The lm350 voltage regulator is 3A DC adjustable power supply IC that high-performance. It can apply voltage output 1.25V to 35V. so have a few components
With the component values shown the circuit is designed to have and output voltage range of approximately 1.25 to 13.5 Volts when measured at the output of the regulator. The calculation on the drawing is used to determine the output voltage for given values of R1 and R2.
Because these regulators have a minimum output voltage of 1.25 Volts diodes D3 and D4 will provide a voltage drop of approximately 1.4 Volts so that zero output to the track can be obtained. These diodes need to have at least a 3 amp rating
The diodes D1 and D2 added to the circuit will prevent damage to the regulator during certain adverse conditions such as the output voltage being higher than the input voltage to the regulator. This can happen if this type of circuit is used as a variable power supply for an electronics test bench and without the added protection of D3 and D4.
A substantial heat sink will be needed for the TTR as quite a bit of heat will be generated when drawing high current with low output voltages. For this reason this circuit is not recommended for use in hand held throttles.
If the less expensive LM 317 regulator is used this circuit makes an excellent test bench power supply just leave out D3 and 4 and add a voltmeter.
LM350 variable power supply
This is a 3A adjustable regulators project using IC-LM350T that look like
In My first Variable DC Power Supply, but this high current higher than 3-Amp. Therefore they cause you can supply more any load circuit. Which we can buy this IC at many store but expensive than LM317.
How LM350 variable power supply works
In Figure 1 you will see that the circuit form same the My first Variable DC Power Supply.
When we apply AC220V or AC110V(for usa) with pressing S1 to turn on this power supply. ACV will flow F1 for protection when overload or too much voltage input.
Then the ACV will flow in to transformer that have the ability to transform voltage and current to lower levels of AC-18V and next to BD1-bridge diode to convert AC to DC.
Next they will through C1-4700uF electrolytic capacitor smooth (filter) the pulsating voltage from a transformer into a steady direct current (DC).
Now we have voltage at this point is 22V to 25V
And then, the current will flow to input lead of IC1-LM350T
Which it is a adjustable regulator IC that designed to many supply for 3Amp output and adjustable over 1.2V to 33Volt, and with current limiting, thermal shutdown, full protection.
This circuit form I see in datasheet,which there are simple detail to do this:
– In normal operation, the LM350T have nominal 1.25V reference voltage (Vref) between the output lead and Adjustable lead (ADJ). This voltage will across R1+R2 (120 ohms+120 ohms = 240 ohms as datasheet) and,
– since the voltage is constant, a constant current then flows through the output set resistor VR1. To adjust voltage output.
-C1,C4-0.1uF is input bypass capacitor, it need if devices (IC1) more than 6 inches from filter capacitors.
-C3-47uF is bypass capacitor prevent 86dB ripple rejection.
-C5- 100uF is used to improves transient response. Output capacitor in the range of 1uF to 1000uF of tantalum electrolytic are commonly used to provide improved output impedance and rejection of transients.
When external capacitors are used with any IC regulators, sometimes necessary to add protection diode to prevent the capacitors from discharging through low current point into the regulator.
Although the surge is short, there is enough energy
to damage parts of the IC.
When negative voltage or 20A spikes flow to backward the output it will be absorbs voltage with D3-diode.
And Then D2 to protect Out and Adj lead.
And D1 is protected voltage spikes to Input and output lead.
IC1____LM350T___3 terminal positive voltage regulator 3Amp_____= 1 pcs.
C1_____4700uF 35V Electrolytic capacitors______________________ = 1 pcs.
C3_____47uF 35V_________”_________”___________________________ = 1 pcs.
C5____100uF 50V_________”_________”___________________________ = 1 pcs.
C2,C4___0.1uF 50V_ ceramic capacitors__________________________ = 2 pcs.
BD1_____4A 200V brigde diode__________________________________ = 1 pcs.
D1-D3____1N4007___1A 1000V Diode_______________________________ = 3 pcs.
R1,R2____120 ohms 0.5W resistors______________________________ = 2 pcs.
VR1______Variable Resistors___ 5K(B)___________________________ = 1 pcs.
S1______On-Off or SPST switch__________________________________ = 1 pcs.
F1______0.5A Fuse______________________________________________ = 1 set.
T1______3A 18V to 21V transformer_____________________________ = 1 pcs.
Heatsink, PCB,wires,and others
How to builds
The circuit have a few parts you can assemble devices on perforated Board and wiring as Figure 2
The IC1-LM350T must be mounted to big size heatsink because at work times it very heat.
Before check circuit and wiring for error. Then adjust VR1 to minimum. Next,we test this project with apply voltage output is 1.2V You can watch video below. Then adjust voltage to 12V
And next I try to use 12Volts 50watt lamp as load. The voltage will must not lower than 12V and I measure current of lamp of 3.5A.
This projects is good as we need. We happy. Thank for watching.