51

The 10k\$\Omega\$ resistor is there to pull down the gate when the input is floating (thus avoiding an undefined/uncontrolled gate voltage). On the other hand, the 100\$\Omega\$ resistor is there to limit gate charging/discharging current (due to the presence of gate capacitance) and to prevent oscillations. But, as you have already detected, both resistors ...


16

It is a gate with an inverted and a normal output. The idea is that the two outputs switch exactly at the same time. There is hardly any delay between them. The symbol as shown in your diagram is rather awkwardly made. More often the following symbol is used for a combined buffer + inverted like that: You will find these used with differential line ...


9

It depends very much on the appliance. But most appliances with power supplies are really taking in power in pulses at a rate of 120 Hz (in a 60 Hz system), so they're already designed to "ride through" gaps of at least 8.33 ms. 10 ms is so close to this that they won't even notice.


8

That is a very ambiguous symbol. It probably means that the buffer has both an inverting and non-inverting output. As Tom Carpenter comments it goes to the input of another gate so it must be an output. There are two common variants of buffers. A buffer with tri-state output and an enable pin. A buffer with true and complement outputs. The symbol is ...


7

There's what's called "analog switches and multiplexers", which exactly fulfill the role of your pushbuttons here. Basically, all large silicon chip manufacturers have them (for example, Texas Instruments). However, the currents from photodiodes are really small and sensitive. You don't want to switch these, if it can be avoided. These switches can easily "...


5

Here that 10K Ohms acts as a pull down resistor only ie., to make the gate low normally. in otherwords , we are defining the FET to be in OFF condition normally.


3

Why do main on/off switches in electronic circuits/appliances (any small, low power electronics) are always placed on the positive wire (in case of DC circuits) between the power supply (batteries, adapters, etc.) and the circuit regardless of the low operating voltage? Mostly, convention. You connect ground to the negative supply, because that's what all ...


3

Each switch position is represented by a vertical line. Contacts are arranged horizontally. Contact closure is indicated by a black dot. Figure 1. Position A: 5 is connected to 6. Position B: 3 is connected to 4. Position C: 3 is connected to 4. 1 is connected to 2. This is rather like a car's ignition + starter switch with 5 - 6 closed in the off ...


2

24 V switching may damage the switch contacts if you are switching high current and particularly if the load is highly inductive. 24 V will probably pass too much current through the LED so you would require an external series resistor on the LED terminal. simulate this circuit – Schematic created using CircuitLab The safest option is to measure the ...


2

All switching circuits will require some current flowing through the switch. It can, however, be kept small so that a miniature switch can control a larger switch rated to connect and disconnect the load. simulate this circuit – Schematic created using CircuitLab Figure 1. Simple relay circuits. Figure 1a shows a simple 12 V relay switching the load....


2

I believe this should work for what you want. simulate this circuit – Schematic created using CircuitLab


2

This is usually done with a relay. The relay coil is fed by the ignition switch and the contacts by a supply, with a suitably sized fuse, directly from the battery.


1

Another option is using ORing diodes instead of the relay. simulate this circuit – Schematic created using CircuitLab In the circuit above the transient between the solar panel and the 12V DC power source will be smooth and at about 11.0 V (assuming a 0.5 V voltage drop). You could play with the transition point by adding more Schottky ...


1

Indeed, relays have a bit if hysteresis. They will require a certain voltage to turn on, but will not turn off again until the voltage is significantly lower. Operating a relay in this "mid-region" of coil voltage is not recommended, as the contact force may not be enough to provide a good connection. This may significantly affect the relays current ...


1

I think the complete circuit for the image you show should look like this. simulate this circuit – Schematic created using CircuitLab The circuit works in two phases, \$\phi\$ and \$\overline{\phi}\$. The figure indicates the phase when the switches are on. In the \$\overline{\phi}\$ Phase The input voltage is sampled on the capacitor \$C_2\$. Note ...


1

What does it mean "high impedance point"? In our switch I just see a capacitor at the output node, but the impedance of a capacitor is not necessarily high The impedance of the capacitor indeed depends on the frequency of operation. The author is implying that we are working at sufficiently low frequencies (i.e. much below \$\frac{1}{2\pi R_{in}C}\$) that ...


1

1) when the capacitor is charged, there is no current flow, thus appears as high impedance 2) the assumption is that the capacitor is charging to be equal to the input voltage, thus there is no voltage across the FET, and with zero voltage across the FET the operation is certainly in TRIODE 3) when the capacitor is fully charged, the left and right ends of ...


1

You are right to be thinking of safety. I offer the following observations: A safety switch shouldn't rely on a spring to driven into the off position when the guarding is open. That way if the switch spring fails then the system fails in a safe mode. This often relies on the use of a cam or similar mechanism to operate the switch. Safety-rated magnetic ...


1

I think the component that you are looking for is an analog multiplexer such as this one: http://www.ti.com/lit/ds/symlink/ts5a3357.pdf But even thought the component has low ON resistance it will change the behaviour of your current design since it will add about 5 ohm in serie with the LED (from the datasheet). Thus you circuit is not a "perfect" ...


1

Either mount the button to the chassis and use flexible wiring to the PCB, or mount the PCB so it is supported at one corner by the switch and can move sufficiently to absorb the maximum deformation that can be passed through the switch.


1

Have a look at MAX14589E. It guarantees the switches are in OFF state when there is no supply. https://www.maximintegrated.com/en/products/analog/analog-switches-multiplexers/MAX14589E.html


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