In one of our design we have manufactured LED strip PCB.
In the design, there is LDO to regulate input voltage to 7.5VDC(V_MBI) from 24VDC(VIN) to feed an IC.
F4 : 48V 120mA OZCJ0012FF2E as poly fuse,
D9 : SMAJ26A as TVS.
C7 : CC0603KRX7R9BB104 0.1uF 50VDC 0603 MLCC capacitor...
I am designing a custom pcb with a 3.3V and 5V LDO power regulator that converts 24V DC input to 3.3 and 5V to power a microcontroler , ethernet modbus transcievers, and some sensors. The whole system drawns maximum 200mA of current during operation
5V LDO = L78M05ABDT-TR
I'm trying to understand how a LDO topology works. It's in NXP IC's datasheet:
It has the control circuitry inside the IC with external transistor.
VDD output = 1.8V
Collector supply = 2.367V
NPN: Nexperia PBSS4620PA
L1: ferrite bead 26 Ohm @ 100MHz
C1 = 470nF
Rp = 1 Ohm
Rs = 0.15...
I need a 5.0V-3.3V for 2-2.5A load consumption, do I need a switching regulator or LDO just fine? I prefer the simplification of LDO design but am not sure if 2-2.5A load current needs better efficiency from a buck regulator.
I am, for several reasons about building a adjustable LDO. Some of the specifications are:
- min input voltage 4.4V
- max output voltage 3.6V
- 500mV dropout, due to sense resistor
- wide current output range
- at least 20kHz bandwidth @ 10uA load, ideally 100kHz
Pass element shall...
I am planning to design a current amplifier using an NPN BJT Transistor with an Emitter Follower configuration. On top of that, instead of using batteries, I am using 2 separate LDO voltage outputs as the input voltage(Vin/Vbb) to the BJT as well as the supply voltage...
I'm just now getting into creating PCB's as a hobby, and created a little board that simply consists of a BLE module powered by a tiny Lithium polymer battery (40mah). Since the max operating voltage of the module is 3.6V, I also have an LDO that will drop the voltage down to 3.3V...