I have a transformer rated 120 input and 10 v output with center tab so it is 120 to 20 volt out. Let us say it is rated 1 A
My source is 240 V ac, so how can I reduce voltage to 120 V efficiently?; so I can use it safely.
And while we are at it; how I can get 9 v DC BOTH positive AND negative to use with an OP. amplifier that need both.

 

You need a 220V rated transformer.

 

If you want to use your 120 VAC transformer on 240 VAC you need something like this:



It might make more sense to just buy a new transformer, as crutschow said, with a 120 VAC primary and the secondary you need.

To get positive and negative voltages, for most cases the best configuration is that shown below.



A common secondary that you can use to get 9VDC is a secondary that is 24 VAC volts center-tapped (240-0-24). If you can find an 18 VAC center-tapped transformer (18-0-18) your power supply would dissipate less power. All voltages are in VAC (RMS) in which the peak voltage is 1.414 times the RMS value.

 

You need a 220V rated transformer.

I have few power adapters that indicate input voltage FROM 110-240 Volt. They all use single transformer.
So, what you are saying is the transformer is CAPABLE of 240 V and the circuitry inside can take it to 110 V.
So I need to have transformer that takes 240 V. And what I have will not accept 240 V.

 

o get positive and negative voltages, for most cases the best configuration is that shown below.

I noticed that configuration has incorrect diode polarities when I previously tried to reference it in another post.
Unfortunately it's the first circuit shown in Google's image listing.

Here's the correct configuration:

 

If you want to use your 120 VAC transformer on 240 VAC you need something like this:

View attachment 143461

It might make more sense to just buy a new transformer, as crutschow said, with a 120 VAC primary and the secondary you need.

To get positive and negative voltages, for most cases the best configuration is that shown below.
View attachment 143462
So, from efficiency point of view I must get a transformer that get me CLOSE to the dc voltage needed as possible. You simple circuit is a perfect solution Thank you.

A common secondary that you can use to get 9VDC is a secondary that is 24 VAC volts center-tapped (240-0-24). If you can find an 18 VAC center-tapped transformer (18-0-18) your power supply would dissipate less power. All voltages are in VAC (RMS) in which the peak voltage is 1.414 times the RMS value.

 

You need a 220V rated transformer.

I did some experiments using RC circuits to act as AC voltage divider !!! ( if such thing is true !!!) And I managed to reduce the Input to 110 V ac that the transformer can take. What is wrong with that ???

 

Thank you @crutschow for catching the bridge being incorrectly connected.

I did some experiments using RC circuits to act as AC voltage divider !!! ( if such thing is true !!!) And I managed to reduce the Input to 110 V ac that the transformer can take. What is wrong with that ???

You will obtain vert poor voltage regulation on the secondary. That means that your power supply would not be able to supply much current over a wide range of input voltages and loads.

It might give ok performance if you do not need much current for your load and use high large capacitors for your divider, but most likely you will find the cost for achieving acceptable performance that way to exceed the cost of a new transformer by a large factor.



If you go the voltage divider route on the primary, please use capacitors designed to be used across the power line and use a fuse in series with the line. The most efficient divider will not contain any resistors other than bleeder resistors. Bleeder resistors should be able to bleed down the voltage at the AC input to less than 45 volts in 100 milliseconds (off the top of my head, not quoting a standard) to prevent dangerous electric shock.For the purpose of calculating the bleeder resistors consider that the capacitors are charged to their peak voltage (312 VDC) at the time the AC source is disconnected.

By the way, not long ago I ordered some 1.1 VA dual primary (120/240) transformers from the U.S. for 4.28 each.

 

Thank you @crutschow for catching the bridge being incorrectly connected.



You will obtain vert poor voltage regulation on the secondary. That means that your power supply would not be able to supply much current over a wide range of input voltages and loads.

It might give ok performance if you do not need much current for your load and use high large capacitors for your divider, but most likely you will find the cost for achieving acceptable performance that way to exceed the cost of a new transformer b a large factor.

If you go the voltage divider route, please use capacitors designed to be used across the power line and use a fuse in series with the line. The most efficient divider will not contain any resistors.

Thank you for clearing these misconceptions that I have. For all practical purposes, you outlined the rules by getting a transformer that accepts 240 V and output an ac close to the targeted load. These are good design principle one should adhere to.

 

You may have missed the revision to my earlier comment:

If you go the voltage divider route on the primary, please use capacitors designed to be used across the power line and use a fuse in series with the line. The most efficient divider will not contain any resistors other than bleeder resistors. Bleeder resistors should be able to bleed down the voltage at the AC input to less than 45 volts in 100 milliseconds (off the top of my head, not quoting a standard) to prevent dangerous electric shock.For the purpose of calculating the bleeder resistors consider that the capacitors are charged to their peak voltage (312 VDC) at the time the AC source is disconnected.

By the way, not long ago I ordered some 1.1 VA dual primary (120/240) transformers from the U.S. for 4.28 each.

 

You may have missed the revision to my earlier comment:

If you go the voltage divider route on the primary, please use capacitors designed to be used across the power line and use a fuse in series with the line. The most efficient divider will not contain any resistors other than bleeder resistors. Bleeder resistors should be able to bleed down the voltage at the AC input to less than 45 volts in 100 milliseconds (off the top of my head, not quoting a standard) to prevent dangerous electric shock.For the purpose of calculating the bleeder resistors consider that the capacitors are charged to their peak voltage (312 VDC) at the time the AC source is disconnected.

By the way, not long ago I ordered some 1.1 VA dual primary (120/240) transformers from the U.S. for 4.28 each.

Thank you for the update. The 120/240 primary is the ideal way to cater for both; safely.
I am still intrigued by the design that ACCEPT safely voltage from 110-240 Volt. It is becoming common practise now that these voltage adapter accept this wide range which means they found a cheap solution. How it is done ??? I am trying very hard to learn how they did it.

 

How it is done ?

By using switch-mode-power-supply principles, not a conventional mains-frequency transformer. The input mains frequency is first rectified, the resulting DC is chopped at a very high frequency (tens or hundreds of kHz), the chopped result drives the primary of a high frequency transformer having a low-voltage secondary and the secondary output is rectified.

 

Designs that accept 110 to 240 volts input with be switch mode power supplies. The first thing thay do is to rectify the input to give between about 140 and 340 volts DC This is then converted to a square wave with a variable duty cycle and a frequency in the tens of hundreds of Khz range. The allows the use of a much smaller (And cheaper.) transformer to convert down to the output voltage which is then rectified. The use of feedback via an optical coupler allows the duty cycle to be controlled to give the correct output voltage with a wide range of input voltages. You will not find any of this type of power supply with an AC output. (There are so called electronic transformers for low voltage halogen bulbs that give a "sort of" AC output. This is a mixture of the input frequency and the switching frequency.)

Les.

 

Designs that accept 110 to 240 volts input with be switch mode power supplies. The first thing thay do is to rectify the input to give between about 140 and 340 volts DC This is then converted to a square wave with a variable duty cycle and a frequency in the tens of hundreds of Khz range. The allows the use of a much smaller (And cheaper.) transformer to convert down to the output voltage which is then rectified. The use of feedback via an optical coupler allows the duty cycle to be controlled to give the correct output voltage with a wide range of input voltages. You will not find any of this type of power supply with an AC output. (There are so called electronic transformers for low voltage halogen bulbs that give a "sort of" AC output. This is a mixture of the input frequency and the switching frequency.)

Les.

Let me tell you what I understood so far The magic of accepting this wide range is done by the optical coupler......So no "mains" transformer is used . Is that true ?????? Is it possible to see a block diagram how it it is done... I love to create such circuit.

 

No the "magic" is AC to DC, PWM generator, HF transformer, HF AC to DC rectification. The opto coupler provides a means of provideng isolated feedback from the output side to the input side of the power supply. (The HF transformer provides isolation in the forward direction.) This is a link to some information on switch mode power supplies. It includes a block diagram. To create one you will need to learn a lot about electronics and high frequency transformer design. It is MUCH cheaper to buy a ready made one than build one. Also in most cases it is cheaper to buy a new one than to repair a faulty one.

Les.

 

RE:""how I can get 9 v DC BOTH positive AND negative to use with an OP. amp""
First method - as most of opamps are able to work with +/- 4,5V then put gnd to middle point (may organize with means of two equal resistors), and apply - to - and + to +.
Other method:
make generator (multivibrator, 555, 39063 or anything else) what makes an AC. Then rectify it to the direction of outgoing minus.

 

No the "magic" is AC to DC, PWM generator, HF transformer, HF AC to DC rectification. The opto coupler provides a means of provideng isolated feedback from the output side to the input side of the power supply. (The HF transformer provides isolation in the forward direction.) This is a link to some information on switch mode power supplies. It includes a block diagram. To create one you will need to learn a lot about electronics and high frequency transformer design. It is MUCH cheaper to buy a ready made one than build one. Also in most cases it is cheaper to buy a new one than to repair a faulty one.

Les.

Very valuable info...and practical advice on top of that. Let me then explain where my passion lies. I have an AC from 100-240 V and I want to INTERFACE it with something that ACCEPT it. According to your explanation it is the PWM that does the trick. OK, then the duty cycle is the controlling factor to safely provide specific voltage to feed the RF circuitry that we BOTH agree to stay away from. My dream is to build THAT part ONLY. Of course life is not that easy and most probably you CANNOT separate that part from the rest ( feed back etc...).
So, in short I cannot build the PWM part only. How true is that.

 

Please take their word for it that you cannot make a mains switchmode supply. They require a lot of design expertise.
You do not have that, and it is extremely dangerous to play with.
Besides, as has been mentioned, you can buy a ready made one very cheaply.

And the "universal input " switch mode supplies have a trade off in that the power out is less for the same size if they were just 110V or 240V. A compromise is made to allow the wide input voltage range.
A"normal" mains transformer will need 2 primary windings. They are in series for 240V and in parallel for 110V. And the connection polarities of the windings is important. Connect then together the wrong way around and you get smoke, or at least a blown fuse.

 

RE:""240 V ac, so how can I reduce voltage to 120 V efficiently""
One method is SCR in phase-drive regime aka `thyristor power regulator` (just type in the google). But it still makes a galvanic contact with network, thus in the case of mismatching one may happen to come under the 120V tension. Eletrocution is great thing, so the nul indicator or at least the differential current fuse is obligatory then. And more over - that motor, trafo or WhatYouHaveElse bit dislike a wrong formfactor of non-sin signal, thus the correct Voltage ought be adjusted by weighting a thermal data of working machine. I have a similar very high class 3kW Bosch grinder produce for US markets, thus I used it with transormer with success, but trafo of 3,3kW is weighty enough, so I bought a Chineese SCR regulator for even twice larger current. It burned in one minute, and spare was burned again. Thus, inductive load is demolishing the system, but if one use a good enough anti-backlash diodes parallel to load, seems system may be organized, if SCR I(max) is much more current-able (may be 4 or 10 fold) as those current You need.

 

RE:""240 V ac, so how can I reduce voltage to 120 V efficiently""
One method is SCR in phase-drive regime aka `thyristor power regulator` (just type in the google). But it still makes a galvanic contact with network, thus in the case of mismatching one may happen to come under the 120V tension. Eletrocution is great thing, so the nul indicator or at least the differential current fuse is obligatory then. And more over - that motor, trafo or WhatYouHaveElse bit dislike a wrong formfactor of non-sin signal, thus the correct Voltage ought be adjusted by weighting a thermal data of working machine. I have a similar very high class 3kW Bosch grinder produce for US markets, thus I used it with transormer with success, but trafo of 3,3kW is weighty enough, so I bought a Chineese SCR regulator for even twice larger current. It burned in one minute, and spare was burned again. Thus, inductive load is demolishing the system, but if one use a good enough anti-backlash diodes parallel to load, seems system may be organized, if SCR I(max) is much more current-able (may be 4 or 10 fold) as those current You need.

That I have noticed quite often. I mean using THYRISTORS when dealing with 240+ V ac. There are Even IC that deals with such range of AC input. See att