what is loading effect

[vikram789]

can anyone explain me loading effect in detail and thereafter explain the statement that "loading effect cumulatively decreases the voltage supply"

 

[abbeyromy]

Suppose you have a common emitter transistor amplifier driving the load of say 4 ohms.

The gain of a transistor is directly proportional to the collector resistance that means higher you keep it better the gain ( keeping in mind that increasing RC too much can lead to a low IC sat and thus drive your transistor in saturation if biasing arrangement is not taken care of).

Generally this collector resistance is in K ohms range for example say 4.7K

So what happens is that this load of 4 ohms is attached to this. It comes in parallel with the RC=4.7K and drastically reduces the effective collector resistance and hence reduces the gain.

I am not sure the next question is being asked in which scenario but i will try to imagine and explain

Now what has happened is that your output voltage is decreased due to the loading effect.
Suppose you are using a feedback mechanism in which the output voltage is fed back to input side then due to loading effect even your effective input is decreasing. Since output voltage is proportional to input voltage also ( Vout = Gain*Vin) so loading effect us cumulatively decreases the voltage supply.

Another explanation can be that if this common emitter stage is driving another stage( output of this is input to other) then the loading effect has causes to decrease the coltage supply for further stage.

More the cascading you do more the gain is reduced. That is why for low load cases we prefer darlington pair arrangement. Refer text books for ac analysis of BJT to get the equations.

Hope this helps!
-Hemanshu

 

[vikram789]

thanks for explanation.......can you explain loading effect considering Digital Domain and Mos. The scenario is that a Low Voltage detector(LVD) is detecting when the o/p voltage of a particular unit goes low so that it can trigger Voltage Regulator. i came to know that o/p voltage that is detected by LVD goes low due to LOADING EFFECT, so i wanted to know what it is

 

[abbeyromy]

@vikram789

This is what is my thinking, please check with seniors!

- suppose you have a circuit with 10V supply and 2 resistors R1= R2=5 ohms connected in series.

Now voltage across each is 5V and current is 1A
For the purpose of measuring this voltage you used a voltage detector/meter across R2 having a ressitance R3 =1 ohms. This comes in parallel to R2 and reduces the effective R2 to 5/6 ohms. Apply KVL and you will see that the voltage across R2 is now just 10/7 rather than 5V. Thus we see that due to loading effect your output voltage decreases, voltage across input resistance R1 increases.

Therefore it is always recommended that a voltmeter should have a high resistance compared to the circuit being measured, to minimize the loading effect.

So I think your LVD is having low input resistance which is causing the reduction in effective output resistance of the Voltage regulator causing it to trip..

Does this sound logical?

 

[vikky]

i think im facing a load problem.my input is 3.3V.but when i connect to PCB its getting only .5V.how to avoid this?

 

[aminoacid]

i think im facing a load problem.my input is 3.3V.but when i connect to PCB its getting only .5V.how to avoid this?

Is it possible to show us a diagram of your circuit?
Source of problem could be:
1): your 3.3V source current is not enough for your circuit
2): Your circuit has a faulty component or design error

 

[vikky]

Is it possible to show us a diagram of your circuit?
Source of problem could be:
1): your 3.3V source current is not enough for your circuit
2): Your circuit has a faulty component or design error

thanks for your reply.
i am attaching my circuit.i am trying to give 3.3V to pin no:10 of AP89010 IC.is it enough to connect a series resistor from source to load?

 

[aminoacid]

A quick look at your circuit, reveal a problem, according to the AP89010 data-sheet the IC operates at 3.3V, but then you're connecting a 5V dc to pin 15 SBT..!
and it should be 3.3V too.

pin number 10 is a select pin and yes you can stick a 3.3V via a push button for example to trigger a specific sound
please have a look at the bottom of the data-sheet for an application example.

Best regards

 

[vikky]

thank you.point taken.
for triggering to be removed SBT should be disconnected and pin 10 should be given 3.3V according to datasheet.But when i give 3.3 V to pin 10 due to loading effect it gets reduced to 1.8V.how to solve this problem.

 

[aminoacid]

Hi, I think this is due to a lack of enough current available.
Could you tell me what's the rating of your supply at the power connector input?

 

[vikky]

3.3V ,200mA.how much current will it be necessary.if i give it from a different source with high current rating,will it solve my problem?

 

[aminoacid]

If you connect 3.3V at J2 power connector you'll get 1.9V at the input of your 5V regulator!

 

[vikky]

If you connect 3.3V at J2 power connector you'll get 1.9V at the input of your 5V regulator!

The circuit is actually getting a supply of 12V 500mA to the input of 7805.Now i am trying to give a external trigger of 3.3V directly to S1(pin 10) to trigger a voice.when i give this i am actually getting 1.8V in pin 10.

 

[aminoacid]

Since S1 has an internal pull-down, the external trigger source appears to have a large output resistance, try for testing purpose to connect S1 to the local 3.3V source..and see if you have the same reading..

 

[vikky]

Thank you sir for the reply.i will try it out and keep you posted.

 

[yogeshkothiya]

Thanxxx!!!!!!!
-Hemanshu

 

[Syncopator]

An ideal voltage source has zero internal impedance.

All practical voltage sources have some, measurablle, impedance even though it may, in a well designed circuit, be very low indeed.

The output of a voltage source, measured when it is not supplying power to anything, i.e. it is unloaded, will be the design value.

When the source is called upon to deliver current to a load the voltage at the output tends to decrease due to the drop across its internal impedance. A well designed source will compensate for this to a large dregree.

On poorer designs, or unregulated ones, the drop can become quite noticeable, even unacceptable.

The drop in voltage is called, since it is due to increasing load current, loading effect.