I'm studying the following circuit   The image is explicit but as you can see the input measured in the upper terminal is a square wave with \$E_c\$ and \$-E_c\$ as levels.   Now I also had no trouble understanding that if \$R_B \leq \beta R_C\$ when the square wave is in the positive half-cycle we have saturation and when it is on the negative half-cycle we have cutoff situation.  Now in saturation we will (because the junction voltages are zero) an output voltage of zero. That is correct with the figure. However when we are in cutoff, applying KVL to the output mesh leads us to   $$U_{CE}=-E_c$$  because currents can be neglected.  That however is not concordant with the figure which shows $$U_{CE}=E_c$$ Why is that? What am I missing?
122684
I'm studying the following circuit   The image is explicit but as you can see the input measured in the upper terminal is a square wave with \$E_c\$ and \$-E_c\$ as levels.   Now I also had no trouble understanding that if \$R_B \leq \beta R_C\$ when the square wave is in the positive half-cycle we have saturation and when it is on the negative half-cycle we have cutoff situation.  Now in saturation we will (because the junction voltages are zero) an output voltage of zero. That is correct with the figure. However when we are in cutoff, applying KVL to the output mesh leads us to   $$U_{CE}=-E_c$$  because currents can be neglected.  That however is not concordant with the figure which shows $$U_{CE}=E_c$$ Why is that? What am I missing?
84471
I need to implement the following circuit in LTSpice but I don't know if my schematic is ok. Also, I don't know what  values for I should use for Cinf and R2.  The following constraints are imposed:   VCC= 10V VCM = 1.5V Vgis a sine wave with a frequency of 1 kHz. Its amplitude will be chosen so as to ensure minimal distortion of the output signal (i.e. v0 should also be a sine wave) R1= 15kΩ.    I need to choose a value for R2 and Cinf in order to have a (theoretical)gain value Av=1500.   How can I compute the theoretical expressions of the DC collector currents of bipolar transistors Q1–Q8, considering VBE≈ 0.6V and the theoretical expression of the small-signal voltage gain Av= vo/vg?   The first picture is the schematic and the second is the LTSpice implementation.  What I found is that:   ICq5=ICq6=ICq7=ICq8=(Vcc - Vbe)/R2 ICq1=ICq2 = [ICq7 / 2 ] * [Beta/(Beta+1)] ICq4=iCq2-current of Q5 ICq3=ICq1-current of Q3-current of Q4  Beta in my case is 182.1 because I'm using a BC847A transistor. A=Vo/Vg = Gm_q2 * Beta_q5 * R1Gm_q2=ICq2 / Vt that is approx 40*ICq2   How can I find the value of R2? This is what I have now, an sine like wave for Vg
89260
If we connect an AC source with a pure inductor when AC source is going through 0 volts we get the following graph of the voltage and current.  There are few things in the graph about which i am confused.  In the first graph/diagram "A" of the picture which shows the voltage when it starts to flow and the corresponding current, the current doesnt seem to be lagging behind the voltage by 90 degrees. What is the reason behind that? Is the statement "current lags behind the voltage in an inductor BY 90 degree" is valid just for the steady state voltage shown in graph "B"?  Second confusion is that for the first half cycle of the voltage in graph "A", current is rising both for increasing and decreasing voltage. How is this possible that current is rising in the inductor for both increasing and decreasing voltage?
18346
I am a licensed radio amateur, and find bewildering the many different explanations, which range from folksy urban myth to Maxwell-Heaviside Equations, of what happens at the termination of a transmission line or feeder. I realise that they all come to the same thing in the end (or should do, pun perfect), but none of them give me a gut feeling for what is going on.  I like diagrams, so an answer in terms of (graphical) phasors for the currents and voltages at the load would suit me best. How, for instance, does a step pulse down the line cause twice the voltage at an open circuit termination? Similarly for current at a short circuit. And how is the reflected step generated by the inductance and capacitance of the line?  Can anyone help, without getting all mathematical, and not telling any "lies to children"?
122291
I have square waves of different frequencies (1KHz to 20KHz), and I need to convert them to a sine wave of the corresponding fundamental frequency.  A RC ladder Low-pass filter was the first thing which I tried, It gave good results but the output peak-to-peak voltage (Vpp) varied a lot with the frequency. The square wave has a Vpp of 3.6V. But after filtering, the output Vpp of the sine wave varies from 3V to 2V as the frequency increases.  Is there any other better way to get a pure sine wave from square wave of same frequency without this voltage drop?  Thank you
123059
In the power amplifier circuit shown if the input is an RF sine wave, with VDD=1.8V and Vgs = 1V, what will be the waveform at the drain of the MOSFET - sine or square wave? Is it possible to get a sine wave at the drain?
67477
According to various sources, the ripple voltage of a full-wave bridge rectifier is   $$ V_r = \frac{V_s}{2}fCR $$  where \$V_s\$ is the peak of the input voltage source, \$f\$ is the frequency, \$C\$ is the capacitor and \$R\$ the resistor.  I have simulated this circuit on PSPICE, circuitlab, and built it in real life:   \$R\$ is \$4.7K\$ and \$C\$ is \$10 uF\$. Voltage source is sine wave with \$0 V\$ offset, \$5V\$ amplitude and \$40 Hz\$ frequency.  All 3 experiments give me a ripple voltage of approximately \$0.65V\$, but the formula gives me \$1.330 V\$ (twice as much).  What am I missing?
113583
The image of the circuit is shown below and it required to find \$V_0\$,  My first attempt at solving this problem is by changing the current source into a voltage source with 1-V and 2k\$\Omega\$ resistance. The fact that the inverting and non-inverting terminals aren't grounded make this problem look difficult.To the point which, I don't know how to proceed with this question or where to start. I would appreciate any help. Following some thought and another schematic from a hint suggested by Alfred, I produced a schematic representing our work.  And  my solution for the problem is below, Using node equation at nodes A and B we have, $$\frac{V_A-V_B}{1k}=-0.5 \text{mA}$$ $$\frac{V_B-(2+V_A)}{1k}=-0.5-x$$ where x is the current that is sent in the output of the op-amp. Using KCL, at the bottom node near the current source we see that the same current that goes through the op-amp also goes through the \$2\text{k}\Omega\$ resistor. Hence, we have, $$\frac{-V_A}{2k}=x$$ Replacing this in the second equation, $$\frac{V_B-(2+V_A)}{1k}=-0.5+\frac{V_A}{2k}$$ And and solving the equations yields \$V_A=-2 \text{-V}\$ and \$V_B=-1.5 \text{-V}\$
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?
101703
I have some problems with analyzing this pretty simple circuit (or more precisely with understanding the solution):  The task is to find the characteristic curve for $$|U_E| \le 5V$$     My first thought was that when the input voltage is below the threshold of the diode, the output voltage is only dependent of R1 and R2 as the diode doesn't conduct. When we are above threshold we have something like R1 + (R2 || R3) - U_D.      The first step would be to find the breakpoint between diode is conducting and not conducting. Well.. The solution says that for the first case (diode not conducting): $$ U_{R2} = U_S = R_{E,S}\frac{R_2}{R_1 + R_2} $$ $$ U_{E,S}=U_S\frac{R_1 + R_2}{R_2} = 2,1V $$   The explanation for this equation is that at the break point, the diode current was zero and following from this the voltage at R2 equals UD = 0.7V. And here's the problem. When the diode is not conducting, why R2 is dependent of the threshold voltage of it? I'd expect that the voltage at this point was only dependent of R1 and R2.     For the output voltage the solution confirms my thought:  $$ U_A = U_E\frac{R_2}{R_1 + R_2} = \frac{U_E}{3} $$
79830
Suppose a series RLC circuit has its resonance frequency f_r. If we apply AC current of frequency f_r to this series RLC circuit then circuit has maximum current as output. My question is little bit hypothetical. What will be the output of the series RLC circuit if we apply AC current of mixed frequency where one of its frequency is f_r? For simplicity we can assume AC current made up of two frequencies, (say f_r and f_s) where f_r != f_s. PS: I am very sorry for any incovenience. Please comment if still there is ambiguity.
542
it once happened that when I touched the positive terminal of the Oscilloscope input probes letting its negative terminal open, I observed sinusoidal voltage signals with 0.5V peak to peak and of frequency nearly 50Hz. I don't understand the exact cause for these signals. When I touched the other, negative terminal output signal reached zero with a very little noise...(with the positive terminal connected to the body..) Is it because the power lines near the place I observed these are 50Hz and those induced a potential difference across me with respect to the supply ground which is also the ground of Oscilloscope... or something else.. When I touched a wire or a resistor with positive terminal of Oscilloscope.. there was no change in the Output.. Its because the equivalent model of our body includes capacitance and inductance with resistance.. If so then actually how we are being induced with such voltage signal..? Actually the discontinuity in signal is because of the difference between the frequency of actual signal and sampling frequency of my camera.
66049
The standard as I learned it is that the flow of current is from positive voltage to negative voltage. The direction of current flow is opposite the direction of electron flow. This is what I learned in school, the assumption made by almost every engineer I've worked with, and the standard in every simulator I've ever seen.  I've encountered two exceptions. One is this , which a colleague has on his shelf from his associates degree training. The other is a pair of former navy personnel I've worked with, both of whom were taught electron flow in the navy and had to unlearn it when they left.  I understand that both are valid mathematical models of the same phenomenon. What I don't understand is why anyone would purposefully and knowingly train people in a fashion contrary to a near-universal standard. (It took a lot of work to make that textbook!) Is the standard as I understand it not as standard as I think it is? Is there some analytical advantage to the electron flow model that I'm not seeing, perhaps in realms of EE I don't visit often?  Why is anyone, anywhere, ever taught that current flows from negative voltage to positive voltage?
114067
I'm studying the circuits which are used to implement operational amplifiers. For example, I've studied the two stages amplifier (in which the first stage is a differential amplifier with differential to single ended transformation, the second stage is an inverter with active load), the telescopic cascode, and other circuits. Here a picture of these circuits:      I noticed that in all of these circuits the book I read always assumes a differential input, that is two (small) signals with the same DC value and amplitudes which are equal and opposite. As a consequence of the superposition principle and of the symmetry of these circuits, we can divide the circuit into two parts and all the nodes on the axis of symmetry become ac grounds. This simplifies the analysis, in particular it becomes easier to find the differential gain.  The question is: who says that, when I close these circuits with feedback, I will have at the inverting and at the non-inverting terminals a perfect differential input? It appears as if the book assumes that I will get for sure this situation. In other words, who says that, when I close the circuit with feedback, I get the same differential gain (if I close the circuit with feedback, in general I don't have two perfect differential signals, and as a consequence I am not allowed to divide by symmetry the circuit and to consider all the nodes on the axis of symmetry as ac grounds)?  For example, let us assume that I use one of these circuits to implement the classic inverting configuration:    You can see that the non-inverting terminal is fixed at ground, then it is impossible to have a differential input for the op-amp.  Thank you  Image added for the comment:
40154
I was trying to solve this unbalanced Wheatstone bridge and found that the current $i$ in the mid wire is 9A .This result was achieved by using node voltage method.    Then I rearranged the same circuit to the form shown in the second image.  and at this point I flipped the right half of the circuit so as to obtain the equal resistors on the same side.    now I again rearranged the circuit as shown.    Here, the final circuit  is a balanced Wheatstone bridge and thus the current i must be 0.  So clearly something went wrong in there.  I saw a post  before this one just to make sure that there is no mistake in rearranging the circuit in the way I have done.   It was answered by the user PhillipWood as being correct and was unfortunately put on hold.  So can someone please point out what exactly went wrong with the above method.
116486
In my intro E&amp;M class, we were given an example problem to determine the voltage within a cylindrical capacitor as a function of radius. A diagram of such a cylindrical capacitor is below.  The solution from my instructor is as follows.  Let the outer, negatively-charged surface be denoted by “b” and the inner, positively-charged surface be denoted by &quot;a”. So, we have $$\begin{align}\Delta V &amp; = V_a - V_b\\&amp;= -\int_{r_b}^{r_a} \vec{E}\cdot\text{d}\vec{s}\\ &amp;= -\int_{r_b}^{r_a}E\cos\theta\text{ d}s\end{align}$$ Now we note that $\cos\theta = -1$ in this case and then make the substitution $\text{d}s = - \text{d}r$. $$\begin{align}\Delta V &amp;= -\int_{r_b}^{r_a} E\cos\theta\text{ d}s\\&amp;= \int_{r_b}^{r_a} E\text{ d}s\\&amp;= -\int_{r_b}^{r_a} E\text{ d}r\end{align}$$   When we actually substitute in the expression for the electric field as a function of $r$, we get the correct answer. However, my concern with this solution is in the nature of the order of substitution. When I approached this problem, instead of first making a substitution for $\cos\theta$, I first made the substitution $\text{d}\vec{s} = -\text{d}\vec{r}$. To me, this vector substitution made more sense than the scalar substitution of differentials since this would be more indicative of the opposite direction between the &quot;direction of integration&quot; and the radial vector. So, using this approach, we get $$\begin{align}\Delta V &amp; = V_a - V_b\\&amp;= -\int_{r_b}^{r_a} \vec{E}\cdot\text{d}\vec{s}\\ &amp;= \int_{r_b}^{r_a}\vec{E}\cdot\text{d}\vec{r}\\&amp;= \int_{r_b}^{r_a} E\cos\theta\text{ d}r\end{align}$$ Now, since the differential radial vector and the electric field are parallel, $\cos\theta = 1$, so we get $$\begin{align}\Delta V &amp;= \int_{r_b}^{r_a} E\cos\theta\text{ d}r\\&amp;= \int_{r_b}^{r_a} E\text{ d}r\end{align}$$ This is the negative of my instructor's solution, which obviously leads to the wrong solution. Where do I go wrong? Why can I not simply reverse the order in which my teacher makes substitutions?
12961
Here is what I know about the Ideal Op-Amp.   The open loop voltage gain is infinite The output voltage is given by the following \$v_o = A(v_+ - v_-)\$ Only with a negative feedback loop is \$ v_+ = v_-\$   My query is with regards to the negative feedback loop as shown below:   My queries are as follows:   Clearly, \$v_o = A(v_+ - v_-)\$ should still apply and since \$ v_+ = v_-\$, shouldn't the output voltage \$ v_o = 0 \$ always? Since \$v_o = A(v_+ - v_-)\$ should still apply, is A still the open-loop voltage gain which for an ideal op-amp is infinity. Thus, would the output voltage always be infinity?