convert wall wart to regulated power supply

From the datasheet:
It is necessary for the bypass diode to turn on before the parasitic element in the IC. As the voltage to turn on the internal parasitic element is approximately 1 V for the PNP output type regulator, a low forward voltage of VF is required.
When the value of the reverse current is large, a considerable amount of diode leakage current will flow from input to output, even if the output is off during shutdown. For this reason, a small value (around 1 µA or less) must be selected. Select an inverse rated voltage that is larger than the input/output voltage difference (80% derating or less) to be used. Select a forward direction rated current that is larger than the reverse rated current value (50% derating or less) to be used. From the above conditions, we recommend a rectifier diode or Schottky barrier diode; but as the inverse current of many Schottky barrier diodes is generally large, select one with a small value.

Search Mouser et al for a rectifier diode with at least:

Vf lower than 1v - Need the diode to turn on before the IC
Vr greater than 16v - Needs to withstand upwards of ~2x the input voltage
Ir lower than 1uA - Needs to have a low leakage current when reverse biased
If greater than 2A - Needs to be capable of feeding your load when the output capacitor is the power source via the diode to the IC input

There are plenty of SMD options that would make it easier with your space constraints.
 
A linear regulator needs a higher input DC voltage than the desired output DC voltage.
How much higher depends on factors like: AC line variation, regulator drop out margin,
input voltage ripple at maximum loading, tolerances of all components, etc.
 
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Thanks everyone. I received the wall warts and took them apart to figure out what to get.
5V 2A wall wart: OC 7.4V, inside full wave bridge 1N54XX, with 2, 5300 µF caps 10V
6V 2A wall wart: OC 9.6V, inside full wave bridge 1N5401 with single 16V 3300µF cap, fuse on one leg of bridge
7V 1.85A wall wart: OC 7.2V, inside does not look like class 2 even though it is marked as such. Input is not multivoltage,but no large transformer. Has the typical coils and small xformer that a switching supply has.

I will try with the 5V and 6V wall warts. The small PC board on the 6V is not efficient in layout as the diodes are not flat to the board, so that takes up room. Since I have an extra broken 5V wall wart, I will just pull the PC board off and make 2 identical PC boards. A LDO regulator should work with the 6V wall wart, not so sure with the 5V. Posting parts list next
 
The diodes might be above the circuit board so they can get the heat out around them. Check the AC voltage on the input to the diodes under load. Expect around 1.4x the AC voltage when converted to DC. As Rayma suggested, put a diode from output to input of the regulator to protect it. Put a small film cap right on the input pin to ground and a small electrolytic from output to ground. There are lots of diagrams on line showing this setup. Remember, if it has been done like you want it before there is a reason for that!
 
OK. Decided to start doing some calculations. Filter C= Load current/4fVripple = 1.5A/4*60*.62 = 10000µF in the US f=60. So a single cap works to have ripple of .62V

FWBridge current = 1.8 x load current (1.5A)= 2.7A So 1N5401 works with 3A current

US voltage is 120V +- 6% or min 112.8V

Vsecondary transformer= (Vout+Vreg+Vrect+Vripple)/.92 x (Vmain/Vmainmin) x .707
= (5 + .7 (LDO BAxxd0) + (4 x .7) + .62)/.92 x (120/112.8) x .707
= 7.5 V secondary

I took my 2 wall warts, and measured the Vac output of the secondary
5V: 6.24V AC
6V: 7.8V AC

So it seems the 6V wall wart will work, saved by the use of a LDO regulator .7V max, vs 2-3V standard.
The 5V one wont work, unless I play some games: use 2 diode rectifier so diode loss is 2x.7 (albeit with less current), increase the cap to lower ripple to .5V. Then the secondary V requirement becomes 6.2V AC, barely works with the 5V wall wart.

Since I have the 6V wall wart, I think I will start with that one.
 
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I have a 5V 2A class 2 wall wart. I carefully opened it up and there is a transformer followed by a full wave bridge and then 2 5600 microfarad 10V caps in parallel. All on a small PC board.

I would like to convert it to a regulated power supply. It seems all I need to do is remove the 2 large caps and replace one of them with a .33 microfarad cap, then in place of the other cap, put a L78S05C with the input and ground legs soldered where the other cap was, clip a heat sink on the TO-220 device, and then move the 5V output lead from the PC board to the third output leg of the L78S05C device. Finally, solder a .1 microfarad cap between the output leg and ground. Then glue the wall wart case back on. Or should I leave one of the 5600 microfarad caps on the board to provide some filtering? (Need to remove the other one to get space for the L78S05C)

Is it just that easy to convert an unregulated 5V wall wart to a regulated 5V linear power supply?

I am trying to use this to power a Wiim Pro + and replace the cheap switching power supply with a linear power supply. The requirement is 5V 1.5A and the input is a USB C connector. I tried plugging the 5V 2A class 2 to power it via an adapter from the 5.1mm plug to USB C but it wouldnt power up. Thinking the unregulated wall wart didnt provide the correct V. Any advice would help. There are a lot of small audio devices that could benefit from a better power supply, but didnt want to spend a whole lot of money on them. This seems a pretty straightforward and cheap solution.
I had a similar need where noise from an SMPS was too high. I used an 8V
switching supply with a 7805 to drop down to 5 V. The SMPS alway puts out
8V so the thermal load on the 7805 is constant regardless of power line Voltage.
It got me a 20dB reduction in analog noise from "iffy" to just fine.

The problem with analog supplies is you need to work properly at low AC Voltage.
On a hot summer day in southern California I've measured the AC line Voltage as
low as 98 VAC. This is not an issue with SMPS.

 
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YouTuber "Kiss Analog", who is a switching power supply designer, calls transformer + rectifier + cap supplies "low frequency switching supplies".

The rectifiers are the switching devices, and they switch at 50/60Hz.

Your choice of power supply therefore becomes either a low frequency switching supply, or a high frequency switching supply.
 
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I disagree on the transformer/diode/cap being a switching supply
as there is no correction if the line Voltage changes. It just follows
whatever the power line is doing. Mr. YouTuber should check out
the power supplies in the Ampex broadcast VTR AVR-1. The rectifiers
in the supply are SCRs fired at the right time to be 3 Volts higher
than the final supply. Then a final series pass linear regulator is used to
drop the 15 Volt raw supplies down to 12 Volts. To make it more linear
they make a 50/60 Hz cosine ramp locked to the zero crossing of the
AC power so that the Voltage change is linear. This was designed in
the early 70s before we had SMPS units. This REALLY is a 50/60
Hz switching supply. BTW they are EXTREMELY reliable. That machine
runs around 25 KiloWatts. They were fun to work with/on.

 
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I disagree on the transformer/diode/cap being a switching supply
as there is no correction if the line Voltage changes.
That's regulation. Hypex power supplies are high frequency unregulated switching supplies.

Look at the choke input power supplies in the "Tubes" forum.

It's an L-C filter, the same as the L-C filter in a high frequency power supply, but the component values are much smaller in high frequency power supplies.

When the transformer secondary voltage is higher than the cap voltage the rectifier switches on, when the transformer secondary voltage is lower than the cap voltage, the rectifier switches off.
 
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I do have a question about fuses. The 6V power supply has a fuse right after the secondary on the PC board. The 5V supply does not have a fuse on the PC board, but possibly on the transformer near the 120V input side. I was going to use the PC board from the 5V supply for the 6V one due to the more efficient layout to be able to put the regulator and 2 caps. But the 5V PC board cannot accomodate a fuse. How important is it to include a fuse?
 
That's regulation. Hypex power supplies are high frequency unregulated switching supplies.

Look at the choke input power supplies in the "Tubes" forum.

It's an L-C filter, the same as the L-C filter in a high frequency power supply, but the component values are much smaller in high frequency power supplies.

When the transformer secondary voltage is higher than the cap voltage the rectifier switches on, when the transformer secondary voltage is lower than the cap voltage, the rectifier switches off.
I don't dispute any of that. I just don't believe they run it open loop.

 
Given you already bought 3 wall warts, none really suitable, still need to add linear regulation, heatsinking, a proper case, build it safely, etc. but worst of all no happy ending, either present or future, I would forget the wall wart route and straight build or buy a properly designed 5V 2A regulated linear supply.

Forum members will certainly suggest a proper schematic, parts, maybe even a case, some might go as far as to simulate it once we all agree on a final design.

Otherwise you will be running in circles for a long time.

jm2c
 
Another option is to pick up a used regulated linear PS such as the ubiquitous Power-One/Condor types. These are robust devices used in industrial machine tools, etc., and they show up on ebay when they get retired. The regulation is great, and the output voltage is adjustable. Bigger than a wall wart, but and you can hide it behind the cabinet...

New ones are still offered on Digikey, etc., for a price, but if one is patient you can get a clean one on ebay for $30-35 or so. I just saw this one, a Power-One 5V/3amp unit with built-in overvoltage protection for $25 with free shipping in the US.

https://www.ebay.com/itm/323774838852
 
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Given you already bought 3 wall warts, none really suitable, still need to add linear regulation, heatsinking, a proper case, build it safely, etc. but worst of all no happy ending, either present or future, I would forget the wall wart route and straight build or buy a properly designed 5V 2A regulated linear supply.

Forum members will certainly suggest a proper schematic, parts, maybe even a case, some might go as far as to simulate it once we all agree on a final design.

Otherwise you will be running in circles for a long time.

jm2c
Of course, you are right. But I am doing this more out of curiosity and for learning about linear power supplies. Early on, I decided that if I just wanted one, it would be far cheaper to buy it, since they go for about $50 and it will cost as much in parts, not counting my time. What I have discovered is that about 15 years ago, you could buy a regulated linear transformer power supply wall wart for around $10! Now that switching power supplies have taken over the market, they are non-existent. Now I am figuring the layout and parts list, and decided I really need to use the original PC board with the fuse, since safety is paramount.
 
From the datasheet:


Search Mouser et al for a rectifier diode with at least:

Vf lower than 1v - Need the diode to turn on before the IC
Vr greater than 16v - Needs to withstand upwards of ~2x the input voltage
Ir lower than 1uA - Needs to have a low leakage current when reverse biased
If greater than 2A - Needs to be capable of feeding your load when the output capacitor is the power source via the diode to the IC input

There are plenty of SMD options that would make it easier with your space constraints.
Would a standard 1N5401 diode work? It meets all the requirement you list. The Vf is 1V though. The lowest Vf I could find was .82V, not sure if that would make much of a difference. 1N540X are plentiful.
 
You can build such a supply using a 2A 9V transformer, bridge and main filter cap and something as simple as a power Darlington, say TIP142, heatsinked of course, with base connected to a 5.6V Zener fed through a 220 ohm resistor.
Place a 100uF cap across Zener.
Suggesting this because your load is beyond what an LM 7805 can supply.
That's as basic as you can get, you can search for better circuits of course.
My point being there is no way this can cost $50 or anywhere near. 🤷🏻