Post your work here to see if it is correct.
thx for the replies...The easiest way to calculate this is to apply Thevenin's theorem across RL. After calculations, the result is:
IL = Vin * (K2 / (K2 * R1 + RL))
where
K2 = (1 - K1) / (1 - K1/K), and
K1 = R1 / (R1 + R2)
K = R4 / (R3 + R4)
Yes, it is, I verified it. However, it is difficult to see how the circuit behaves, when the result is presented in this format.according to wat i hv done, i used nodal analysis and i get a formula of
IL =( Vin*(R2/R1)*y)/(1-x*y*RL)
where x = 1+ (R3/R4)
and y = R1/(R1R2+R2RL+R1RL)
is it correct??
I will post the solution soon. Stand by.bwt, adriann how u use Thevenin's theorem to solve?? u open circuit RL then find Vth and Rth??
I assume you refer to the resistance seen by Vin. If the current flows into Vin, the resistance seen by Vin is negative. For this, all you have to do is prove that the voltage drop across RL is greater than Vin. In other wordsanother thing... how to verify that the circuit is a negative impedance converter(VDCS)? can i just use different voltage to test the circuit and measure the current then plot the graph to so than the V-I graph having negative gradient (negative resistance) to verify it?? can anyone suggests any better method??
my lecturer wan us to verify the circuit capability(it is to produce voltage dependent current source?) and its limitation(for practical only) with using circuit analysis and experimental results.Yes, it is, I verified it. However, it is difficult to see how the circuit behaves, when the result is presented in this format.
I will post the solution soon. Stand by.
I assume you refer to the resistance seen by Vin. If the current flows into Vin, the resistance seen by Vin is negative. For this, all you have to do is prove that the voltage drop across RL is greater than Vin. In other words
IL * RL > Vin.
In this case, R1 current flows into Vin. To see this, let’s look at the K, K1 and K2 ratios in the result I posted earlier. This is the reason why I like to leave formulas with resistor ratios like K and K1. That way I can better see what is going on as resistors change. Here is how I would analyze the result I posted earlier, aside from plotting I1 versus Vin.
1. Assume that R2 is infinity (not connected). In that case, K1 is zero and K2 is 1, so IL becomes Vin / (R1 + RL) and the whole circuit is a regular non-inverting amplifier.
2. Assume that R2 is connected. As R2 decreases, K2 increases and IL*RL increases.
3. This circuit presents a negative resistance as R2 decreases furthermore. As R2 decreases K1 (which is sub unity) increases and, at some point 1-K1/K becomes negative, since K is a constant. As a consequence, K2 becomes negative, and IL*RL becomes greater than Vin. Therefore, R1 current flows into Vin and increases as R2 decreases. From Vin point of view, the circuit is a negative resistance.
4. As R2 decreases more, at some point Vo saturates to the op amp supply and the current supplied by op amp through R2 into Vin cannot increase any more.
I do not know your resistor values and what you are trying to accomplish. This was just a sample of the analysis you can make. Whenever you see a difference at a denominator (like K2) you should assume that you will see some instability for some values.
I posted the solution here:bwt, adriann how u use Thevenin's theorem to solve?? u open circuit RL then find Vth and Rth??
I think you have plenty of info now to come up with an analysis.my lecturer wan us to verify the circuit capability(it is to produce voltage dependent current source?) and its limitation(for practical only) with using circuit analysis and experimental results.
thx for the solution...so the circuit capability is a voltage dependant current source right ? for practical test i just need to measure the current IL with using different value of Vin right??I posted the solution here:
http://masteringelectronicsdesign.c...eorem-to-solve-a-negative-resistance-circuit/
I think you have plenty of info now to come up with an analysis.
Again, you can analyze it yourself. Look at the IL expression. As long as it depends on RL, the circuit is not a current source. It can become a current source when K2 is large. In that case IL = Vin / R1. So, yes, at RL connections, the circuit becomes a voltage dependent current source. For example, if R1,2,3,4 = 10k, and Vin = 1V, IL is 0.1mA and does not depend on RL.thx for the solution...so the circuit capability is a voltage dependant current source right ? for practical test i just need to measure the current IL with using different value of Vin right??
by Jake Hertz
by Jake Hertz