Can you help for Current Step Up?

Just buy the correct supply.

The 4.7 μF capacitor on the input side seems too small for the 1 A rating, which goes to show how cheaply the power supply is made.

That reasoning is completely wrong. The transformer increases the current. You haven't a clue how the power supply works.

The power supply only delivers 1 amp when the receiving item "wants" 1 amp. If it "wants" 2 amp, the resistance of the receiving item is too small and the voltage out of the power supply drops and the current reduces. It is a form of overload protection. You can see the black thing with 4 legs. It is an optocoupler and when the LED on the output stage of the power supply goes dim, the circuitry reduces the amount of time the transformer receives pulses and the whole circuitry starts to close-down. It's called self-preservation.

The 4u7 and bridge produce DC at a voltage up to 40% more than the AC incoming voltage.
The supply is 5 watt so the supply current is somewhere between 10mA and 25mA.
The drop for 4u7 with 25mA is only a few volts. Certainly nothing like 20v to 40v.

You can put 2 supplies in parallel and get 2 amp.

Colin said:

The power supply only delivers 1 amp when the receiving item "wants" 1 amp. If it "wants" 2 amp, the resistance of the receiving item is too small and the voltage out of the power supply drops and the current reduces. It is a form of overload protection. You can see the black thing with 4 legs. It is an optocoupler and when the LED on the output stage of the power supply goes dim, the circuitry reduces the amount of time the transformer receives pulses and the whole circuitry starts to close-down. It's called self-preservation.

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I don't think that the opto-isolator does that. The opto-isolator tells the input to stop or slow down when the output voltage is high enough.

The output current is probably limited by the current limit on the transformer input that stops the transformer saturating.

I agree about the voltage dropping when more current is taken from the supply than it can manage.

Diver300 said:

I don't think that the opto-isolator does that. The opto-isolator tells the input to stop or slow down when the output voltage is high enough.

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You are correct (and Colin was wrong), the opto-isolator is the voltage feedback to keep the output voltage constant. The current limiting (over-current protection) will be done separately on the primary side of the PSU, usually by monitoring the voltage drop across a resistor in the emitter (or source) of the output device.

Are there any components on the bottom side of the board?

ak

Nothing on the bottom side.

Rexkh said:

Nothing on the bottom side.

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Then I think I know this power supply. Most of the parts are used near their maximum rating. The transformer is used at the edge of core saturation. This type of PWM oscillator uses core saturation to end a cycle.

To get 2X more current you will need to rewind the transformer with larger wire. To get the same turns you will need a larger transformer (bobbin). The core of the transformer must have a larger cross section area to handle the 2x more amp turns. (major rework of the transformer) This alone kills this project.

As stated many times the input and output capacitors will need a 2x increase.
The output diode will need to be doubled in size. (probably moved up to a TO220 type)
Because the current in the large transistor is 2X the base current probable needs a 3X increase so the small transistor will have 3X more current in it.
We don't know if the big transistor will survive.
R5? needs to be much larger. (wattage rating)

Many of the parts that need to be replaces are physically much larger. So it will not fit in the old box.

D1 through D4 and D7 can be the same. The opto-isolator and Zener are OK. But that is about all you can keep.

This circuit is built to provide 1A output but it provide only 130mA so I guess there might be a piece is a bottle neck. Can you tell me how to test to find the bottle neck?

You are overloading it. Just buy the right supply.

Mike.

Rexkh said:

This circuit is built to provide 1A output but it provide only 130mA so I guess there might be a piece is a bottle neck. Can you tell me how to test to find the bottle neck?

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How are you measuring the 130 mA? If you just connect an ammeter to the power supply, you are shorting it out and it might provide far less than its rated current. You need some load that takes 1 A at 5 V, such as a 5 Ω resistor, to test that the power supply can supply the rated current.

You could just connect it to the load that you intend to use. If you do that, and if the voltage is 5 V, then the power supply is adequate for that condition. If the load isn't working correctly, and is being supplied by 5 V, the problem is not as simple as an inadequate power supply.

What is the load?

Is it connected using USB?

Rexkh said:

I guess there might be a piece is a bottle neck.

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No there are many bottle necks.

Lets talk power supplies.
1) Constant voltage with out current limit: The output is about 5V at currents from 0 to 1A. At 1.3A it blows up.
2) Constant voltage with current limit: (CV/CC) At current from 0 to 1A the output is about 5V. Put a 2 ohm resistor on the supply and it will output 2V because it is regulating at CC (constant current). Short the output = 0V, 1A. Put a current meter across the outputs and 0V, 1A. This type is often used to charge batteries because it will hold at 1A (at any voltage below 5V). As the battery reaches charged it switches to a constant voltage.
3) CV with fold back: {because most CC supplies over heat at 0V} Probably you supply will output 5V at any current from 0 to 1A. As soon as you try to get more than 1A it will drop the voltage just like a CC supply but: At 2.5V the current limit is 0.5A and at 0V the current limit is 0.13A. (simply put, at a short circuit, it will drop back to a much smaller current to save its self)