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Is it a good idea to use rectifier diodes in series to reduce a power supply voltage?

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JohnJohn20

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Hi. I have a 12V PSU which I want to reduce to 4V 2A PSU. I have used a NPN power transistor (BD137) plus a 270Ω resistor and a 5V zener to make a regulator which gives me a 4.4V output which is still too high.

Is it a good idea for me to just put a silicon rectifier of some sort in series with the regulator output to give me a further 0.6V drop thus giving me a 3.8 V regulated output?

Would that be reliable?

Ideally I should use a 4.6V zener which I can't get.

Thanks. Chris.
 

I have used a NPN power transistor (BD137) plus a 270Ω resistor and a 5V zener to make a regulator which gives me a 4.4V output which is still too high.
Click to expand...
BD137 isn't suited for 2A load, neither in terms of rated current nor power disspation.
 
FvM said:
BD137 isn't suited for 2A load, neither in terms of rated current nor power disspation.
Click to expand...

Ooops. My mistake. It is a BD437. Thanks. But the same question still applies.
 

Either use a real adjustable voltage regulator circuit (involving at least a second transistor) or a combination of zener and rectifier diodes that best fits 4.6 V. 4.7V is a standard zener voltage, by the way. Not to mention voltage regulator ICs...
 
Hi,

You dissipate a lot of power..twice as much as you use.
--> not efficient, a lot of heating, you need a big heatsink

Use a switch mode step down curcuit.

Klaus
 
KlausST said:
Hi,

You dissipate a lot of power..twice as much as you use.
--> not efficient, a lot of heating, you need a big heatsink

Use a switch mode step down curcuit.

Klaus
Click to expand...


Why would there be more heat dissipated dropping 0.6V over a rectifier and dropping the same voltage using a regulator? 0.6V @ 2A either way.

By switch mode you mean a PWM setup. Right? That would be unnecessarily complicated for me.

- - - Updated - - -

FvM said:
Either use a real adjustable voltage regulator circuit (involving at least a second transistor) or a combination of zener and rectifier diodes that best fits 4.6 V. 4.7V is a standard zener voltage, by the way. Not to mention voltage regulator ICs...
Click to expand...

To me simple is good (and simple). Don't think I will find quite a suitable 4V 2A IC. But I will go looking for a 4.7V zener tomorrow. Cheers.
 

Hi,

Why would there be more heat dissipated dropping 0.6V over a rectifier and dropping the same voltage using a regulator? 0.6V @ 2A either way.
Click to expand...
It seems to me you want rather drop 8V instead of 0.6V...

That would be unnecessarily complicated for me.
Click to expand...
There are ready to buy step down modules. It is not more complicated to buy this than to buy any other electronic device..

Klaus
 

The power dissipation on a linear device, any linear device, will be (12-4)*2 = 16 watts.

To dissipate that much power you will require a large heatsink. I'm sure it would cost more than a SMPS that you purchase on ebay.
 

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