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Something weird about this Rectifier

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One of my friends said that he recently conducted an experiment on rectifier to calculate its regulation. The experiment was fine until he observed something:
He observed that the No Load Voltage is less than the Full Load Voltage... Everything was checked thrice and the apparatus was found to be perfectly alright...atleast that's what he told me...
Now, how can the No load Voltage be less than the Full Load voltage?
Then it would turn out that the %Regulation is negative!!!

And just in case: %Regulation = [(V(noload)-V(full load))/V(full load)]*100

Can anyone explain what's happened...

Why is there a decrease in voltage when the load has been removed and why is it less compared to the voltage when there is no load... It should be the other way around right...
 

There are some possibilities.

1. Instrumentation error. Use an oscilloscope to measure the levels. There may be waveform related measurement errors using a multimeter.

2. The circuit may be designed for driving long lines to the load. This voltage change is to make the IR drop have less effect. I have seen a petrol engine powered portable electric generator that did this.
 

Possible cause for such behaviour can be bad wiring. If voltage drop caused by load current adds to reference voltage output voltage can be higher than with no load situation. Such solutions were used in old days for output voltage drop compensation.
 

You didnt say if you were testing just a rectifier or a rectifier followed by a regulator.
If it is only a regulator. I agree with flatulent on a instrumentation error. Since that with no load you are meauring pure DC and with load a ripple appears. The voltmeter may not like the ripple and give a false reading.
Only another reason I can think about is a strange behaviour of the transformer with increased current (and flux). I never saw this but you can measure the AC voltage on the secondary to check it.
If you are using a regulator. I would check the circuit for proper biasing specially with no load.
 

Thank you everybody... Those were some logical answers... The instruments were perfectly alright, he said.. Even their professor had the instruments checked... Not one, but many students got the same eror.. Any more explanations..??

The apparatus required for the experiment were:
1. Diodes 1N 4007
2. CRO
3. Voltmeter(0-30V)
4.DC Ammeter(0-20mA)
5.Multimeter
6.Transformer(9-0-9)
7.Connecting wires
8. Bread board

I guess that should give some more ideas...
 

Parts and equipment seem to be adequate for this experiment..

However, that was what your friend did tell you.
Sorry, mate, I don't thing he told you the truth...
 

Was the filter after the rectifier an inductive input type? This type requires a minimum load current otherwise the output is higher.

Can you post the schematic of the circuit?

Try measuring the transformer output with just resistive loading. It could be the B-H curve of the core changing the coupling with load.
 

What is CRO? Cathode Ray Oscilloscope or Croatia or Capacitor or what?
 

CRO is Cathode Ray Oscilloscope... And the schematic consists of a power supply given to a step down transformer...The trans secondary connected to a diode and then to a load resistor...

I think the diode is the culprit here... The diode's non-ideal characteristics are very crucial here...Maybe the Forward and Reverse resistances along with the Voltmeter's internal resistances... Maybe the parallel combinations and the series combinations are doing something here....
 

It has nothing to do with diode "nonideal" characteristics or with any element of the circuit. Have you used CRO and what is visible on the screen in both cases (no load and loaded rectifier). All what hapens should be visible on the scope. Voltmeter or multimeter are reading average value of voltage and it is higher when rectifier is not loaded. Are you doing experiment in presence of strong RF field (emmision)? This case can be cause of such problems.
 

Surpirising that it has nothing to do with the non ideal characteristics of the diode... bu then why do we connect a bleeder resistor? And i think there is not much RF emission in the lab... I performed the same experiment an year ago with the bleeder resistor without any problem... However, i never tried without the bleeder resistor...
 

U/I characteristic is a sort of hiperbolic function. To avoid significant changes of DC voltage when load currents are relatively low we add bleeder resistor to move working point of rectifier above the "knee" on to the more linear part of U/I characteristics.
 

Could it be something simpler?

If reading by meter (unloaded) the meter is the load.
The meter has High resistance and impedance.

Now given this thought, when the load is connected,
the meter and load are in parallel. I think the truer reading is made due
to total load is less then resistance.

Maybe this helps to point in another direction.

My imputs to help
wa
 

HI WA, that was exactly what i was thinking and you can even see my previous post.... i was talking about some parallel combinations... I think what you say can take us in the right direction...

When the diode is forward biased... and there is no load... the whole current has to flow through the voltmeter as there is no other path and also the diode's forward resistance comes in parallel with the high resistance of a voltmeter... Now when the diode is reverse biased, we are taking the reading just after the diode and the ground....Now the reverse resistance of the diode is very high, much higher than the voltmeter... and the reading now changes...

Until here, i could guess... If this is right...can someone summarize the whole thing please....
 

is the transformer has CT or not?
 

After simulation results verify your thinking. It is diode reverse resistance or leakage current causing this behaviour of rectifier. Voltmeter or multimeter measures average voltage. When load is not connected and only instrument is connected with it's 10Mohm resistance it shows average value which is a summ of voltage when diode is forward biased and plus when it is reverse biased. Because of diode reverse resistance during reverse biasing, summ voltage is smaller as with ideal diode. When you connect load, voltage can be higher as .5V depending on diode reverse resistance and load resistance. Just imagine, with infinite load resistance there is no rectification and the average output voltage is 0V (AC signal is appearing on infinite load resistance).
 

Can you give me the parameters for simulation please...I mean the diode forward and reverse resistance and the voltmeter's resistance...

And, yes the transformer is centre tapped... 9-0-9 model...
 

Simulation was as follows: Vac=40Vpp, Rload=10k-20MEG, diode=1N4007. I used PSPICE model for diode where Is parameter Is=3.115n. I do not know what kind of simullation are you intending to do but if diode reverse current is set to 100uA or reverse resistance to 500kohm you wil be able to demonstrate behaviour. I used half wave rectifier (only one diode) connected to AC voltage generator and Rload connected in circuit.
 

I was planning to do the simulation in MultiSim... Thanks for the values though... And yes i meant a half wave rectifier only...thanks borber
 

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