Hot answers tagged

9

Given that the pins are open drain, the following circuit should work quite nicely: simulate this circuit – Schematic created using CircuitLab You'll get some wasted power as one of the resistors will be pulled straight to ground when an LED is turned on, however if you go for a low-current LED, you can increase the resistance to reduce the amount of ...


8

The link that you provide and the comment by John D say what must be done. The armature and field connections must be reversed with respect to each other. In the diagram, you disconnect the black lead and the red lead from the two brushes and reconnect each to the opposite brush. The two red wires going to the field assembly are two wires, not one ...


4

Here's the relevant part of the system diagram from the datasheet you linked: A Schottky diode typically has a forward voltage drop of 0.2 to 0.3 V. So if 5 V is applied at "V+", then only about 4.8 V is available to supply the Richtek switching converter or any other circuits that might be connected to that same circuit node. The reason it's called a "...


4

No, it will not. The circuitry of a boost converter is not set up to take in power, only output power. For that you will want to search for a buck converter, instead of a boost converter. https://www.amazon.com/s/ref=nb_sb_ss_c_1_8?url=search-alias%3Daps&field-keywords=buck+converter&sprefix=buck+con%2Caps%2C233 however, most don't accept ...


4

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: -


4

I am pretty sure that the IR carrier is just on-off keyed with the output of an UART. It's 8-bit asynchronous serial, probably at 38400 baud, with a single stop bit. There are 6 bytes of data in a transmission. It can't be 7-bit serial with a parity bit, as the last bit matches up with neiter odd nor even parity. The data being transmitted is ...


4

The short answer is you can't because physics. The long answer is you can't because voltage is a difference of potential. What you're looking for does exist. Thermoelectric generators translate a difference in heat into a difference of potential through the reactions between different materials. A resistor is generally a fairly uniform material exposed ...


4

In general adding heat to a resistor will change it's resistive property. A negative temperature coefficient resistor will have it's resistance go down with increased temperature. On the other hand a positive temperature coefficient resistor will see an increase of resistance as temperature rises. The change of resistance does not alter the fact that the ...


3

In LTspice, setting the rise and fall times of a pulse source to zero will not mean that they will be null, as that will be both a physical impossibility, and a machine problem -- a low and a high cannot coexist in the same time. So, LTspice circumvents this by setting them to 10% of the Ton. In your case, you have Tperiod=10u and Ton=5u, which means that ...


3

At that frequency you have this... simulate this circuit – Schematic created using CircuitLab


3

The approximation is good enough because \$V_{be}\$ is a lot greater than \$V_t\$ (thermal voltage at a room temperature of \$27^o\$C). \$V_t\approx 26\text{mV}\$ at \$27^o\$C. Even for a value very close to zero, the approximation works well because the exponential term grows much larger than 1. Say \$V_{be}=100\text{mV}\$ (which is very close to zero), ...


3

I have ordered a Wi-Fi switch named Sonoff. I have a doubt whether the device will work if the polarity is reversed or will it get damaged? It will almost certainly work It will almost certainly not be damaged IT WILL LEAVE YOUR APPLIANCE LIVE WHEN TURNED OFF. If I'm correct, AC doesnt have polarity right? Most AC supplies have one side referenced ...


3

It looks like you have parity bits in the transmission (or maybe some extra ninth bit): - The red arrows show the start bit positions so your payload data looks to me like: - 10010010 11110111 10001100 P = Parity (ninth bit) S = Stop It's not Manchester encoded else the idle period would be 101010101 etc..


3

The thing that i need to make this work is that the ignition lock needs to be open (let eletricity pass trough) when the key is turned. First some terminology: When you close a switch electrical current can flow. When you open a switch the load is switched off. You must have measured the switch incorrectly. All ignition switches will isolate and turn off ...


3

The motor is a single-phase induction motor of the permanent-split capacitor type. It is called permanent-split-capacitor because it is a split-phase motor with the capacitor permanently connected rather than connected for starting and disconnected after the motor comes up to speed when started. The diagram on the label that shows the windings and capacitor ...


3

Actually I think that none of the components you've circled in all the schematics are responsible for the reverse voltage protection at all (as you correctly pointed this out yourself in all three cases: "diode D1 blocks excessive positive voltage from the input supply", "R5 to block excessive currents from passing to the GATE pin in the event of excessive ...


3

An ideal diode would be the best solution with a suitable low RdsOn FET. D4185 RDS(ON)< 20mΩ(VGS= -4.5V) $3.19 simulate this circuit – Schematic created using CircuitLab


2

Assuming the -24v is a control input, so doesn't need much current, and only needs approximately -24v, then a cheap, quick and dirty way is to use a charge pump. Like this. simulate this circuit – Schematic created using CircuitLab What's drawn as an opamp is any circuit capable of making a rail to rail square wave. The output is unregulated, and ...


2

Yes, the ESC you bought will drive the motor. The motor is 3 phase BLDC. The only problem you may have is that the ESC is low torque startup (it's designed for a Heli/quad copter. However given the large 10:1 gearbox you may be ok. If you need high torque at low starting speeds you really need a BLDC with Hall Effect position sensors To add from your ...


2

While synchronous (two MOSFET) boost converters can indeed convert bidirectionally, your board isn't necessarily of the synchronous switching topology (it could use a diode instead). In addition, that board certainly won't be regulating the input side so don't expect 12 V at the input. If it's set to produce e.g. 60 V and you backfeed 61 V it'll try to ...


2

If you need the voltage just for indication or measurement then a simple DC differential amplifier with unity gain will do the job, because the output voltage is 10V minus input voltage: \$V_{out} = 10V - V_{in}\$. : simulate this circuit – Schematic created using CircuitLab Note that this circuit is not easy to build as it looks even for an ...


2

A BJT is obviously more complicated than your equation(s) provide. But those equations are often good enough when just considering the forward active region. To get a feel for the simplest DC model that was developed, see my answer to Why is Vbc absent from bjt equations?. Today, the term \$I_S\$ is usually implied to take on the meaning used in the Ebers-...


2

A resistor produces an output voltage that rises with temperature. It's called "thermal noise" and is calculated as: - \$\sqrt{4\cdot k_B\cdot T\cdot R\cdot \Delta F}\$ Where \$k_B\$ is Boltzmann's constant, T is the temperature in kelvin, R is the resistance in ohms and \$\Delta F\$ is the bandwidth of the noise you might be interested in. See this wiki ...


2

It's not presented. You might be able to extract it by curve fitting Figure 4, which shows the typical relationship between \$V_{BE}\$ and \$I_C\$. Even easier, you can find a SPICE model for 2n2222, which will likely include a reasonable value of \$I_{ES}\$ determined by a similar curve-fitting technique.


2

1) Are they ceramic capacitors? The capacitors in the photo of the link appear to be 0603 ceramic capcitors. 2) is correct the procedure of unsolder and measure? Yes, usually the best way is to unsolder because the other parts of the circuit are in parallel with the capacitor. By taking it off the rest of the circuit is not measured. I do this all the ...


2

the -100V is a 5A inductive pulse, whereas DC survival spec is only -30V +50V. The reason for this -30V is in case a alternator charged 24V vehicle is used to jumper cable another with the cables reversed. (momentarily) The 2nd simple circuit is missing parts and will not survive all the test conditions for ISO-16750-2 tests 1 thru 5b . Keep this date in ...


2

It sounds as though you are connecting your ohm-meter to a live circuit. This could destroy your meter. simulate this circuit – Schematic created using CircuitLab Figure 1. Equivelant circuit of ohm-meter. Most budget digital multimeters use a 0 - 199.9 mV (0.2 V) voltmeter circuit in them. All of the voltage and current ranges on the meter switch ...


1

A common test circuit for measuring the reverse recovery times of diodes is shown here: If you want to compare two different diodes, do not put them both in the same circuit. Put in a second power supply, pulse generator, etc. and only have the oscilloscope common to both circuits (each test circuit gets its own channel/trace). You may also have to adjust ...


1

It is the presence of holes and extra electrons in the semiconductor material that make it conductive. At the join between the layers in a PN junction the holes and electrons join together and cancel each other out creating that depletion layer which is an insulator. When you attach a reverse voltage you force the holes and electrons to go towards each ...


1

This current is due solely to carrier generation within the depletion region due to phonon interactions creating an electron-hole pain every once in awhile. Once this hole and electron are generated in this region of high electric field, they are swept in opposite directions. Since there's so few carriers being generated due to thermal interactions generally,...


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