# Tag Info

1

Be aware of the capacitance to ground on each pin of the digital pot. The spec you embedded in your question suggest a range in the order of about 100pF. I don't know the exact switching frequency of your proposed application but, if it were 200kHz, 100pF would have an impedance of about 8 kohms (reactive) and this might be significant compared to the ...

1

You should be fine with a DPOT in the voltage divider. However, do not replace the entire divider with a DPOT. The voltage on the DPOT pins cannot exceed the power rails. I would suggest replacing only the lower half of the voltage divider with the DPOT; this will ensure that the maximum voltage the DPOT can see is the reference voltage of 1.2 volts. In ...

-1

No you cannot. Intern there is only one 12V source. If you try to put in series 2 of the +12V outputs you created a short circuit ( +12V to ground). ? WARNING ? As you can see in the comments below @Passerby: sees some dangers in the rest of the answer which I cannot confirm yet. But inside the power supply there is also a -12V. You can use +12V and ...

0

If you want the processor to switch things off and have them stay off until some further stimulus, the approach given by Mr. Lathrop is a good one. In some other cases, however, what is required is to have a signal from the processor which can kill power to everything for some (typically short) period of time and have it automatically come back on. That ...

2

If you don't need much current, you can maybe use an LED? A red LED has a voltage drop about 1.5V, which will drop your 3.3 down to 1.8V. It seems weird to use a diode or LED to regulate power, but this FPGA development board which I've used, the XSA-50: http://www.xess.com/shop/product/xsa-50/ (click on manual to see the schematic) uses a 1N4148 to do ...

6

A simple way to do this is to put a P-FET in series with the processor's power. A high value resistor pulls the gate high, which keeps the FET off if nothing else happens. A pushbutton shorts the gate to ground when pressed, which powers up the micro. The micro then holds the gate low thru a transistor. The micro should be able to power up and actively ...

3

This seems to be a problem I've seen a few times on stack exchange. Consider an op-amp with localized negative feedback - The manufacturer designs the op-amp so that under the very worst case situations it is stable. The worst case situation is unity gain - this has the biggest chance of being unstable. Anyway, each year the boundaries get pushed a bit more ...

0

I found that the easiest, but not the cheapest, way to determine the inside diameter of a 'barrel' connector is to use a set of "pin gages" which are hardended and ground to precise external diameter steel 'rods' that come in sets. A lot of times sets are available cheaply at discount tool supply places, or if you have a machine tool supply store available, ...

1

This is the same problem as any "how do I get DC from AC?" question. You rectify it: simulate this circuit – Schematic created using CircuitLab You can perform the rectification with four discrete diodes, as shown here, or you can purchase pre-made bridge rectifiers that are exactly this, in an integrated package. C1 is optional, and serves to ...

0

setup the voltage of the power supply to the level required by the output luminosity. From what I can see, the required voltage of commercial strips of 5050 SMD LEDs is generally 12 V. Example. You do not alter luminosity by varying the voltage. You need a 12 V DC supply designed for driving LED strips. include a rectifier to keep "no-power" time as ...

7

There is a reasonable possibility you made a simple error. Many bench power supplies provide +/- dc outputs but also have an earth connection. The earth connection is of course earth BUT, the power supply's dc outputs may well be isolated from this earth stud/socket. Many power supplies are configured like this. What you might have done is connect the scope ...

2

Is it dangerous to step down from such a high voltage to reach 5 V DC? No, this is very common. All "wall-wart" adapters do this. Use a properly certified product to do this. Are there potential stability issues? No, but you'll have slight imbalance between usage of phases, this won't cause any issues. If your application requires a 3-phase ...

0

Assuming the three phase outlet contains a neutral pin ... Conversion from three phase with neutral to single phase is a matter of wiring. Use an electrician to prepare a three phase plug to single phase socket assembly. This can be done off site. The electrician is to choose one of the three phase active terminals on the plug, connecting it to the ...

0

12 volts battery: max voltage is around 13.1V - 13.3V & min voltage is around 11.1V - 11.3V 7 AH: Battery can deliver less than 7 Amps for an hour or 1 Amp for 7 Hours. If the battery is fully charged it contains max voltage and 7 AH capacity. As u go on discharging the voltage and capacity go on reducing.

2

A true digital signal is a list of numbers. Consider digital signal processing which has, as input, a list of numbers and, as output, a different but related list of numbers. When one uses a spreadsheet to find a moving average of the DJIA, one is doing digital signal processing. Also, remember that, abstractly, the contents of computer memory is just a ...

2

Power Conversion The input voltage to a computer power supply is 120V AC in the US, where the "AC" stands for Alternating Current. It alternates 60 times per second, or 60 Hertz (Hz). You could say that this is an "analog" source, because viewing it on an oscilloscope or a graph over time, it's a sinusoidal wave. If you measure the instantaneous voltage at ...

1

I'll try to go through the question section by section and explain mistakes shown in the logic, if there are any. We know that the 120v from the wall is analog And the digital one that the components will use (transistors, diodes, etc.) will look like this: Yes, that could be considered correct. That would imply that an ADC circuit must be ...

4

Your logic is somewhat sound in assuming that an Analog to Digital conversion is happening, but there is a big set of steps that you are missing. Suffice it to say that this answer section is not sufficient to describe fully how a computer takes a 120VAC signal coming out of a wall and turns it into the webpage you are looking at. Turning the 120VAC signal ...

2

Theoretically a cable having a twisted pair for data and two other conductors for power and return is OK - induced currents from the power conductors will tend to equalize on both data wires (because they are twisted) and only produce a small common-mode corrupting signal that can be dealt with by using appropriate components on the CAN bus receivers. ...

1

Power over Ethernet (PoE) can provide up to 15, 25 or 51 W over twisted pair data cables by passing power over unused pairs or by applying a common mode voltage over the signalling pairs. Using more pairs obviously helps allow greater current. PoE tends to use higher voltages (up to 57 V) which may also help overcome resistive losses. So, supplying power ...

5

A possible negative effect could be the resistance of the wire. That's not caused by the twisted pair, but rather the wire diameter. If they're really signal wires the diameter will be rather small, and have a non-negligible resistance. The resistance will cause a voltage drop, which may become noticeable at longer distances. If the wire is thick enough ...

3

The datasheet is INCORRECT! but your understanding is not so far off... The battery capacity of 7Ah, x its voltage (12V), indicates its stored energy not power. If you need 7A for 1 hour you need to read the fine print, to see if that was the capacity at its "1C" rate (1x its capacity, or drain it in an hour) Oh wait, there is no fine print on that page, ...

1

Yes you understood it correctly, a 12V,7Ah batteries is supposed to provide 7A @12V during one hour. Regarding regarding consumption in Watts, it is the current multiplied by the voltage the device is supposed to run on. If your 60 watt bulb is supposed to work on 250V it will draw 0.24A when powered that way. That's the basic theory. There are some ...

0

This is a custom Li Tai power transformer: their standard offerings. You will need to contact them to get more information.

2

Here are a couple of buck-boost converter offerings from Linear technology that would fit the bill. The second one is the preferred solution to me: - Here is the LT page where you can enter your own parameters and it provides you with the chip options.

3

You basically need a switched mode power supply that is capable of both stepping down and stepping up the input voltage to 12V because your input voltage range is greater than and less than the required output voltage. Therefore a boost converter will not suffice; neither will a buck converter. The simplest power electronic converter that could do this is ...

1

A SMPS (switched mode power supply) is a good choice here. (also called DC/DC) Several typologies allow the input voltage to be either lower of higher than the output voltage. Have a look at: Flyback architectures SEPIC architectures And many others If you are not comfortable with them, you can always do a multi-stage approach with the drawback of ...

0

The output voltage is stable when you connect your beaglebone to the normal DC power supplies. While the output of solar output need to be regulated by Maximum Power Point Tracking (MPPT) algorithms to keep stable. Normal buck IC like UBEC or FAN2002 do not provide MPPT capability. As most solar converters aim to generate large power, there are rare ICs ...

-1

In terms of efficiency, this is about as good as it will get. The most efficient method of transforming the voltage from 12V to 5V is to use a buck converter as opposed to a linear regulator. The UBEC is a buck converter. For power levels in this range, an efficiency of over 90% is considered good. If you were willing to search around for a custom buck ...

3

I think your answer may make intuitive sense but there are some other ways you can get speed control of your electric motor. I've drawn one here as an example: simulate this circuit – Schematic created using CircuitLab I am not proficient in circuitlab but you can see I have an EMF-absorbing diode, a motor (I represented this with an inductor), ...

0

I was able to achieve the circuit I desired. I ran into some issues along the way, but ultimately this circuit has allowed me to use an Arduino to control how many batteries are in series and are used to power a fan. The Q1, Q2 or Q3 transistor has to be closed, then the Q4 transistor has to be closed and the motor will be spun at a variable speed based on ...

3

You are confusing first order approximations or rules of thumb with exact reality. Yes, the emitter voltage of a emitter follower with a fixed base voltage stays constant to first approximation, but when you're pushing the limits you have to look more carefully than that. There are a number of problems with your circuit. The zener voltage won't be exactly ...

0

With the addition of an op-amp you'll get a lot more performance. The emitter, under load conditions will always be less that thbase. The BJT output will sag with load currents because it has only finite current gain whereas if you add an op-amp to drive the BJT you'll get better performance all-round: - This is just an example circuit from the internet. ...

0

Check the DC current gain in the datasheet for D1047. For a 1A collector current, the DC current gain is 60 worst case. This means that the base current will be $I_B = \dfrac{I_C}{h_{FE}}=\dfrac{1}{60} \approx 17\text{mA}$. This means that your zener circuit will be loaded with 0..17mA, depending on the output load. Now check how your zener circuit ...

3

First of all, I'm going to assume your complete circuit looks like this: [BTW, you should post your complete circuit if you expect to get any meaningful answers.] Secondly, the unity voltage gain of the common collector refers to AC, not DC. From the image above, you can see that the output voltage will be $V_Z-V_{BE}$. And $V_{BE}$ will have some ...

0

There is the WISP http://wisp.wikispaces.com/ They use a large dipole and a cascaded rectifier to generate voltage.

2

First of all, before answering your questions, it is important that you understand the difference between regulated and unregulated supplies. Regulated supplies will put out close to their rated voltage regardless of the load. Unregulated supplies will put out a much higher voltage with no load or a small load, and only approach their rated voltage under ...

11

All in- and outputs on most digital devices today have clamp diodes to the power rail. These diodes are there for protection of the device, to prevent a pin from having a higher or lower voltage than the supply rail. What you are experiencing here is that you actually power the DAC through its data pins and the respective protection diodes. Although the ...

1

As the other have posted, PWR_OK comes after PS_ON so what you propose won't work. However, there is no reason you cannot just leave PS_ON grounded all the time. For as long as a computer is turned on, the motherboard is keeping it grounded. You just want to make sure that you turn off the power supply before you connect or disconnect any load.

1

The way a PWR_OK works, or at least the ones I have worked with, is by signaling the stabilization of the power output with a certain tolerance. This is used for sequencing purposes (e.g. When 5V line is ready enable the 3.3V PS). This means you won't be able to use PWR_OK unless PS_ON is high first.

1

As far as I understand how ATX power supplay works, you have to pull PS_ON to turn the power on, and then you get PWR_OK eventually. Without pulling PS_ON, you'd wait forever for PWR_OK.

0

I saw a novel solution to this same issue once. An AC motor spins a flywheel to a few thousand rpm. The opposite shaft of the flywheel from the motor is a reduction gear drive that feeds a genset head. When AC power fails, the flywheel stores enough energy to run the genset for several minutes. An led on one side of the flywheel and a phototransistor on the ...

0

Since you said cheap........ Any car repair place can get you a used car battery that has one weak cell. Sometimes just for the asking. Car batteries get 12V from 6 2V cells in series. So charging it at 13.8V yields a 10V output. A simple buck converter fed from this setup will get far beyond 20 minutes, even with full load demand. The buck converter will ...

1

The voltage and current adjustments work differently depending on the mode of operation (assuming the power supply supports both modes of operation). (1) For constant voltage mode (power supply acts as a voltage source), the voltage adjustment sets the output voltage and the current adjustment sets the current limit. This means that the power supply will ...

0

You are correct. Setting of the power supply to 2a, only sets its max limit.

0

The batteries can handle a high current load for less than a second even a dead short. If the battery is only a 2AH battery nothing will happen other than shorten the life of the battery. This is ok as long as the motor starts turning and current drops to no more than twice the AH rating of the battery, for a lead acid battery, the surge current is higher ...

2

The LM317 is a linear regulator, which is really inappropriate for significant power, like what a model train requires. You say the train draws 300 mA. That means at 3.3 V drop accross the regulator it would dissipate 1 W, which would require a heatsink. I'm not sure why your regulator is dying. I thought (haven't checked) that the LM317 has integrated ...

0

Ok so the reason that there is low resistance between +V and ground is merely because there is a capacitor between them and my multimeter was charging it. The reason for only the negative wires frying then was probably because the ground on the arduino/pc had a much lower potential than of the PSU and a very large amount of current then then went from PSU ...

0

The following earthing configurations are quite common in DC power supplies: Positive output (i.e. +12V): low-side connected to earth Negative output (i.e -54V): high-side connected to earth Note that in both of these cases it's the reference rail that's earthed. A +12V supply is +12V with regard to the earthed rail; a -54V supply is -54V with regard to ...

1

If your PSU looks like this 12 Volt (single voltage output), 30 amp power supply: The adj label probably refers to an adjustment screw, not to any of the terminals. The -V label does not mean -12V, it is equivalent to COM on the PSU in your reprap link. I would not expect the ⏚ terminal to have a low resistance to the +V terminal with all cables ...

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