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# Can a voltage divider act like a voltage amplifier?

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#### gayu

##### Junior Member level 3
Yes. If a voltage divider uses a resistor, why cannot we use a negative resistance components like tunnel diode to get the voltage amplification.Clarify me.

GD

gayu,
You cauld also use a negative immitance converter (NIC) to produce a negative resistance. However, implementing a NIC requires the use of a gain block, such as an op-amp. If you're going to do this, you might as well use the op amp in a more conventional amplifier topology.
Regards,
Kral

Note that you cannot implement a negative resistor without having some kind of additional power supply. And that is where the amplification is coming from. Sorry -- no free lunch

Sorry v_c, I donot get your point. Why do we need an additional supply.Can you throw some more light into it.

GD

Consider that you have a negative resistor (two terminal device). Now we know that the power dissipated in a resistor is Ri² and this is positive if R>0. But since we have a negative resistor Ri²<0 and the power is not dissipated, but rather generated. So this device cannot be passive and must contain a power source.

Say you have a 10V source and a 2Ω resistor. The resistor should draw 5A from the positive side of the source. Now if we had a -2Ω resistor, a current of 5A would flow into the positive side of the battery. So in this case the negative resistor is generating power and the source is absorbing the power. How can this be? Well the only way is to have some kind of power source inside the two terminal device that we are calling a negative resistor.

Does this help out?

Best regards,
v_c

v_c,

I accept your idea.But I have one more doubt. In the case of micro wave solid state devices we talk about negative resistance with no relation to the power generation. That is when the voltage increases the current across decreases.. some thing related to the phase shift of 180 which gives rise to the negative resistance. If that is the case, where is the logic of power generation comes..Please clarify

GD

I think you can have devices that exhibit negative resistance characteristics over small regions of operation. Over these regions the increase in voltage may result in decrease in current. But this would then be characterized as a incremental negative resistance. I think a tunnel diode would be such a device (echo47 already mentioned this in his earlier post). Take a look here https://www.allaboutcircuits.com/vol_3/chpt_3/11.html

For a positive resistor if we plot the i-v characteristics we get a straight line going through the origin. The slope of this line is 1/R, with R being the resistance. For a tunnel diode the slope is different depending on where you are on the curve. Not only that but the slope may be positive or negative. So as you can see on the above link, if you operate between Vp and Vv your incremental resistance will be negative.

However, note that this is still a passive device. Why? Because the the power dissipated in this device (the product of i and v) is still positive. Since the device is always operating in the 1st quadrant where i>0 and v>0. I think the device also operates in the 3rd quadrant (not shown in the figure) where i < 0 and v < 0 but the product i*v is still positive. This devices always dissipates power.

Does this help?

Cheers,
v_c

well to be able to do that you would need something which is capable of producing negative impedance

Yes it is possible to do that. It is called a Negative Impedance Converter (NIC) -- this was already mentioned by Kral (see above). The problem is that the NIC is implemented using opamps so it needs to have power supplies to run it. So the circuit can act like a voltage divider but the amplification is coming from the power supplies that are running the opamps.

Best regards,
v_C

I hope volat divider is different fro an amplifier which makes use of active devices to strengthen the ac signal making use of dc supply..
Voltage divider is one which is used to divide or else decreasing the voltage w/o using any ac signal..ok

Consider a passive LCR circuit, this circuit is a voltage divider at most frequencies but at the resonant frequency w=1/√(LC) the voltage across L or C would be an amplified voltage. So this is like a voltage divider at 1 frequency and a voltage amplifier at 1 particular frequency.
Consider the values of components:
R = 10Ω, L=1mH, C=1nF and Vac=100V

then calculate the voltage across the inductor in steady state at a freuqncy of 1 KHz and 1MHz you will see how that happens.

aryajur said:
Consider a passive LCR circuit, this circuit is a voltage divider at most frequencies but at the resonant frequency w=1/√(LC) the voltage across L or C would be an amplified voltage. So this is like a voltage divider at 1 frequency and a voltage amplifier at 1 particular frequency.
Consider the values of components:
R = 10Ω, L=1mH, C=1nF and Vac=100V

then calculate the voltage across the inductor in steady state at a freuqncy of 1 KHz and 1MHz you will see how that happens.

the voltage amplification effect would certainly happen in the LCR circuit. but the input should be AC at the resonant frequency. also, the voltage increases but the current decreases so we still can't get power amplilfication. voltage amplification of dc signal using passive component is something I never heard before. maybe someday there will.... who knows?

Voltage divider is a passive circuit.I cant understand how it can work as amplifier.Wont this be against law of conservation of energy?

I think that the negative resistance property of the tunneling diode occurs when a certain bias is established and so it is called a -ve differential resistance.
I mean as others said before to get such a -ve resistance a bias is needed (not for free)

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