Circuit Stories

  I  answered  this  SE EE question  just now. OP asked: " Why do we need a resistor in a Zener diode circuit...?" We can answer this question by looking at the problem from a few points of view: "Dynamic resistor" From this functional point of view, a zener diode can be considered as a non-linear resistor that has the property of keeping the voltage between its terminals constant as the current through it varies. It does this by changing its "resistance" depending on the current. This can be illustrated by a simple experiment: Imagine that a current I flows through a variable resistor with resistance R so a voltage drop V = I.R appears across the resistor. Our task is to keep this drop constant. We can do it by a simple trick - when the current, for example, increases, we decrease the resistance to the same extent and vv. As a result, the product of the two quantities (the voltage drop) will not change. The role of the constant resistor in this arrangemen

   I  answered  this  SE EE question  yesterday. My answer How to compare voltages To detect if a voltage is below/above a given value, you need to compare it to that value. This can be done in two ways: ... by a threshold element First, we can apply (a part of) the input voltage to an element with a fixed voltage threshold, for example a diode, LED, base-emitter junction, zener diode, etc. If the input voltage is higher than the threshold, the current will be diverted to the threshold element. This idea is implemented in the OP's circuit where a part of the input voltage is obtained by the voltage divider and the base-emitter junction of the first transistor serves as a 0.7 V threshold element. When the threshold is too small (as in the OP's case), it can be increased by adding another or more voltage threshold elements (diodes) in series. In the OP's circuit, another diode can be inserted between the emitter and ground to "lift" the emitter with another 0.7 V. B

  I  answered  this  SE EE question  yesterday. Yes, Vout value can exceed the rail voltage in the case of an AC amplifier with decoupling capacitors. These capacitors are charged through the galvanic load to some bias DC voltage (usually, Vcc/2) that is subtracted/added depending on where the load is connected: If the load is grounded, the bias voltage is subtracted from the output voltage and the latter can be below ground. If the load is connected to Vcc, the bias voltage is added to the output voltage and the latter can be above Vcc.

I  answered  this  SE EE question  yesterday. The positive feedback in a Schmitt trigger is "self-reinforcing" ie, the output voltage rises uncontrollably like a snowball in an avalanche. Here, this effect is useful since it accelerates the transition and, in this way, it converts an analog circuit into digital one. In amplifier circuits like Armstrong's regenerative circuit, the positive feedback is "dosed" so that there is no continuous increase in the output voltage, only some additional amplification. It is hard to imagine how this amplification happens, especially when we make a follower amplify (I suppose there is some effect of accumulation), but it is a fact. It is interesting to compare the two types feedback circuits: In positive feedback circuits, from a low-gain amplifier we get a high-gain amplifier. In negative feedback circuits, from a high-gain amplifier we get a low-gain amplifier. Historically, in the past they used positive feedback to make am

I just  answered  this  SE EE question . OP said: "Why then, when the current stops, does it produce a kickback with inverse polarity?" Charging. As the voltage source charges the coil, current flows out of the positive end of the source and then enters the coil creating a positive voltage at its input. The coil thus acts as a load ("chargeable current source"). Discharging. When we open the switch, the coil begins acting as a current source that must produce the same current in the same direction. For this purpose, the coil reverses its polarity and begins to increase its voltage until it achieves its goal. But because the switch is open, it never succeeds, and its voltage theoretically reaches infinity. RL differentiator. Usually, there is a resistor in the circuit that limits the current. Here, for example, is my illustration of such an RL differentiation circuit, which is a snapshot of a Flash movie frame (you can watch it by downloading the exe file ):

I answered this SE EE question yesterday. My initial answer You have connected a real ammeter to an ideal voltage source. The problem of this arrangement is that the voltmeter inside the ammeter actually measures the entire source voltage... and this voltage does not depend on the shunt resistance Rsh (here we ignore the fact that the current will be significant). Thus your ammeter will act as a voltmeter. The problem with this 22nd century meter is that it is not a "true" ammeter such as a 19th century "coil ammeter" which directly measures current through the magnetic field it creates. The modern one measures current indirectly by voltage across a resistance and that is why it is "mislead" in this situation. We can observe two cases: 1. Known resistance. If the meter "knows" what the resistance Rsh is and uses it to calculate the current (this is the situation when we switch ranges and Rsh changes stepwise), the reading is true (in the sense

 I answered this SE EE question yesterday. Short explanation Non-inverting configuration. Simply put, the problems of the non-inverting configuration come from the op-amp inputs being at some voltages and, even worse, those voltages are changing. And when there are voltages, all kinds of parasitic leakages and stray capacitances "attack" and change them. The differential input has also been a problem in the past and they used a single-ended input. In addition, the differential input cannot be fully balanced and a voltage appears at the op-amp output even when the two input voltages are exactly the same (the so-callel "common-mode error"). The good things are that the differential voltage (between the two inputs) is close to zero and the input resistances are high. Inverting configuration. In contrast, all input voltages in the inverting configuration are zero and parasitic leakages and capacitances (to ground), even if they vary, do nothing. But this configurati

I  answered  this  SE EE question  on Dec 18, 2022. My answer How to compare voltages To detect if a voltage is below/above a given value, you need to compare it to that value. This can be done in two ways: ... by a threshold element First, we can apply (a part of) the input voltage to an element with a fixed voltage threshold, for example a diode, LED, base-emitter junction, zener diode, etc. If the input voltage is higher than the threshold, the current will be diverted to the threshold element. This idea is implemented in the OP's circuit where a part of the input voltage is obtained by the voltage divider and the base-emitter junction of the first transistor serves as a 0.7 V threshold element. When the threshold is too small (as in the OP's case), it can be increased by adding another or more voltage threshold elements (diodes) in series. In the OP's circuit, another diode can be inserted between the emitter and ground to "lift" the emitter with another 0.7 V

I  answered  this  SE EE question  on Dec 18, 2022. My answer How to understand diode switching circuits Understanding circuits starts with looking for familiar circuit building blocks. For this purpose, we can use all possible points of view on the unknown circuit. Here, considering this specific (OP's) example, I will share in a more general way my experience in understanding and explaining such diode circuits (diode limiters, diode logic gates, etc.) where diodes act as switches. I think this will be no less helpful to OP and readers than just some specific explanations. Here is my heuristic procedure illustrated by the simplified picture below (D1 is removed and D2 is replaced by a "piece of wire"). Customizing the circuit diagram Diode circuit visualized Drawing the diagram in a "geometric" form. It is preferable to draw the circuit diagram in a stripped-down form, placing the elements with positive voltage at the top and those with negative voltage at the

  I  answered  this  SE EE question  on Dec 18, 2022. My answer Basic ideas Perfect current integrator To make an integrator, we need a storage ("accumulating") element. In electronics, we usually use a capacitor for this purpose. When we charge it with a constant current, the voltage across the capacitor linearly increases as we want. Thus, the capacitor gives us an idea of ​​time. Imperfect voltage integrator In electronics, we usually work with voltages. So we would like to integrate voltage. The capacitor integrates current; so we need to convert the voltage to current. The humble resistor can do this work. Thus we obtain the simplest voltage integrator. But a problem appears - the voltage across the capacitor is subtracted by the input voltage. As a result, the current decreases and the voltage slows its rate of change. Perfect voltage integrator We can solve this problem by a clever trick from life - to compensate the "undesired" voltage drop, we add the same

 I answered this SE EE question on Dec 18, 2022. Here is my answer. There is a risk of damage only if "ideal" voltage sources (circuits with very low resistance) are connected to the two op-amp inputs. Current is what damages devices. In practice, there is almost always some resistance in series protecting the input.

I answered this SE EE question on Dec 18, 2022. My answer Maximum but undefined open-loop gain At first glance, the idea of ​​negative feedback is absurd because you take a very good amplifier with (almost) infinite gain and make it "bad" with gain on the order of tens and hundreds of times... and sometimes (in the case of a voltage follower) even a unit. Maybe that is what the patent specialists thought when they refused to issue a patent to Harold Black for something like a "perpetual engine". Now everyone knows the benefit of this "worsening" - from a high but undefined gain we get a relatively low but precisely determined gain. We achieve this by forcing the "undefined" amplifier to act as an "inverted voltage divider" made of two precision resistors. Therefore, as a rule, in negative feedback circuits, amplifiers operate at their maximum ("open loop") gain that the manufacturer was able to achieve. Moderate but fixed op

 I answered this SE EE question on Dec 16, 2022. My answer The challenge It is always a big challenge to understand a circuit that is new to you. Let's then try to do it with the OP's "topology"... Analyzing the circuit I suggest that we analyze their circuit from left to right (as input signals flow). First we see two input sources - one is constant (3.5 V) and the other is adjustable. Non-inverting amplifier We can drive the non-inverting input of the op-amp through the first voltage source; then the circuit name would be "non-inverting amplifier". Inverting amplifier With the same success, we can drive the inverting input of the op-amp through the second voltage source; then the circuit name would be "inverting amplifier". Differential amplifier We can even drive both op-amp inputs through both sources; then the circuit name would be "differential amplifier". Indeed, it is quite imperfect since the two input gains are not equalized.

I answered this SE EE question yesterday. Here is my answer. Despite the remarks, there is still something rational in the OP's question and I will try to use it to explain in the simplest possible way the idea of ​​a shunt voltage stabilizer... Current-to-voltage converter As OP has noticed, using just one resistor of resistance R with current I flowing through it, we can create a voltage V = R.I (Ohm's law). This resistor can be (and usually is) connected to ground ("grounded"... or "pull-down"). If the ground wire has any resistance, we might think of it as a resistor... but it gets pretty complicated; so let's separate the resistor from ground. When the current changes, the voltage will also change in proportion to the current. This device is very useful and has many applications... Voltage stabilizer ... but in other cases we (and the OP) want the voltage to not change when the current varies. How do we achieve this? Let's give the floor to OP

I answered this SE EE question yesterday. Here is my answer. I rely on the OP's claim that the circuit solution in question is an "improved Howland current source" and offer my explanation of the idea on which this famous circuit is built. The idea We can figuratively call it "dynamic voltage source". In some (usually transistor) current source implementations, to keep the current constant, we vary a resistance RI connected in series to the load RL while keeping the supply voltage V constant. With the same success, we can vary the voltage V while keeping the resistance RI constant... i.e., to supply the RI-RL network by a "dynamic voltage source" (see Fig. a on the left below). For example, if RL increases its resistance, the voltage drop VL = I.RL across it will increase... but the voltage source will increase its voltage V with the same value... and the current I = V/(RI + RL) will not change. Figuratively speaking, the voltage increase removes the

I answered this SE EE question  10 hours ago. Here is my answer. Looking for the idea This circuit solution is much more clever than it seems at first glance. It gives the illusion of being a very imperfect half-wave rectifier ... and has led some to improve it (including a second diode to maintain a virtual ground on the inverting input) to make it a perfect half-wave rectifier . But I think the idea here is different: This circuit is intended as a "slightly imperfect full-wave rectifier", and the "improvement" only destroys this idea. Operation Here is how the circuit works shown in a slightly different way than generally accepted. Negative input voltage The diode is forward biased ( on ) and the circuit is simply an inverting amplifier. In this configuration, we can see (with a little more imagination) two "voltage sources" - negative and positive, connected in parallel to the load. The first of them consists of the input voltage source and the two 2k

I answered this SE EE question on Nov 19, 2020. My answer Razavi's book is great but I cannot agree that "when v(input)=0 op amp raises Vy such that it barely turns on the diode." OP is right - if the op-amp is ideal, its output voltage will be zero when the input voltage is zero. The diode will be turned on only if the input voltage rises at least a little. There is a lot of philosophy in this specific circuit that illustrates the unique property of negative feedback systems to overcome disturbances. It is not only a particular circuit phenomenon; it is a ubiquitous phenomenon that can be seen everywhere around us. I have revealed this idea below in three consecutive steps to show its evolution. Prerequisite to understand what the op amp does in such a negative feedback circuit is to think of it not as a fast-acting amplifier but as a slow-acting "being". Undisturbed follower First replace the diode by a piece of wire (i.e., connect the op-amp output directly

I answered this SE EE question an hour ago. Here is my answer. Looking for the idea To see the idea in this circuit, it should be drawn more clearly in the following order: op-amp -> disturbance (diode) -> load.Then we can see that, in this "disturbed op-amp follower", the negative feedback is closed after the disturbance. Here is my (and my students') Wikibooks story about this phenomenon observed in negative feedback circuits. To understand it, try a simple experiment - apply some negative input voltage (e.g., Vin = -1 V) and put yourself in the place of the op-amp. What will you do to make the voltage of the inverting input equal to the voltage of the non-inverting input (the "golden rule")? What will be the op-amp output voltage? The classic "golden rule" The so-called "golden rule" is introduced in the famous book of "The Art of Electronics". It says that an op-amp with negative feedback does its best to maintain equal