Reducing Switcher Output

[Krankshaft]

This one should be simple for you guys.

I have an electronic device that oddly enough wants a 3.0 volt adapter I need this supply in a wall wart package.

Rather then go out and buy one I had another idea that has worked before.

I have a wall wart switcher with a 6 volt output a pretty wierd voltage if you ask me and I have no use for it with it's current odd output voltage.

I cracked it open and it uses a KIA431 shunt regulator. The reference voltage is controlled by two resistors in a voltage divider formation between the anode and cathode with the reference pin of the regulator connecting between the two resistors.

If I modify the divider so that the reference voltage is 3.0 volts I should be set.

Here is the datasheet for the regulator:

http://www.alldatasheet.com/datashee...EC/KIA431.html

Now here is my problem the last time I did this it was on a wart it used a Texas Instruments shunt regulator and the equation for resistor calculation was alot simpler.

Like this:

Vo=Vref*(1+R10/R14)

Nice and easy the way I like it but this regulator has an equation like this (it's on page 3 of the datasheet).

VKA= Vref (1+ R1/R2) + I ref X R1

What is I ref? Reference current? The reference voltage of the regulator is 2.5 volts by the way.

So how would I use the equation on this new regulator to get my resistor values to make Vref = 3.0 volts?

Thanks for the help in advance.

[Gianni]

Re: Reducing Switcher Output

Hi Krankshaft

Iref is the reference current flowing from the resistor connected from output and pin Ref.
All the time I have to calculate the resistors to obtain an output voltage I use a simple way, ignoring the Iref because it is very small (few µA).

The Vref is 2.495V (Typical value) so if you place a resistor (Rref=2700ohm) between Pin Ref and ground, the current flowing trough the resistor is:Ir=Vref/Rref=2.495/2700=0.000924074mA.
Now you must calculate the voltage across the resistor placed between Output and pin Ref: Vr=Vout-Vref=3-2.495=0.505V.
At this point you can calculate the resistor from output dividing the voltage Vr by the current flowing in Rref: R=Vr/Ir=0.505/0.000924074=546ohm
546ohm is not a standard value, the closest standard value is 510ohm so if you use it you obtain the following output: Vout=Vr X Ir + Vref =510 X 0.000924074 + 2.495=0.47127774+2.495=2.96627774V

In other words, the KIA431 (it is the same as TL431) works to keep the voltage at 2.495V on its Pin Ref so you have to calculate the correct divider to obtain this voltage.
You can use a trimming potentiometer instead of a fixed resistor so you can adjust the output voltage.
The resistors used for regulation should be 1% tolerance to obtain a better and stable voltage; consider also that KIA431 and TL431 have tolerances so in the worst cases with 2700ohm and 510ohm you can have 2.9V (for Vref=2.40) and 3.03V (for Vref=2.55).

For your wall wart switcher pay attention to the power dissipation because reducing the output voltage by half at same output current means to have the double power dissipated if it is a linear regulator, and probably the heat sink it is not big enough to handle it.

Hope this help
Gianni

[Krankshaft]

Re: Reducing Switcher Output

Thanks for the help I get it now.

The wall wart is rated at 2 amps the device I'm using it on is only going to draw a maximum of 100 ma so I don't think dissipation will be a problem.

[Krankshaft]

Re: Reducing Switcher Output

Sorry my edit time ran out so what you are saying is if I attempted to draw the switchers full rated current after reducing the output voltage by half I could have Chopper FET overheating problems? That makes perfect sense.

Or are you saying that the reduction in output voltage could cause the FET to overheat due to the additional heat load placed on it due to the voltage drop?

[Gianni]

Re: Reducing Switcher Output

What I mean is this: if it is a linear regulator, the KIA431 drives a POWERtransitor or PowerMos to regulate the output, we can assume the unregulated voltage on it is at least 7Vdc. So if you load it at maximum, 6V 2 A, the voltage drop on PowerDevice is 1V and the power dissipation is 1V X 2A=2W.
Now if you reduce the output voltage at 3V and you put a load of 2A, the voltage drop on PowerDevice is: 7 - 3=4V.
In this case the power dissipation becomes 4V X 2A=8W, four time of the original value.
If your load is low (100mA) than probably you have non problem at all.

If the regulator is a switching type, than you should not have problem with power dissipation but maybe with ripple and stability.

Ciao
Gianni

[Krankshaft]

Re: Reducing Switcher Output

Ok thanks again.