# Tag Info

16

The answer is clear: Buy a new supply and find something else much simpler to "learn" on. Switching power supplies are outrageously complex and not something you're gonna be able to wrap your head around without a good understanding of electronic theory.

7

Powering up with a random component removed is a bad idea unless you know its function and understand what will happen without it in the circuit. It could be part of a clamp that prevented overvoltage damage to your switching FET for example. So by removing it you could cause damage to other components. (Which it sounds like you did.) Diodes don't create ...

6

They weren't originally designed for it, but I have seen applications that use bidirectional radio chips such as the TI CC1101 to pass control/status data in both directions through a coax alongside DC power and another RF signal. The chip handles a lot of the details and is relatively easy to set up with a microcontroller. In fact, in one design, a very ...

5

If you terminate your current transformer with a resistor to get a voltage signal out, a normal AC current transformer would saturate very quickly if you apply any DC current due to Vt=NAB where V is positive value (or negative, same effect) and as t goes on, B just keeps building until saturation. For AC, V keeps changing and nets out to zero. In a zero ...

5

First of all, consider that fiddling with mains voltage can kill you should the discharge go through the heart. SMPS have large caps inside which will keep deadly voltage for hours after you pull the plug. Are you sure you can work safely with it? If yes, you should check if the hot (high voltage) side of the PSU still works. My guess would be that by ...

3

You are on the right track, in thought process at least, not procedure. Know that fundamentally, there is no such thing as a DC motor. All motors operate based on changing (alternating/rotating) magnetic fields which are produced by changing (alternating) currents. They are all "AC" so to speak at the lowest level. What goes into the coils is ...

3

Before the capacitor (transistor side), the average DC output voltage will be half the total supply rail of the amplifier hence, it your amplifier supply rails are 40 volts and 0 volts you will see a quiescent DC voltage of 20 volts.

3

It's not possible to measure a DC current with a transformer. In the case of AC the changing magnetic field associated with the current in the primary induces a current in the secondary winding which is what you can measure. As mentioned in the comments above if you need to detect a DC current you can use a Hall effect device which does not require a ...

2

A Vienna rectifier is pretty much exclusively used for making power-factor corrected (PFC) three-phase DC supplies. The working principle (as with any single-phase PFC) is that boost control is used to produce a DC voltage that is (usually) significantly higher than the peak AC voltage. Only in this way can proper PFC be achieved. This rectifier circuit ...

2

You have a DC source trying to drive it positive relentlessly and quickly towards the positive rail, so it spends some time saturated there every cycle. When the 400Hz AC input overcomes the -400mV source it struggles off the positive rail and integrates resulting partial half cycle. Integration, if you recall your basic maths, always has a constant. You've ...

2

If this can heat up a liter of water in under 10 minutes that would be great, but it is not necessary. $$P = \frac {m \times \Delta T \times SHC} t$$ where $P$ is power (W), $m$ is mass (kg), $\Delta T$ is the temperature rise (°C), $SHC$ is the specific heat capacity of water and t is time (s). So boiling water from 20°C requires: $$P = \frac {1 ... 2 On a regulated adapter, presumably what you have, the output voltage specification is a nominal output voltage for any load within the normal range of operation (usually starting from zero). The output current specification is a maximum. If you connect the adapter output across an ammeter you are basically shorting it and it will try to protect itself, often ... 2 9 volt batteries are designed for low current long life applications. Even if you could draw 1.5 Amps at 9V in an ideal setup, with a typical 500 mAh 9V alkaline battery, you would get less than a 3rd of an hour life on it. In reality the battery chemistry will reduce the voltage and current capacity as the current draw increases. From https://www.... 2 9V/6 ohms= 1.5A which will kill the very cheap low power batteries extremely quickly. Why kill the battery with a parallel resistor? Rayovac does not show performance graphs so I show one of an Energizer cheap old 9V Zinv battery like what you asked about. 2 What will happen if the resistance is lower than that? The vehicle is moving and is generating plenty of mechanical rotation on the motor shaft, we apply PWM2 100%. I suspect the motor is going to generate more than the rated 27A? Yes. If my assumption is correct then why is that? Because the rated current is not the maximum current the motor can actually ... 1 I'm guessing from the details you give that this was a 3-wire fan? the basic details others have not expressly given, but rather hinted at, are that PC fans don't run on just voltage. you cannot just put 9 volts on a 12 volt PC fan coil to run it or slow it down. it needs 12 volts, and data from a sensor showing it's speed and position to itself so it to ... 1 Try again with a working fan and then follow the instructions. The internal fan electronics commutates the DC with the Hall Effect magnetic rotor sensor. Instead of guessing use, Red=+, Blk=- , yellow= tach out or don’t care 1 What will happen if the resistance is lower than that? There will be a higher current when the MOSFET is on. I suspect the motor is going to generate more than the rated 27A? If my assumption is correct then why is that? The motor will not generate more current than the load is drawing. To do that it would need to produce more voltage. I mean it is clear ... 1 frequency devision multiplexing: With consumer satellite hardware signalling is done using voltage modulation and 22 kHz tone bursts if the data rate is low I'd send the signal up the line by modulating the supply voltage (eg switching between 13V and 15V ) and signal back down using 22kHz bursts. Are there existing components or ICs to facilitate such a ... 1 How would you combine DC power, bi-directional serial data, and an RF signal on a $\color{red}{\text{single}}$ 50-Ohm coax feed-line? Let's not forget what the question says regards cables: -$$\color{red}{\boxed{\text{the single coax feed-line}}} Regular UART level serial data comms would interfere with RF reception but you could certainly modulate ...

1

Single rail push-pull stage would typically have half of the supply voltage at output stage before the DC blocking capacitor. This way, the output can swing about half-supply up or down from the midpoint.

1

Power adapter are rated by their maximum current and voltage, a 9V, 1.5A rating means it will output 9V max and 1.5A max. The exact current that will be output depends on the circuit. You can think of it like this: the adapter outputs a specific voltage and then the circuit draws whatever current it wants till the adapters reaches its maximum. Kinda like ...

1

You can invert the signal with another FET. Drive a N channel FET with the contacts. When grounded, N FET is off, and does not turn the P FET on. When contact is released, N FET can get gate voltage and turns on, pulling P FET gate low to turn on the connector voltage.

1

You have just described an electric kettle. They are cheap and readily available. You haven't said anything about the voltage or current. European ones are likely to run on around 220 to 250V, and require around 10A. The heating element won't care if the supply is AC or DC. But if the kettle has an automatic switch to turn it off when boiling, then this ...

1

Try gas discharge lamp oscillator. Connect a lamp in parallel with a capacitor and feed it from DC through a resistor. When the capacitor charges to the ignition level of the lamp the arc in the lamp discharges the capacitor and the arc vanishes. The capacitor start again to get charge. These oscillators were common before the era of transistors. Read this ...

1

Some additional general comments meant as a complement to what's already posted. High-frequency 'squealing' noises coming from power supplies generally implies incorrect / abnormal behaviour. Magnetorestriction caused by high pulsed currents (i.e. the power supply hiccupping energy into a shorted part) can cause acoustic noise. When you hear this, ...

1

It is a DIODE 1000v 10A10 , cathode connected to + for reverse voltage protection.

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