (I_in V_in ) I_out ≈ I_in for a series linear regulator (as opposed to shunt regulators, which are a minority). Input and output currents cancel. η ≈ V_out / V_in In your case when V_out is 1.8V, and V_in is between 3V and 5V. The efficiency of the LDO will be between 60% and 36%. Any half-decent buck converter would have at least 80% efficiency at the expense of parts cost, switching noise, PCB real estate, engineering risk. [prettier calligraphy for the equations forthcoming ]