Need Help in making a 15V Battery eliminator

[rishiraj]

Hello all, I want to make a 15 V Battery Eliminator. Now I know that to make it I need the following-:

1. Transformer (greater then 12-0-12 rating)
2. Rectifier (Bridge Rectifier)
3. Filter.

Now, for the Rectifier I'm quite sure that I need to use the Bridge one because of its Ripple factor and greater Transformer Utilization Factor.

But I'm clueless about the Filter part, Now I need 15V as output, so what combination of Capacitor and Resistor should I use?? How do I determine that?? And I only know that 12-0-12 transformers are available, is a output voltage of greater then 12-0-12 transformer available too?? Please help.

Thank you

 

[nitishn5]

A 12V transformer with a Full Wave rectifier would give an Output Peak voltage of 12*√2 V Which is about 17V.
A regular Capacitor filter following the rectifier stage would give a DC value of 17V at no load.
With load you would see the sawtooth type waveform typical of capacitor charging and discharging.
You can use a Low Dropout Regulator to get 15V.
If your load current is high your regulation might be poor since the sawtooth at the capacitor filter might go below the minimum voltage for the regulator and you might have to go for a different approach.
If you are looking for efficiency you can go for an SMPS such as a Buck or Boost Converters.

You can transformers with other turns ratio also. 15-0-15 is also available.

 

[rishiraj]

A regular Capacitor filter following the rectifier stage would give a DC value of 17V at no load.
With load you would see the sawtooth type waveform typical of capacitor charging and discharging.
You can use a Low Dropout Regulator to get 15V.

Suppose I use 1 12-0-12 transformer and a bridge Rectifier then I get output V(Peak) as 16.92V and V(DC) as 10.77 Volt(2V(Peak)/pi) . I don't know anything about Voltage Regulator but I thing would like to ask is, can I make that 10.77 Volt output to 15 volt DC using Capacitor and Load Resistor??

Sorry, I may sound too immature in all this since am new into it.

I don't need a high current as output, just 0.5A will do, the only thing of importance in my circuit is that I need a 15V DC as ouyput

 

[nitishn5]

Dude, it's OK to be new. Everyone is at one point in time. No need to feel sorry.

A Capacitor filter at the output of a bridge rectifier would give a DC Voltage of value equal to V_peak of the input waveform. In this case for a 12V transformer it would be 16.92V.
The Vdc that you have as 2V_peak/pi is the average voltage of the rectified (but not filtered) waveform.
Since your load current requirements are not that high, and you need a fixed voltage, I suggest that you go for the following setup.

Get a 15-0-15 Transformer.
Make a Center-Tapped or a ridge Rectifier to get a V_peak = 15√2 = 21.21V
Put a Capacitor Filter, an electrolytic capacitor of value 100µF would be fine. Put a ceramic capacitor of say 100nF in parallel. This has better high frequency response.
Now you have a 21.21V DC Voltage with some ripples.

Get the Voltage regulator IC LM7815. This is a fixed Voltage regulator for 15V. It requires a minimum of 17.7V at the input which is why you need a transformer with higher secondary voltage. This IC is easily available. It has a maximum current of about 1A which should be sufficient.

Refer to the Datasheet of the IC at it will tell you about the capacitors required at its output. You can go for the same 100µF + 100nF at the output too.

Now you have a 15Vdc fixed Voltage capable of supplying up to 1A.
You should note that the IC might get hot if you are supplying 0.5A or more for a long time. You can attach a heat sink to the IC to provide for heat dissipation. Any normal size heat sink would do for your application.

Your circuit should look something like the one shown here.

This is one of the simplest method of achieving your target. Note that this technique is very inefficient. The regulator has an input of about 21V and an output of 15V. A difference of 6V is lost on the regulator itself. If you are consuming 0.5A, this would mean an output power of 0.5A * 15V = 7.5W and input power of 21V * 0.5A = 10.5W. The difference of 3W is lost in the regulator itself.

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Another option to improve efficiency is to use a 15-0-15 transformer + Rectifier + Capacitor Filter followed by a Low Dropout Regulator such as the following
https://www.linear.com/product/LT3085
I'm not sure of its availability. If not available, you can easily build one of your own using an Opamp, a PMOSFET, a Zener and some resistors though.

 

[rishiraj]

Dude, it's OK to be new. Everyone is at one point in time. No need to feel sorry.

A Capacitor filter at the output of a bridge rectifier would give a DC Voltage of value equal to V_peak of the input waveform. In this case for a 12V transformer it would be 16.92V.
The Vdc that you have as 2V_peak/pi is the average voltage of the rectified (but not filtered) waveform.

This is a good lesson I learnt.

Well, You gave a complete guide to me. Thanks a lot. Now my next step is to check the local market for the products, and then choose the right option accordingly. I think I'll follow your step by step instructions.

Thanks a ton.

 

[nitishn5]

Mark this as solved once you are able to get what you want.