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35

The pointer should have balance weights to counteract the pointer's weight, so the meter reads the same regardless of position. It’s likely that those weights became dislodged, so you’ll need to take on the delicate task of rebalancing the movement. The zero adjust screw won't fix this, it only changes the preload on the meter's spring. You can see this in ...


18

You will find a variety of arms projecting from the moving coil, some with wire wrapped round them as weights (copper or possibly even solder as a field repair), often forming a cross shape. First picture here shows the left and bottom arm, the bottom arm has a sleeve weight on it. One or more of these weights may have fallen off; if you are very lucky they ...


17

The Question You write at the end: As capacitance is dependent on plate area dielectric material and distance, with two point wires in an open circuit the capacitance simply becomes infinitesimally small but non-zero. If so what is it about capacitors that the voltage can pass from left to right plate but in an open circuit it can't. Is there some arbitrary ...


14

When I was at school (about 1980) we did a number of experiments using a gold leaf electroscope, which I found very useful at an intuitive level and I would recommend following that up if that's the kind of thing that works for you. In those days it was described like this: you apply a positive voltage to the first capacitor plate, and this has the effect ...


11

One of the capacitor ends is floating so there is no completed circuit for current to flow, so no charge builds up. So there will always be 0V over the capacitor terminals. So if left terminal is set to some voltage by some square wave or other signal, the right terminal will also have that same voltage, because there is always 0V over the capacitor and it ...


10

The reason you see a voltage in your first picture is because in SPICE, dangling nodes have a hidden (very high) resistance to ground. That is because the matrix solver has problems with elements that don't have currents, and that resistance ensures a DC path to ground with a small enough current not to influence the rest of the circuit, but enough to make ...


9

Analogue meters might have a preferred measuring orientation. If the meter is DIN certified, the orientation will be indicated on the faceplate. See top right, under Nennlagen. (Example found here: https://www.eevblog.com/forum/testgear/exploring-the-symbols-on-a-analog-multimeter-quiz/)


9

simulate this circuit – Schematic created using CircuitLab The 74HCT1G125 buffers and level-shifts the 3.3V to 5V (Vdd should be +5) and the RC filter converts the 5V PWM to analog with a sub-mV p-p ripple and rise time less than 20ms.


9

simulate this circuit – Schematic created using CircuitLab the conversion is not perfectly linear, but this improves the more that R2 has higher resistance than R1. M1 is some mosfet that switches at less than 3V For < 20ms time constant satisfy the inequation (R1+R2)C1 < 20mS


8

It's not an issue. Analog meters have an orientation they are designed to be used in (or are designed to be used in any position but need to be zeroed specifically for that position). That needle is a lever with weight on one end. We're spoiled now in that we can balance our digital multimeter onto a nearby pile of junk and take readings with it.


6

In the first picture the open circuit on the right hand side of capacitor can be considered to be a very large resistance. As the capacitor's left plate voltage rises, the current through this very large resistor (capacitor charging current) is equal to Vout/(very large resistance). Therefore the charging current is negligible, the capacitor hardly charges ...


6

To answer the question directly, in physical space there is no difference between an open circuit and a capacitor; by a capacitor is meant a two-terminal device whose capacitance is relatively large compared to the stray capacitances around it. In SPICE, there is no physical space, so it cannot even define the capacitance of an open circuit, the model simply ...


6

Since there is 5V supply available, one solution would be to use a logic gate to convert 3.3V PWM signal to 5V PWM signal, and then RC filter the 5V PWM signal to analog with the requirement of the 20ms time constant. The logic gate would have to be suitable for accepting 3.3V levels at input while being powered at 5V to provide 5V output. One suitable type ...


5

Mathematically I understand that through KVL that the output in the first image will match the input square wave. I am trying to understand physically what is happening, how is the voltage transferring across from left plate to right plate? Your simulation is just a mathematical model. Nothing is physically happening in it. Its behavior is solely determined ...


5

The connection of the diodes looks weird to me. I'd say "they are clamping diodes" but the connection should be anti-parallel across the inputs: simulate this circuit – Schematic created using CircuitLab also put some light on why Capacitor C3 is used. Without C3, the circuit will be a non-inverting amplifier. With C3, the circuit will be ...


4

Here's the circuit of a Clapp oscillator:- simulate this circuit – Schematic created using CircuitLab "That's not my circuit", you say. "I don't have L1, C1 and C2!". But you do. L1 is the parasitic inductance of the wires and components making up C0 (your C1 and C3) between the FET Gate and Ground. C1 is the FET's internal Gate-...


4

Think of the capacitor as of a voltage source with zero voltage. The capacitor does not (significantly) change its voltage when a small current flows through it; here even there is no current flowing. So the capacitor will transfer ("shift") the input voltage variations. This arrangement is used in coupling capacitors of AC amplifiers. This circuit ...


4

Your basic understanding of the RC differentiator is correct... Its basic operation comes from the current in the capacitor as a result of a driving voltage: \$ I_c = C \times {{d v_{in}}\over{d t}}\$. So far, no resistor appears...but we need one to convert the capacitor's current into an output voltage. We must choose a small-value resistor, so that all of ...


4

Use the .param command .param C 10u


4

The 0.9MHz voltage source at the bottom is what I want to use to model common mode noise on the signal. Is this the correct way to think about this? No it's not quite right. With this setup, the common mode input voltage would be $$v_{cm}(t) = (50\ {\mathrm mV})\sin\left(2\pi(0.9\times10^6) t\right)+(0.5)(0.2)\sin\left(2\pi(6\times10^6)t\right)$$ That is, ...


4

The meter needs to be calibrated to zero by rotating the spring balance adjustment (which is the thing that looks like a slotted screwdriver head at the bottom center of the meter face). It's not really a screw, it is an adjustment mechanism for setting the "zero volts" position of the meter. Because it's a mechanical spring system, it is sensitive ...


3

One application would be peliters, it would be nice to control the power to the device. Minus the resistive heating it's the power that determines how much cooling/heating the peltier gets. When you start talking about controlling power, it usually means there is a V/I curve that is not constant like a resistive load. Solar panels, peliters, diodes and many ...


3

One other application I'm aware of is in electrosurgery. There are certain modes where you want to deliver constant power to the scalpel so that as tissue impedance changes the power delivered and "feel" of the scalpel don't change.


3

The circuit calculates (V7 * V6)/1 The division by 1 is caused by the -1V of V5. The reason for needing to divide by 1 is to reduce the output voltage at the output of the adder so that the input to the anti-log amp is in the correct voltage range to be converted back to a meaningful result. Without the division by 1, the voltage at the output of the adder ...


3

The analog readings for pins are obtained using the ADC peripheral, not a GPIO port. Additionally, you might not get the performance benefits you are looking for, since the relative overhead of analogRead is not as large as the overhead of digitalRead, and most of the time is spent actually waiting for the ADC to do a conversion. The exact registers and ...


3

As we don't have a schematic, I'll take a guess. Looks like a simple triac-diac "dimmer" type circuit. In such a case, the potentiometer is connected as a rheostat. In such a case, the center and the left pin (from the front, pins down) of the pot are either connected to each other or the left pin is unconnected. So you can try the 1M pot with 1M 1/...


3

I have no idea what a "supply modulator" is but if the load has decoupling capacitors on it, then the voltage slew rate will be pretty low. Besides that, the opamp has a slew rate of 6.5V/µs, so it will need 300ns to swing 2V (from 1.6V to 3.6V). It has to swing twice per period, so with 1MHz square wave the best you can hope is 300ns ramp up, ...


2

Both the ADS1263 and HX711 have programmable gain amplifiers in front of their ADC units. The PGA on the ADS1263 can be set to x1, x2, x4, x8, x16 or x32. The PGA on the HX711 can be set to x32, x64 or x128. Other TI chips that have PGAs include: ADS111x, ADS1018, ADS124x.


2

To not leave this question without an answer, here's the solution that I commented earlier: My first suspicion would be the incredibly large 1.6kOhm resistors (and the 4.7kOhm ones even more, of course). You just need 2pF of stray capacitance to get a -3dB rolloff frequency of 60MHz with a 1.6k resistor. If you want to go to such high frequencies, all ...


2

As a boy, I had a 'rocket radio'. No battery, just an alligator clip to attach to the heating vent. To tune it, I would pull the nose cone in and out. I often wondered how it made a sound I could hear (earphone only) with no battery. Voltage Spike has provided the answer. But let's see what's going on. According to wiki, analog radio signals, by the time ...


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