Hi all in EE stackchange, I wonder how the compensating resistor (\$R_3\$) in integrator with DC gain control equals \$R_3 = R_1 || R_2\$, which is to offset effect by the input bias current. I was trying to derive it by same way in Razavi's textbook but I can’t reach the result of compensating resistor \$R_3=R_1 || R_2\$ (As depicted in below Figure) It would be very appreciated if someone let me understand this. Thank you in advance!
121560
I found the following explanation online on determining the values used for the compensating resistor:  In either case, the compensating resistor value is determined by calculating the parallel resistance value of R1 and R2. Why is the value equal to the parallel equivalent of R1 and R2? When using the Superposition Theorem to figure how much voltage drop will be produced by the inverting (-) input's bias current, we treat the bias current as though it were coming from a current source inside the op-amp and short-circuit all voltage sources (Vin and Vout). This gives two parallel paths for bias current (through R1 and through R2, both to ground). We want to duplicate the bias current's effect on the noninverting (+) input, so the resistor value we choose to insert in series with that input needs to be equal to R1 in parallel with R2.  Although it is a 'crisp' and simple explanation, I don't understand how it is true, since the current flowing through R2 won't directly go to the ground, but will probably enter the op-amp at the output (op-amp as a current sink). Also, it might go through a few more resistances before going to the ground, since Vout isn't directly connected to the ground.
27363
What is the idea of choosing inductor &amp; capacitor value for buck DCDC convertors (no specific part number, in general)?  I've tried random one with low resistance, and it kinda works, but want to know idea behind calculation of optimal value (based on freq, and max current).
109862
Heat produced across a resistor can be computed by the following formulae  H=I²Rt H=(V²/R)t Going by #1, heat is directly proportional to the resistance whereas going by #2 heat is inversely proportional to resistance.  Which is the right formula to be applied to calculate the heat generated across a resistor &amp; why?
81489
Can somebody help me solve this. I need to find the equivalent resistance of these resistors. I really don't know where the trick here is. If somebody could help me I would really appreciate it
97571
I know this has been asked time and time again but bear with me as I am struggling to understand, so say I have this simple circuit here   with 120v (dc for simplicity) as the input voltage and I have an led that runs on 3.4v and max 20ma ohms law states I=V/R so R=V/I, so R=120v/0.02a = 6000 ohms or 6k is this correct? how do I know the voltage drop across the resistor is correct, and what about power dissipated by the resistor? also, Is there an easy way to calculate everything at once?
100870
I'm trying to calculate E12 resistor series myself.  Above is an illustration of the idea where such equation can be written:  x = r3 + r3 * %tol = r4 - r4 * %tol  form that we can derive:  r4 = r3* [(1 + %tol)/(1 - %tol)]  hence in generalized form we can write (%tol = 0.1 for E12 series);   r(i+1) = ri* [(1 + %tol)/(1 - %tol)]  But when I iterate the above formula in a for loop I obtain the following results for the first resistor values:  1.0000    1.2222    1.4938    1.8258    2.2315    2.7274    3.3335    4.0743    4.9797    6.0863    7.4388  Then I take [(1 + %tol)/(1 - %tol)] = 1.2 rounded value instead of 1.222222 and write the formula as:  r(i+1) = r(i)*1.2  But again when I iterate the above formula in a for loop I obtain the following results for the first resistor values:  1.0000    1.2000    1.4400    1.7280    2.0736    2.4883    2.9860    3.5832    4.2998    5.1598    6.1917  But the real values for E12 should be as:  1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8  What am I doing wrong here?
52658
I'm wondering why pull up resistors in text books are told such that they pull up the voltage to \$V_{CC}\$. Here is a diagram:    When the button is open what will be the input voltage to the micro controller? \$V_{CC}\$? No voltage drop at \$R_1\$?
28722
I'm having a bit of trouble understanding currrent limiting resistors in simple LED circuits. I know that I can determine the optimal resistor like so:  \$\displaystyle R=\frac{V_{s}-V_{f}}{I_{f}}\$  But I'm having a hard time understanding how this one value modifies the voltage and the current to the correct values for the LED. For example, if my calcuations for a super bright blue LED (with \$V_{f}\$ being 3.0-3.4&nbsp;V and \$I_{f}\$ being 80&nbsp;mA, and a voltage source of 5&nbsp;V) gives me 25&nbsp;ohms (using the lower bound of the forward voltage), that's fine. So the current throughout should be 80&nbsp;mA and the voltage drop for the resistor and LED should be 2 and 3 volts, respectively.  But what if I used a 100&nbsp;ohm resistor instead? Or any other value—how would I calculate the voltage drops and current? Would I assume one of them stays the same?
69024
I'm quite new to electronics and I'm having a hard time understanding the "pull-up resistor" principle. I have read a lot of articles about that, and I think I've got it but I'm not 100% sure so I have a question. In , after the first image, it says:     When the momentary button is pressed it connects the I/O pin to Vcc and the microcontroller would register the input as a high.   But I don't get it. Where is VCC? From what I see, there is no power source on this schema, just a microcontroller wired to a button that are both wired to ground so how can there be any voltage at all in this circuit?
26320
Shouldn't the current passing through a resistor be lesser than that which passes through a circuit? My understanding is that since Current = Charges/Time. If there exists a resistance to the flow of charges, then that must mean the charges slow down, meaning that more time is required to pass through a point. So, the current should then decrease. But, since this opposition to the flow of charges doesn't exist in the ENTIRE circuit, it should really only decrease the current in the resistor, right? Am I confusing the resistance of the resistor with the resistance of the wire? Any help will be appreciated! Thanks a bunch for lending me some of your time :D
67967
Well, I'm trying to find the Thevenin equivalent of this circuit:    As you can see, the only current source is a dependent source. However, the control variable is in a wire where, if nothing is connected between A and B, is \$i_\Delta = 0\$ A, and that makes the source 0 A. However, if I connect an impedance \$Z\$ to the circuit, I will have current, and a Thevenin equivalent. I don't know how to solve this problem. I've bee searching here for a while, and I found this:   . I can see how is solved that case. However, I still don't know how to apply that method to my circuit.  Any help about finding the Thevenin equivalent is appreciated n.n
78974
I understand that for resistors in series we need to add R for each resistor to get the equivalent resistance, where R is resistance. I also understand the mathematical manipulations that show that for resistors in parallel we need to add the 1/R for each resistor to get the equivalent resistance. However I have a couple of questions that have to do with the physical interpretation of the mathematical manipulations involved.   First, I am trying to understand why in the case of parallel resistors we have to add 1/R's instead of adding R's as in the case of resistors in series i.e. could it be thought that the current in the wire is like a sort of water flow in the pipe that has two horizontal parallel pipes positioned one on top of the other and in between connected by a number of vertical pipes that have a tap each that could be adjusted to represent different resistances. Then at the top of each vertical pipe the water pressure P exists and at the bottom the pressure will be reduced by P*R.   For example, suppose we have three vertical pipes, so then the pressure at the bottom of each (after the tap) is P*R for the total of 3*P*R so why is the equivalent resistance not 3*R*P but 1/R = (1/R1 + 1/R2 + 1/R3) and how can I be able to visualize this relation?
28697
I have the following circuit, where I put R52 = R53 = 2700 Ohms :    Triggering levels are of 1V and 2V approximately according to my computations, that is okay according to the following figure showing input in green and output in blue :    But what I find really strange is that the maximum voltage output is around 3.3V and not at 5V. How would you explain that ? And last detail : how would you explain that the output is not exactly at 0V when the input is high?  Thank you very much for your help.
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Exact question: In electric circuit theory, the combined resistance $R$ of two resistors $R_1&gt;0$ and $R_2&gt;0$ connected in parallel obeys $$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$$  Show that $R&lt;\sqrt{\frac{R_1R_2}{2}}$.  This proves that $R$ cannot be large if the individual resistances $R_1$ and $R_2$ are small.
65138
If I have a configuration of 5 resistors as follows:        &ndash; Schematic created using   How can I calculate the expected total resistance between A and B, given only values for R1-R5?  edit: okay, here's what I've tried so far.  I intended to find the expected voltage drops across each resistor, and from there I knew I could easily find the total resistance of the circuit.  First, I observed three loops in the diagram: R1,R3,R4; R3,R5,R2; and R1,R2,R5,R4.  I remember from  my old electronics class that the voltage drop around each loop in a circuit should always equal zero, as follows: $$ v1+v2-v4-v5=0\\ v1+v3-v4=0\\ v3+v5-v2=0 $$ (Assuming that current is flowing from A to B, I further assume that it is flowing down through r3.  If this is wrong, then the value for v3 would just have the opposite sign, and all would be well).  The sum of currents flowing into a point should equal the current flowing out, so I have these equations as well: $$ \frac{v1}{r1}=\frac{v2}{r2}+\frac{v3}{r3}\\ \frac{v5}{r5}=\frac{v3}{r3}+\frac{v4}{r4}\\ \frac{v1}{r1}+\frac{v4}{r4}=\frac{v2}{r2}+\frac{v5}{r5} $$  So now I have  6 equations in 5 unknowns... and if these equations are correct,  then there should be one redundant equation.  However, plugging them into a cas such as maple cannot find any unique values for v1-v5 in this system (an infinite number, in fact), suggesting that there are at least 2 redundant equations above.  Does this mean that there is no unique resistance for this circuit fragment, or what have I done wrong?
110568
I cannot understand how 3-wire measurement will improve the resistance measurement.    They say that IF lead resistance \$ R_L \$ is the same, it will compensate. How can this happen?  If bridge is balanced (without \$ R_L \$):  \$ \frac {R_1}{R_2} = \frac {R_g}{R_3} \$  If \$ R_L \$ is added: \$ \frac {R1}{R2}= \frac {R_g+R_L}{R_3+R_L} \$  \$ \frac {R_g}{R_3} \$ is not same as \$ \frac {R_g+R_L}{R_3+R_L}\$; is this true only if \$ R_G=R_3 \$ in the beginning? So this means that \$R_L\$ does not get compensated?
32743
We are teaching op amps in a high school class, before calculus.  So we can't use control theory to teach how op amps respond.  All the same, I would like to have an intuitive explanation for how a feedback circuit works.  Take the negative feedback, for instance.  Is there a clear way to show how an initial difference of delta between V+ and V- leads to a very large (G(V+ - V-)) difference in the output, which then leads to ...  I'd like to be able to take that argument and show how the voltage of the output converges in addition to the standard virtual short argument.  Can anyone explain it clearly?