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Capacitor Power Supply

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Answered (Verified) This post has 1 verified answer | 4 Replies | 2 Followers

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14 Posts
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kanaga raj posted on 17 Aug 2009 5:40 AM

Hi

I'm using capacitor power supply to Msp430.In this power supply using three phase,two and single phase supply.In this three and two phase are properly working but in single phase was not properly working.

Input:

240V/50Hz

Output:

5v/30mA  is  required.

In this i'm now using 100ohm resistance,1.4uF capacitor and 5.6V zener diode with 3.3v regulator need 5v input  supply.

Only  getting upto voltage of 3.8v for input to regulator.

Wat can i do for getting 5V supply to the regulator.

I already refereed the document(SLAA024) for design process, from this i cant get the proper solution.

Answered (Verified) Verified Answer

Top 150 Contributor
46 Posts
Texas Instruments Employee
Verified by kanaga raj

The circuit shown is simply a half wave rectifier.  The 100 Ohms and 1.4uF limit the current flow and the zener clamps the voltage to 5.6V max.  The large cap (C21) is used to keep the voltage up during the periods of the AC cycle where the voltage is less than 5.6V.  Since this is a half wave rectification on a single phase, there is a minimum of 10ms (half cycle of 50Hz) where there is no voltage charging C21.  You can calculate the value of capacitance needed using I = C (dv/dt).  The I is the load current or the current used by the system.  dV is the allowable change in voltage on the capacitor 5.6V - 3.6V = 2V (this assumes you need at least 0.3V across the LDO to maintain regulation.  dt is the time between AC cycles which is 20ms.  If I assume a load current of 100mA then  the capacitor would need to be 1000uF.

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Top 150 Contributor
46 Posts
Texas Instruments Employee

It is hard to diagnose without schematics and scope shots.  I assume you used the circuit on page 3-65 of the app note SLAA024.  I am guessing that the ripple voltage is too high and the regulator drops out every cycle.  The fewer the phases, the higher the ripple voltage will be.  You can check this with a scope on the input of the LDO.  If the voltage is dropping out once an AC cycle then you will need to increase the size of Cch that is shown on page 3-65.

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14 Posts
Community Member

Thank u scot

Circuit used for  supply was  from SLAA391 figure2(capacitor power supply) of page 3.

In this I have used the capacitor value of Cm=1.4uF and Rm=100ohm/6W and Breaking voltage of Zener is 5.6V and charging capacitor value was 220uF.

In slaa024 document formula  provided to calculate Cchmin Value.In this what is  the T and Del Vch?

I'm new to this field.If there is any mistakes or very basic things to applied plz

apologies me. I searched lot in internet but i not get the answer.Plz reply

 

 

 

 

 

 

 

 

 

Top 150 Contributor
46 Posts
Texas Instruments Employee
Verified by kanaga raj

The circuit shown is simply a half wave rectifier.  The 100 Ohms and 1.4uF limit the current flow and the zener clamps the voltage to 5.6V max.  The large cap (C21) is used to keep the voltage up during the periods of the AC cycle where the voltage is less than 5.6V.  Since this is a half wave rectification on a single phase, there is a minimum of 10ms (half cycle of 50Hz) where there is no voltage charging C21.  You can calculate the value of capacitance needed using I = C (dv/dt).  The I is the load current or the current used by the system.  dV is the allowable change in voltage on the capacitor 5.6V - 3.6V = 2V (this assumes you need at least 0.3V across the LDO to maintain regulation.  dt is the time between AC cycles which is 20ms.  If I assume a load current of 100mA then  the capacitor would need to be 1000uF.

Not Ranked
14 Posts
Community Member

Thank u scot lester

You have given a clear explanation on my question.Once again thank u scot, we need ur support to continuously ..

regards

kanakNPK

 

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