Author Topic: design of linear regulator efficiency tester under differnent load conditions  (Read 2446 times)

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Offline yogece

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hi,
  I would like to design a liner regulator efficiency tester such as 78xx, LDO.can any one suggest me from where to start
 

Offline digsys

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This has to be the easies assignment ever !
Open up EXCEL - create at least 7 columns - Vin, Iin, Pin, Vout, Iout, Pout, Efficiency
Eff = Pout/Pin x100
That's 100% accurate for Resistive loads, for ANY type of Linear Regulator (or use RMS for changing loads).
Hello <tap> <tap> .. is this thing on?
 

Offline yogece

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No.I thought of designing automated efficiency tester using arduino;I need some suggestions regarding the circuit designing.


Thanks for your reply
 

Offline Bored@Work

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What digsys tries to tell you is that building a linear regulator efficiency tester is a pointless activity. The efficiency can be calculated with good accuracy and need not be tested.

can any one suggest me from where to start

By getting your theory right.
I delete PMs unread. If you have something to say, say it in public.
For all else: Profile->[Modify Profile]Buddies/Ignore List->Edit Ignore List
 

Offline jmaja

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This has to be the easies assignment ever !
Open up EXCEL - create at least 7 columns - Vin, Iin, Pin, Vout, Iout, Pout, Efficiency
Eff = Pout/Pin x100
That's 100% accurate for Resistive loads, for ANY type of Linear Regulator (or use RMS for changing loads).

That's true for switching regulators as well, but how do you know Iin as a function of Iout? For larger loads you can assume Iin=Iout for linear regulators, but not for low loads, since they have ground pin currents from few uA to few mA.
 

Offline digsys

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Quote from: jmaja
That's true for switching regulators as well, but how do you know Iin as a function of Iout? For larger loads you can assume Iin=Iout for linear regulators, but not for low loads, since they have ground pin currents from few uA to few mA. 
It's still a loss DUE to the regulator, therefore included in the efficiency.
Hello <tap> <tap> .. is this thing on?
 

Offline yogece

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What digsys tries to tell you is that building a linear regulator efficiency tester is a pointless activity. The efficiency can be calculated with good accuracy and need not be tested.

can any one suggest me from where to start

By getting your theory right.
I would like to design something like this
 

Offline jmaja

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It's still a loss DUE to the regulator, therefore included in the efficiency.

Of course, but it makes some sense in measuring the efficiency at different loads instead of just calculating it, which you can do very accurately at higher loads.
 

Offline Jon86

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Inrush currents and ripple with an arduino? Good luck...
Death, taxes and diode losses.
 


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