22

FFTs work by treating signals as 2-dimensional -- with real and imaginary parts. Remember the unit circle? Positive frequencies are when the phasor spins counter-clockwise, and negative frequencies are when the phasor spins clockwise. If you throw away the imaginary part of the signal, the distinction between positive and negative frequencies will be lost. ...


15

There are two ways to answer this question: an engineering one, and a physical one. While many engineers prefer to think in an engineering way only (sometimes they can't even explain how the stuff really works), I believe that knowing the concepts is very important. Therefore, I'll try to answer your question from a physical point of view (which is much ...


14

The best thing to do with an LM741 in most cases is to replace it with a more modern single supply opamp. An LM358 dual or LM324 quad is about as good, usually cheap and available. This page from a project by some one at the CIT in CEBU suggests 100 PHP each which is far too dear. Digikey has them for about 20 PHP in 1's prices here and about 13 PHP each ...


12

You can generate a negative voltage with a small current quite simply with a source of AC, and a voltage doubler rectifier. (This shows a doubler, reverse the diodes and capacitors for an inverter) See here for some circuit diagrams. The AC would come from your Arduino, by toggling a single pin regularly. You must add a series resistor too. It might be ...


11

Do not figure it out, junk it. Current cannot flow from the plus 5 to the ground because these 3 terminal regulators do not sink current. In rare cases it can work if the -5 draw is always greater or equal to the +5 draw. Look for voltage splitting circuits. A common approach is to use an amplifier to force the ground to ground, at low currents even an ...


9

Polarity-inverting buck-boost is what it is called.


9

There are many different BJT models, with varying degrees of usefulness in varying circumstances. (See SIDEBAR at bottom.) I'm not going to delve into any of that as it's not necessary in this case. A nice simplification is quite sufficient for your use here. Ignoring Nth order effects that aren't important here, a BJT's collector current is determined by ...


7

First you should be asking yourself if you really need a negative voltage. Most things you can do with a 741 can be done using a virtual ground - a voltage divider between Vcc and ground to give a voltage at 50% of Vcc - this is then used as the ground reference, Vcc becomes Vcc/2 and ground becomes -Vcc/2. Failing that, with batteries you can just have a ...


7

In theory, this should work. I'm not sure how it will behave in practice though. So, in theory you cannot drive a current higher than what the 7810 can supply. Edit: See @russ_hensel's answer below. Will not work. How much current are you driving? Most virtual grounds need to be buffered. See some circuit examples at: http://tangentsoft.net/elec/vgrounds....


7

Like you said, the fact that the two opamp inputs will be nearly equal is a simplification and depends on parameters often not explicitly stated. This is a good question in that it is essential to know the limits of any shortcuts or rules of thumb you use. As clabacchio already said, one place the assumption are violated is if the opamp output is clipped, ...


7

PC fan pinouts can be found easily online: If you're trying to figure out which wire is ground on the fan itself, and you don't have a connector, then just connect 12V to the middle wire and try ground on either of the outside wires. When the fan spins, you've found it. The other, if connected, is a tachometer. Connecting it to ground briefly won't cause ...


6

An op-amp's output pin can only swing between its power rails - many op-amps can only get within a volt or so of the rails, while 'rail to rail output' amps can reach voltages very close indeed to their rails. They can't output voltages outside that range, though - the power still has to come from somewhere! I suspect your simulation used a simplified op-amp ...


6

What you are looking for is a rectifier, if the voltage drop of the diode is of no importance ,then this circuit will do the trick. simulate this circuit – Schematic created using CircuitLab But, if you want to compensate for the voltage loss over the diode you'll need an Opamp . simulate this circuit


5

A 24VAC transformer will give you as much as 40VDC with a light load. That's pretty marginal for an LM317 with zero output. Way too close for comfort, in my opinion. Anyway, if you were to use a third LM317 to regulate the input voltage down to +5VDC, you could use that 5V to power a 7660 charge pump to generate about -4.5V for your JFET current sink. ...


5

Linear tech produce a number of inverting switchers like this: - Check the data sheet to see if it can produce -24V. If not then use the search engine on this page here There's also this one: -


5

The voltage gain of this system, which is currently unity is determined by the opamp and feedback network. The push pull amplifier is placed within the feedback loop of the opamp, and is there to provide current, to drive a lower impedance load than the opamp alone is able to. The push pull stage is a pair of emitter followers, and as such doesn't provide ...


5

You can use inverting charge pump, buck-boost voltage inverter or a virtual ground op amp based solution to split rails. This is one example component which can provide \$-15\,\mathrm{V}\$. It is a simple dc-to-dc converter module. Part number: P7805-Q24-S15-S Do not miss to notice the pin numbering in both configuration.


4

If the input voltage goes negative, then the zener will be forward-biased, and will behave not unlike an ordinary diode. Your \$10k\Omega\$ resistor will limit the current. simulate this circuit – Schematic created using CircuitLab The voltage across D1 will be about 0.6V. The current is given by Ohm's law: $$ \frac{6V-0.6V}{10K\Omega} = 0.54mA $$ ...


4

It will go below ground one diode drop but the 10k resistor will limit the current into the zener and hence into the circuit that may be damaged so, the only question you have to ask is "how much current can my input take before it is damaged". My gut feeling is that on most devices (MCUs op-amps and the like) 10k from +/- 6V is not going to damage anything ...


4

Assuming you're talking about round-type (cylindrical) batteries, such as D, AA and AAA, it's to ensure maximum contact with the flat end of the battery, which is the "negative" terminal as described in the ANSI standard. You'll commonly see leaf-spring contacts as well as coils. The side with the "nub" will automatically provide a solid contact if enough ...


4

Because in telecommunications (phone central offices), things run off of -48 V. So there are plenty of parts and equipment built that way. This link explains why. It has to do with electroplating.


4

First, you should have a voltage suitable with your batteries set up, as Russell said, then, there are several ways to obtain negative voltage referance for your OP-AMP. In one episode of EEVblog, Dave tells 3 options and compares them for a spesific application. Here is the episode where Dave designs a negative voltage referance generator for uCurrent ...


4

As long as the Op-amp can set the inputs equal driving the output at a certain voltage, it will. That assumption falls when it cannot, such as if it has an open circuit in the feedback (positive or negative). Then, it'll saturate to one of the rails, depending on which input is driven higher. Note that the open circuit feedback can be also a reversed diode. ...


4

Some comments from my side: 1.) The stability check in the BODE diagram concerns the LOOP GAIN response only (because once you did mention "closed-loop system" in your text.) 2.) The shown system is "conditionally stable". That means: It is stable - regardless the properties at the frequency A. However, if you REDUCE the gain within the loop until the gain ...


4

I'd rather go with a simple common base level shifter simulate this circuit – Schematic created using CircuitLab R1 is choosen not to saturate Q1 looking for "high speed" turn-off. In this conditions fall time is mostly ruled by R2 and output stray capacitance. Output clock more probably swings from 0V to around 13V w.r.t the -15V line. This ...


4

All signal generators need a reference for the signal. Your old sig-gen likely had the output referenced to mains ground while your PC one may be floating. Assuming, that is, that whatever you are feeding is also referenced to the same ground. Reliance on the mains ground can be problematic though. Using the reference supplied from the signal generator is ...


4

All you need to do is turn the LED around. simulate this circuit – Schematic created using CircuitLab


4

Because the N-channel MOSFET can carry higher current and it is cheaper then its similar size counterpart P-channel MOSFET. N-ch MOSFET can be used also as high side switch, but it requires an additional charge pump. The shunt is placed on lower side, because the common mode voltage would be too high for the operational amplifier. Again, it is possible to ...


3

"Ground" is a much misused term in electronics. Occasionally (mostly in AC Power systems and some radio antennas) "Ground" does mean a connection to the earth. However, in most cases (especially portable battery-operated circuits), "Ground" is simply the point in the circuit that the designer decides to use as a reference point for measuring voltage ...


3

It is quite common to use the physical properties of the NUB to provide polarity reversal protection in a battery. So the NUB will fit between two plastic shoulders and make contact with the plate. If there were a spring this end you would need to more tightly control tolerances in the spring and its mounting compared to a flat plate. For this reason it ...


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