7

Make sure the direction of the current for F1 is oriented the way you want it. LTSpice follows the "passive sign convention" -- namely, the positive current direction through a current or voltage source is from the positive node to the negative node. This means that a voltage source will have positive current when (conventional) current is flowing ...


5

If I have a resistor across which some voltage is provided and a periodic ( non-sinusoidal and non-zero) wave of current is given to the same resistor then what would be the average power. The average power is the average of the square of the voltage waveform divided by the resistance of the resistor. This boils down to calculating the RMS of the voltage, ...


3

simulate this circuit – Schematic created using CircuitLab The 9 ohm (A-D, D-C) and 6 ohm (A-B, B-C) resistors each form a voltage divider with 22.5V at their midpoints. Since the 9 ohm resistor between B-D has the same voltage at both ends, no current flows through it.


3

There already are two answers, I'll tackle the part that confuses you. What you have there is not a "system" -- you have a mathematical equation described, visually, as a block diagram. And because it's represented visually, you're interpreting it as a feedback system. It is not. As I mentioned the \$G\$ and \$H\$ "blocks" are simply ...


3

I believe that this system, where the loop has net gain and the response functions G and H are instantaneous, is a physically-impossible and mathematically ‘pathological’ case. I am speaking a bit colloquially with this term, but will explain what I mean by it later. The upshot is that all we need to do is add an infinitesimally-small amount of memory to get ...


3

There's two causes that are kind of related. The LM358 is not a comparator, but an op-amp. Op-amps can be quite poor comparators, and when inputs are very close to being equal, the output can be in a voltage that is half-on, instead of being fully high or fully low. There is also no hysteresis built into the circuit to have definite thresholds when to switch ...


2

Adding hysteresis is easy. Add a resistor between +input and ouput of LM358. Resistor will be high (some 10k or higher) for a little hysteresis. Must simulate to confirm value. I use microcap12 (free) from Spectrum Software. Ok. Made simulation. 10k -> hysteresis of 3 V approximately. 100k -> 700 mV hysteresis. If you want, use a variable resistor to ...


2

As others have stated, the block diagram you have drawn is physically unrealisable, so our physical intuition does not apply here. What you have found by \$\frac{Y}{X}=\frac{A}{1-AB}\$ is a DC equilibrium point of your "feedback" system. It's true that if we feed in an input X and compute the output Y using the aforementioned equation, we obtain a ...


2

Why can't the equation include that information? I don't see anything wrong with the derivation and block diagram. It won't and it never will, because for you to know the output of any time varying system, you must know two things: The initial condition of the system (t=0) OR the current state and current time the transfer function of how the system ...


2

From the comments: but the diode would be forward biased when the input value becomes greater than 5.6, right? Assumption: The diode is forward biased simulate this circuit – Schematic created using CircuitLab If the diode is forward biased (and is silicon), then \$V_a-V_3=0.7\text{V} \Rightarrow V_a = 5.7\text{V} \$. The current running through \$...


1

The inductor will oppose the change in steady state, hence, at the instant the switch is closed, the current through the inductor will be the same as the steady state value. Therefore \$\small I= \:... \$


1

Because it enters into discontinuous mode (DCM). If the energy has nowhere to go, it is a good feature that this buck converter stops transfering more energy. So you cut-off the load meanwhile the converter dumped the energy from inductor to the output as every period. Since there was no load, the voltage increased over. Due to the loss in capacitor and ...


1

simulate this circuit – Schematic created using CircuitLab Figure 1. Equivalent circuits. R1 and R2 form a 2:1 potential divider so Va can never exceed 5 V. The Thevenin equivalent voltage source is a 5 V peak with 5 kΩ source resistance. The rearrangement of R1 and R2 into R3 and R4 should make the potential divider clear. As shown by the equivalent ...


1

Feedback is an active criterium by which a quantity from the output is taken and fed back at the input, such that the output is dependent on both the input and the output. Think of it as an implicit operation with \$V_o=f(V_i,V_o)\$, not just \$V_i\$. As a simple example, think of the basic inverting opamp with its transfer function \$V_o=-(R_2/R_1)V_i\$, ...


1

Actually the high current pulses will be a few milliseconds long, so you’ll want a capacitor, or perhaps a bank of capacitors that can source 2A for a short period without significant voltage drop. If you have a switch mode converter driving the modem then the output-side capacitor only needs to maintain the supply for as long as the switcher needs to ...


1

You can break the loops at the red crosses as Rob Fox explains it in his article. You can do hand analysis or use Tian's method or Middlebrook's GFT. Middlebrook shows what is an ideal injection point and why it is normally not enough to inject a single voltage or current. See https://web.archive.org/web/20160401041428/http://ardem.com/D_OA_Rules&Tools/...


1

\$Z_0\$ is the surge impedance of an LC circuit. This is impedance seen by the initial transient current. Here is a good link to start on transients. Allan Greenwood’s Electrical Transients in Power Systems has couple great chapters on transients like this and is relatively inexpensive.


1

Your main load supply is 400 volts and your circuit attempts to deliver a constant current feed to the load hence, as you open SW1, the MOSFET is driven hard (at that precise moment) to try and force a current into the load despite the SW1 contacts opening. This creates a spark and, it seems to me, that if it didn't create a spark then it wouldn't be a good ...


1

At a guess, part of your problem is your construction technique. Start by changing your schematic a bit simulate this circuit – Schematic created using CircuitLab Now, about the changes. First, make the ground connection between the supply and source of the FET your primary connection. Every other ground should be outside this connection and not to ...


1

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Well, we are trying to analyze the following opamp-circuit: simulate this circuit – Schematic created using CircuitLab When we use and apply KCL, we can write the following ...


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