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You can estimate this fairly easily using $$ P_\text{IN} = \frac {P_\text{OUT}}{\text{Eff.}} $$ for each device. So, let's say a sine inverter has an efficiency of 75% (I didn't look this up) and the mains to DC converter has an efficiency of 85% then the power required for a 12 V device powered on mains rather than directly will be $$ P_\text{IN} = \frac {...


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Stealing the maths in a comment by Transistor, You're going to need a battery that can deliver about 42A. An 18Ah lead acid or other rechargeable battery may do that for a few minutes only. You also need an inverter than can supply 400W, and probably a short term surge rating above that. Check the rating on your inverter. If it cost $10, it's unlikely to ...


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No, not the capacitance. You are just measuring the effect of leakage current via the schottky diode as voltage when the multimeter completes the circuit with its impedance. If you use the multimeter to measure voltage over the non-conducting schottky diode, you will get very near zero volts, which means that both ends of the diode are at same voltage of ...


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I like the current limiting feature of my bench power supply. It is a very nice safety feature. Not for my safety, but for the safety of my circuits devices. I tend to keep it low. Usually 30-50mA, and the power supply output voltage display will indicate low when the current limits out as an additional warning that current is being limited. I can ...


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Typically, a single supply ADC with the analog and digital ground tied together will automatically require a dc offset input. The best approach is to use an external op amp, that buffers the input and provides the dc offset, like this one below: Image source: Analog Devices - Circuit Note CN0336, Figure 2


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Usually using a bench-top supply to power a known device that requires a fixed DC voltage is not much different from powering it from a wall-wart, except that you can limit the current to a more reasonable value in case of a short or circuit malfunction. This is very useful when the device is under development. A wrong connection or a short can burn out the ...


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The answer to this depends on the load of the Power Supply (PS). If you feed chips (gates, sensors, etc.) you must provide the nominal voltage, which is given by the manufacturer. In this case, the current that the circuit draws is not up to you, but it depends on the circuit impedance. The reason you set the current limiter to some non-zero low value is ...


2

The old style flourescent lights, in which an arc has to re-strike at 120 Hertz, creates dirty (high-frequency content) energy on the local power lines. The fast risetimes of the arc, perhaps 10 microSeconds, (on scopes I've used, holding the scope probe tip up toward the tube some few feet above the lab bench) gets converted into radio waves by the wiring ...


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No they are not .Electric fences operate on a CDI principle and can produce short duration peak currents of amperes .They are not lethal .


12

The language used by the sellers of this device is not particularly scientific. It appears likely to be a placebo device marketed to (self-described) sufferers of "electrosensitivity". There is no scientific basis for these claims, and this device is unlikely to have any real function. The advertisement and sale of such devices has been prosecuted in some ...


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What exactly is "dirty electricity"? This term isn't a technical one. We might say a power line is dirty if its voltage doesn't vary according to the 50 or 60 Hz sinusoidal waveform we normally expect. I understand that certain devices may draw electricity in aberrant ways causing it to spike/surge. What do we mean by electricity "surging"? It's ...


0

I would interpret the equation to mean Ven as the underlocking voltage, then find Ren. So if I wanted the underlocking voltage to start at 5V, \$ R_{en} = \frac{315}{14*(5)+3.8} = 4.26\$ At this point I would assume they mean 4.26 kiloohms, but this is not clear, you could experiment with the device and turn down a lab supply below the 5V limit and see ...


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Before you changed your circuit, you were biasing the input of the I2C switch to your reference voltage, but not the output. This means that with the switch open, the output would drift to wherever it wants -- probably ground. Then when you switch it on, it was seeing the input, which was biased. It should be enough to bias the outputs to your reference ...


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A loud pop suggests a transient in which a large burst of current is being sunk to, or sourced from, your output (or equivalently, a sudden large voltage swing on the output.) This suggests that, when the output pin is connected or disconnected, the voltage on it suddenly swings. This makes sense for connection, because a floating output pin has nothing ...


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A step function will also sound like a pop, you should see a step function in the output because the output cap is not charged prior to the switch being enabled. One solution is to pre-charge the output cap with the same bias. Connect a separate large resistor from the output pin to the same bias voltage used on the input.


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The guitar is connects directly to the grid of the first preamp tube via a pair of grid stopper resistors, the ground of the amplifier and input jack being common with that of the 3 prong power cord. That's OK provided that your power supply is isolated. The first part of your question says, "How could I obtain such a voltage from mains without using a ...


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A switching regulator (Boost topology) or a voltage doubler + linear regulator. Solutions with transformers will generally give better performance than those without though.


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A bit late, but perhaps useful for someone using LTspice on Mac: In LTspice, place an .op data label and save/close the file. Open the .asc with a text editor and locate a line that starts with DATAFLAG. Change the text inside the double quotes to another expression. E.g. exact voltage at the connected net as "$", "round($)" with zero decimals, "round($*1k)/...


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So this looks as if the circuit controls armature current? It controls armature voltage. would it be better if the winding current was controlled? Controlling armature current with an outer voltage or speed control loop would be better, but more complex. By "winding current" I suspect you mean field winding current as apposed to armature winding ...


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Well it depends, but just about every device running off a floating (rather than earthed) DC power supply already has both of its power input potentials relative to earth oscillating at mains frequency and at a good portion of mains voltage, merely staying at a fixed potential relative to each other - and obviously these devices are still operational. ...


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Your approach and understanding seems to be correct. Without a multiplier you don't have the option to reduce the multiplier resistor value to increase the sensitivity. With such a low value resistance it sounds like a moving iron type (although I haven't looked at these for several decades). My old analog moving-coil multimeter was 20 kΩ/V and if ...


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It's depends on what you mean when you write "When it hits about 5A". If the "hit" is very short, say up to 1/10 of a second (and in this case this can probably not be seen with an ammeter but with an oscilloscope), then a capacitor may help. If the "hit" is longer than that, this cannot be solved in this manner (unless you use really huge capacitors, which ...


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Your boost converter is not 600W. It says Output power: = Input voltage * 10A in the eBay listing. Therefore, 10A*12V = 120W output power. The laptop you have can use up to 19.5V*9.23A = 175.37W of power. If it does, the boost converter can't keep up and limit its output so laptop will see voltage reduction and switch to internal battery. Capacitors ...


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