This is the First DC power supply in my life that made to use in many projects. It is ideal for those who want to adjust voltage from 1.25V to 30V and currents up to 1A. Which is sufficient for normal use Such as is power supply instead of an one 1.5V AA battery or when you want to listen to music from a 30-watt amplifier that required voltage of 24V 1A, it can be done easily.
In the days before that we commonly used the transistor that is very difficult, large, and probably more expensive ICs. But this circuit can be created with a single IC is LM317 The LM317 or LM117 series of adjustable 3-terminal positive voltage regulators is capable of supplying in excess of 1.5A over a 1.2V to 37V output range,
And has many special features that I like are :
• Output Voltage Tolerance 1%
• Line Regulation 0.01%
• Load Regulation 0.3%
• Prevent the deposition temperature.
• Short-circuit protection.
• Ripple is eliminated with ratio of 80dB
• Maximum input voltage 40V
The working of circuits
Followed circuits below, the transformer T1 is changed a AC 220V down as AC 24V to the bridge diode rectifier D1(1N4001) to D4(1N4001) there is dc voltage into the filter capacitor C1 equal to DC35V
The output voltage from IC1 Depending on the Voltage Adj pin of the IC. Or to adjust the VR1.
The VR1 is control output dc voltage 0V(1.25V) to 30V(32V) or 37V maximum voltage at 1.5A max all range.
Circuit-of-my-first-variable-dc-power-supply-1-2v-to-30v-1a-by-lm317
Calculate the LM317 output voltage
And we can calculate output voltage equal to:
Vout = 1.25 x {1+ (Rp/R1)
- Vref = 1.25V
- Typically R1 is 220 ohms or 240 ohms as datasheet. I use 220 ohms.
- Normally as datasheet I see them use VR= 5K (Pontentiometer) But I have VR-10K only since it easy to use.
Rp = {(VR1 x R2) / (VR1 + R2)}
Then we test it,Suppose, rotate VR1 to lowest resistance cause Rp = 0 hms. put it in formula above:
Vout = 1.25 x {1+(0/220)}
= 1.25V
But when adjust VR1 to maximum resistance VR1 and R2 are parallel together. Rp = 5.46K = 5460 ohms. test it in formula above:
Vout = 1.25 x {1+(5460/220)}
= 32.2V
Then the capacitor C3 is Better performance filter of IC1.
The diode D5 and D6 ( both is 1N4007) is the protector from external voltage to reverse to makes the damage to the IC1.
This is the First DC power supply in my life that made to use in many projects. It is ideal for those who want to adjust voltage from 1.25V to 30V and currents up to 1A. Which is sufficient for normal use Such as is power supply instead of an one 1.5V AA battery or when you want to listen to music from a 30-watt amplifier that required voltage of 24V 1A, it can be done easily.In the days before that we commonly used the transistor that is very difficult, large, and probably more expensive ICs. But this circuit can be created with a single IC is LM317 The LM317 or LM117 series of adjustable 3-terminal positive voltage regulators is capable of supplying in excess of 1.5A over a 1.2V to 37V output range,And has many special features that I like are :• Output Voltage Tolerance 1%• Line Regulation 0.01%• Load Regulation 0.3%• Prevent the deposition temperature.• Short-circuit protection.• Ripple is eliminated with ratio of 80dB• Maximum input voltage 40VThe working of circuitsFollowed circuits below, the transformer T1 is changed a AC 220V down as AC 24V to the bridge diode rectifier D1(1N4001) to D4(1N4001) there is dc voltage into the filter capacitor C1 equal to DC35VThe output voltage from IC1 Depending on the Voltage Adj pin of the IC. Or to adjust the VR1.The VR1 is control output dc voltage 0V(1.25V) to 30V(32V) or 37V maximum voltage at 1.5A max all range.Circuit-of-my-first-variable-dc-power-supply-1-2v-to-30v-1a-by-lm317Calculate the LM317 output voltageAnd we can calculate output voltage equal to:Vout = 1.25 x {1+ (Rp/R1)- Vref = 1.25V- Typically R1 is 220 ohms or 240 ohms as datasheet. I use 220 ohms.- Normally as datasheet I see them use VR= 5K (Pontentiometer) But I have VR-10K only since it easy to use.Rp = {(VR1 x R2) / (VR1 + R2)}Then we test it,Suppose, rotate VR1 to lowest resistance cause Rp = 0 hms. put it in formula above:Vout = 1.25 x {1+(0/220)}= 1.25VBut when adjust VR1 to maximum resistance VR1 and R2 are parallel together. Rp = 5.46K = 5460 ohms. test it in formula above:Vout = 1.25 x {1+(5460/220)}= 32.2VThen the capacitor C3 is Better performance filter of IC1.The diode D5 and D6 ( both is 1N4007) is the protector from external voltage to reverse to makes the damage to the IC1.
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