10

Current and voltage ratings do not neccessarily apply at the same time and in a single operating point. A switch is a pretty good example for this: it must not conduct a current higher than the rated current. This is relevant when the switch is closed. a voltage higher than the rating must not occur between the switches contacts. This is relevant when the ...


7

That switch is for AC mains wiring at either 50 Hz or 60 Hz depending on which is prevailing in your locale. It is not rated for audio at all. So you just have to try it out and see if it works. In general, lower frequencies are harder on a switch. DC is the hardest. Higher currents are hard on a switch. Higher voltages don't necessarily put a lot more ...


6

In this specific case the way you can tell that the second one is a better choice is that Rds(on) is specified down to 4V, but in the first one, Rds(on) is specified only at 10V. In power switching applications I usually try to stick to MOSFET's that have Rds(on) specified at the Vgs I will be using (or lower). Vgs(th) is only marginally useful in figuring ...


5

During switching there is more arcing at higher voltage than at lower voltage. Switches have a voltage limit, but they also have different current limits at different voltages. There are also wattage limits sometimes, and the rated limits may be different for different loads, depending on if they are resistive, inductive or capacitive.


5

It is simply a test circuit for you to test and observe. It is not a "complete" circuit in the sense you expect - it is not a "switch." The point of the exercise is for you to carry out the experiment and develop an understanding of how to assemble the circuit and test setup, and then to gain an understanding of how the base current of a ...


3

Swap the resistor and switch Nothing is perfect. Your switch has resistance. Wanting "0V" is like saying you want something to be exactly one meter long.


3

There are two issues with your schematic. You need a power P-channel MOSFET to switch the power to U4. You can command it with a small N-channel MOSFET (M2) like this: simulate this circuit – Schematic created using CircuitLab When M2 is off, M1 is also off (as R1 pulls its gate to source). R2 keeps M2 off for the brief period when the ATmega starts ...


3

If the PIR detector output can activate an electromechanical relay (such via an open collector transistor) then, one of the spare contacts in the relay can be used to short out the open collector transistor and keep the relay self-activated. Alternatively, If the PIR detector has a voltage output then it can be diode-orred with the spare relay contact in ...


2

You could use two 24VDC relays as follows: simulate this circuit – Schematic created using CircuitLab The COM is wired through the NC contact on RLY3 to provide "break before make". If you don't need that feature you can avoid using the NC contact.


2

DC motors don't use start capacitors. You want a reservoir to keep the supply from sagging. Use dQ = C * dV, or C = dQ/dV. If you need 10A for 2 seconds, dQ = 20 Coulombs. If you can tolerate a 2V sag in the supply voltage, dV = 2V. Then C = 10 Farads. Double check your specs though : if the no load current is 6A and the rated current is 10A, you probably ...


2

If you have two indipendent power busses you need 4 bilateral switches. Eeach switch needs either two MOSFETs or (highiy deprecated due much higher losses) one MOSFET and a bridge rectifier. In both cases you also need four indipendente floating gate drivers. simulate this circuit – Schematic created using CircuitLab So it all just sums up to 8 ...


1

the grounds may not be connected. I hope it's meant that the main and backup supplies' grounds may not be connected. Here's a concept diagram: simulate this circuit – Schematic created using CircuitLab Main and Backup supplies do not share the grounds here. CONTROL BLOCK is supplied from the Main supply. This block drives the relay and directs the ...


1

An automotive wiper motor will require 12 volts - your two AA cells only provide 3 volts - probably not enough to move the motor. Also, the motor probably requires around 3 Amps - far too much for AA cells, even if you had enough AA cells in series to make 12 volts. Eight C or D cells in series would probably work, but a 12 volt lead-acid or gell-cell ...


1

You can't use an N-channel MOSFET like this, because source voltage will be higher than gate once it opens. You can either use a P-channel MOSFET and reverse the gate logic or put your N-channel MOSFET to a low side (switching the "ground path"). But as we don't know what lies behind the U4 and which type of convertor it is, I'd rather pick ...


1

To power down a whole circuit the logic needs to go something like: Switch feeds a GPIO signal (Sense), and also powers the circuit (through a diode). A power controller (U2) is powered on and activates a Parallel power pathway to the switch (M1). Switch is turned off - Parallel pathway remains active, but Sense is now off since diode prevents Parallel path ...


1

I think the real purpose of this setup is to measure and draw the input IV curve of the transistor - the function Vbe = f(Ib), which is actually the IV curve of a diode (p-n junction, as you noted). "Ib vs Vbias" does not make much sense because practically, it is the IV curve of the resistor Rb. "Sweep the value of Vbe" is not so correct ...


1

For baud 115200 below 5V, the 30~50MHZ analog switch is enough. FTDI FT232 is up to 8Mbps, or 8MHz. It uses external 16MHz crystal. You can use dual 2-channel SPDT (Vishay DG9415DQ, MPS MP2735), triple SPDT (such as TI 74HC4053/CD74HC4053, Nexperia 74LVC1G53) to include SW1. Or single SPDT + single DPDT as in your schematic. You can check 74HC series MUX IC.


1

You've designed an emitter-follower circuit. The voltage at Q14's emitter will be at least 0.7 V lower than the GPIO and on a 3.3 V supply that's nearly 25% of a voltage loss. This might be OK but we don't know what you're trying to drive with this circuit. You need to specify the requirements of the device you're driving. What is the maximum voltage ...


1

When GPIO_POWER_EN is high, then R1_EN will be at the voltage of GPIO_POWER_EN - 0.7 V, the transistor is acting as an emitter follower. If you need a high side switch then something like the following would work. simulate this circuit – Schematic created using CircuitLab


1

You'll need to design a relatively uncomplicated circuit to achieve your goals. It'll consist of 2 parts. The 'electronic switches' themselves and the interface to the rest of your design whether analog or digital. I'd consider using either jfet's for the switches or analog transmission gates such as the widely used CD4016, 4051 or 4066 from the inexpensive ...


1

Here's the switching scheme. The sliding contact of the switch bridges the fixed contacts (terminals) 1 & C, C & 2 and C & 3 in positions 1,2 & 3 respectively.


1

What you are doing is at left and what you must do is at right: simulate this circuit – Schematic created using CircuitLab


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