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Low Noise Adjustable Linear AC-DC Power Supply

Hesambook

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Features:

AC – DC Conversion

Double output voltages (Positive – Ground – Negative)

Adjustable positive and negative rails

Just a Single-Output AC transformer

Output noise (20MHz-BWL, no load): Around 1.12mVpp

Low noise and stable outputs (ideal to power Opamps)

Output Voltage: +/-1.25V to +/-25V

Maximum output current: 300mA to 500mA

Cheap and easy to solder (all component packages are DIP)


A double output low noise power supply is an essential tool for any electronics enthusiast. There are many circumstances that a double-output power supply is necessary such as designing pre-amplifiers and powering OPAMPs. In this article, we are going to build a linear power supply that a user can adjust its positive and negative rails independently. Moreover, just an ordinary single-output AC transformer is used at the input.



References

Source: http://bit.ly/2Qc9idO

[1] LM317 Datasheet: http://www.ti.com/lit/ds/slvs044x/slvs044x.pdf

[2] LM337 Datasheet: http://www.ti.com/lit/ds/symlink/lm137.pdf

[3]: Schematic Symbol and PCB Footprint for LM317: https://componentsearchengine.com/part.php?partID=248007

[4]: Schematic Symbol and PCB Footprint for LM337: https://componentsearchengine.com/part.php?partID=290650

[5]: Altium Plugin: https://www.samacsys.com/altium-designer-library-instructions
 

betwixt

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Why the 10 Ohm resistors in series with the input? The additional filtering they provide is minimal. If it is to reduce input surge current they would be better placed before the first reservoir capacitors.

Brian.
 

Hesambook

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1.2mVpp is high for a Low Noise Grade Power Supply. It will presumably be higher at the loaded condition.
Look at that..
https://www.analog.com/en/products/lt3045-1.html#product-overview
In additional to, a Filter has to be used at the Input with appropriate High Quality Capacitors.
1mV is high?! Really it is difficult to understand if it is the noise floor of the oscilloscope or not. Besides, in ALL power supplies when we increase the output current, then the noise will increase. even many manufactures don't mention the noise figure! in this design I tried to address many applications with cheap and easy to use components. of course, there are more advanced expensive options.

- - - Updated - - -

Why the 10 Ohm resistors in series with the input? The additional filtering they provide is minimal. If it is to reduce input surge current they would be better placed before the first reservoir capacitors.

Brian.
They are necessary for filtering
 

asdf44

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Any thermal shutdown protection?
 

Hesambook

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Any thermal shutdown protection?
No, you gonna use a heatsink if the difference between the input and output voltage is high. Please check if LM317 or LM337 has introduced internal thermal protection. I didn't check
 

KlausST

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Hi,

I think this is a standard design for a linear power supply.
Not bad.
There's nothing wrong with it. It will work ... for most cases satisfactory
*****

I'd like to comment the voltage specifications. (From the view of an electronics engineer)
In the schematic the input capacitors are rated 25V.
So if one wants to chose a transformer...it's nominal output voltage should not be higher than (25V - 0.6V) / (1.2 × sqrt(2)) = about 14V.
(A transformer rated with 14V (@ nominal load) may have 1.2x higher voltage when not loaded, the capacitors will charge to peak of sine = sqrt(2) × RMS, minus the voltage drop of the diode of 0.6V @ low load).
In detail it depends on transformer specifications, but also in mains voltage stability.
One should also consider some headroom for the capacitor voltage rating.

So let's use this 14V transformer....but now at full load of 500mA output current.
The nominal output voltage now is 14V RMS. The capacitors now are pulse charged once per fullwave and maybe just in 20% of time.
This means every 20ms (@50Hz) there need to be a pulse current of 2.5A during 4ms. (Estimated).
This causes a higher voltage drop in the diode, let's assume 1.2V.
And additionally the sinewave gets a flattened top. It won't go up to 14V x sqrt(2). Let's assume a 2.5V reduced top (assumed 1 Ohms source impedance).
The peak value at the first capacitor will then be 14V x sqrt(2) - 1.2V - 2.5V = 16V.
The capacitor will be discharged (100% -20%) of 20ms with 500mA, this gives a ripple of 16ms x 500mA / 2200uF = 3.6Vpp
The second capacitor will reduce this peak-to-peak ripple, but let's assume it will be about 2.0Vpp at the second capacitor.
Additionally there will be an average voltage drop of 10 Ohms x 500mA = 5V at the resistor. (@2.5W power dissipation!!)
Thus the minimum voltage at the second capacitor will be 16V - 2.0V - 5V = 9V at full 500mA output current.
If one now considers a 2V voltage drop at the voltage regulators one can expect just 7V of clean, regulated output voltage at 500mA load.

Note: All the voltages are estimated, they may differ +/- several volts in the real circuit.

How to improve the situation:
* using input capacitors with higher voltage rating
* then you are able to use a transformer with higher output voltage
* using diodes with less voltage drop
* using full bridge rectifier
* using a transformer with more stable output voltage vs load current
* using larger capacitors
* reducing lower ohmic series resistor (now 10 Ohms), or replacing it with an inductor
* reducing specification for output voltage
(When you buy a power supply with rated 25V output voltage, then it should be able to supply this at full specified load current)

Klaus
 

FvM

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Another point to consider. Asymmetrical output load may cause transformer saturation due to the single wave rectifier configuration.
 

Hesambook

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In the schematic the input capacitors are rated 25V
That is my typo mistake in the schematic, they are rated at 35V in the bill of materials
 

AbleMan

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IDEA ON THE POST: Different transformer output voltage can be used in the design and the variable resistor be used to fine turn the required voltage. Do not forget to add fuse at the output and test with DMM to be sure of the voltage
 

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