# why we need Load resistor?

Discussion in 'General Electronics Discussion' started by sarabjot singh, Apr 13, 2014.

1. ### sarabjot singh

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Apr 10, 2014
Hello guys,
I am new to electronics and i want to ask why do we apply a load resistor in a circuit as the potential difference will always be same across the output terminal and how to calculate the value of a load resistor?

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010

The output voltage may well change if the source has an impedance significantly different from zero.

Presumably you calculate the load resistor to dissipate a certain energy or to allow a certain current to flow. Typically the choice of the value (and power rating) is done by a simple application of Ohm's Law.

3. ### Harald KappModeratorModerator

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Nov 17, 2011
A resistor can have different functions in a circuit. Only one of them is a load resistor - and even then load resistors can be applied in different ways.

A simple example of a load resistor is an audio amplifier where the load resistor is the speaker.

4. ### Y2KEDDIE

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Sep 23, 2012
When working with vacuum tube power supplies, a bleeder (load resistor was often used as a load to discharge high voltage off the filter capacitors when the supply was turned off .

A load is necessary to have power tranfer. Otherwise a circuit does no work, nothing happens.

To get maximum power from a battery you also need to match the load to the source impeadance (the internal resistance of the battery). This is typically very low. (milli ohms). The voltage will decrease at the output as the load reistor increases past the value of the source resistance. This is why a battery voltage check is useless without a load.

In a vacuum tube or transistor circuit, the plate or collector respectively has a charateristic load impedance/ resistance it needs to work into to give maximum output. Typicaly you want to match to that resitance/impeadance. (Load) You can look up the plate resistance in a tube manual, transistor maual.

I hope these examples help.

5. ### Ratch

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Mar 10, 2013
What you say about matching the load with the source is true, IF the source impedance cannot be changed. BUT, if the source impedance can be lowered, then more power will be transferred than if the source were equal to the load.

Ratch

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Maximum power, yes. Maximum energy, no.

7. ### BobK

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Jan 5, 2010
Of course some chemistry batteries will catch fire if you try this.

Bob

8. ### Y2KEDDIE

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Sep 23, 2012
Yes, more power will be delivered if the source impedance is lowered, this is why the load resistance in important. If you have a 5 watt source and you try to match a 10 watt load to it, it will burn up the source..
Its a case of having too little or too much. If the load resistance is too high ,to little power will be transfered, if the load resistance is too low the source will be taxed too heavily and burn up. If it's matched equally you will have maximum power transfer.

9. ### Ratch

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Mar 10, 2013
The ability of the source to provide sufficient power, or the ability of the load to dissipate all the power it receives, has nothing to do the the Power Transfer Theorem. That is an unrelated consideration analogous to making sure a resistor's dissipation is sufficient for the current through it or the voltage across it.

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Dec 18, 2013
I have seen people get confused with this and state for maximum power transfer to the load they match the source to the load? Their theory is if you have an 8Ohm impedance speaker then you must make the source 8Ohms for maximum power transfer. In this case they are calling the source the output of the amplifier and not considering the power supply which is the source of the power.

11. ### BobK

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Jan 5, 2010
Yes, I challenge anyone to show how their audio amp with an 8 Ohm output impedance transfers more power to an 8 Ohm speaker than my 0.1 Ohm output impedance amplifier.

Bob

12. ### duke37

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Jan 9, 2011
If the amplifier has an output impedanceof 0.1R, then maximum power will be obtained with a 0.1R load. It is a good way of producing smoke for a very short time.

If there are reactances, then a conjugate impedance will be required.

13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
In fact the argument is correctly stated the opposite way.

If you have an amplifier with a 2 ohm output impedance, the maximum power can be delivered into a 2 ohm load.

However, in an audio environment we need to consider the rated output of the amplifier. If it is rated as delivering 100W RMS into 8 ohms, we should not expect that it has an 8 ohm output impedance. In face, as Bob has suggested, the output impedance will be substantially lower.

The lower output impedance improves the "damping factor", and this is essentially a benefit of the impedance mismatch. Indeed an amplifier with an output impedance equal to the load impedance would have quite poor performance.

14. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Here's another angle to look at this. If the source and load have the same impedance then whatever power is dissipation in the load will also be dissipated in the source. If you think about this it's hardly a desirable condition.

Chris

15. ### Ratch

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Mar 10, 2013
Perhaps not under some circumstances. But the Power Transfer Theorem (PTT) does not address whether the source is capable of supplying the needed power, or the load is capable of receiving all the available power, or whether the damping factor is OK, or whether the frequency range is sufficient. The PTT simply gives the conditions where the most power is transferred from the source to load. If the load is fixed, then the most ower is transferred when the load impedance is the conjugate of the source impedance. If the load is fixed, then the most power is transferred when the source impedance is at the lowest possible value. The transfer efficiency also increases when the source impedance is lowered.

5,164
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Dec 18, 2013
I remember some guy who wanted to build an amplifier and not buy one. He thought because he had a 8Ohm speaker he needed an output of 8Ohms for maximum power.

17. ### BobK

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Jan 5, 2010
Yes, in fact, he argued that if the output impedance was less than 8 Ohms, putting a resistor in series with the speaker would increase the power delivered to the speaker!

Bob

5,164
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Dec 18, 2013
That's the one Bob. I remember you answering it.

19. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Perhaps some of this muddled thinking can be traced back to vacuum tube power amplifiers, whether it be in audio or RF applications. In the case of audio power amplifiers it was good practice to match the load with the typical 8R output impedance of the output transformer. Cranking up the gain of those amp's with no load on the output could generate voltages on the primary side that could destroy things. Satisfying the Maximum Power Theorem was good practice in those designs.

The same HV concerns were adhered to in RF power amp designs with the additional element of dealing with the characteristic impedance of transmission lines thrown into the mix. The advent of low impedance solid state hasn't changed the transmission line element of transferring power to the load though.

Chris

20. ### duke37

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Jan 9, 2011
I think Cdrive is a little wrong.
The dynamic output impedance of a pentode could be as high as 1M but the optimum load resistance will be of the order of 10k.
As Cdrive says, running a valve amplifier without a load will generate damaging voltages. Similarly running a transistor amplifier into a shorted load will give unacceptable currents.
Audio amplifiers often have negative feedback to reduce the output dynamic impedance and reduce distortion. The maximum permitted power will not be afected greatly.

The mains supply into your house will have a source resistance of less than 1 ohm. You would not wish to try to get maximum power out of this !

If the source and load resistences are the same, the power dissipated in the source will equal the power dissipated in the load. This may be OK in a low power signal circuit but will be disastrous in high power circuits.