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Using a regulated current source to light them, wire the LEDs in series and short out the segment which you want to be dark. simulate this circuit – Schematic created using CircuitLab You can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage. Here's a simple way to build one using a LM2596S module: ...


17

First, "zoom in" to that rising edge by adjusting the time base. When you start getting close, you will start to see the rising slope of the signal. As you do this, you will start to lose resolution on your captured signal. You can capture new samples of that rising edge using the scope's triggering mechanism. Once you can see the rising slope, capture a ...


14

This is an issue with scope setup and misunderstanding of how to interpret scope captures. You must capture the rising edge of a single pulse at a reasonably small resolution by using a single trigger. Good news is that this is exactly what oscilloscopes are designed to do The generic procedure is: Set trigger to edge (up) and trigger level at ...


11

Here is a test I did with my 200MHz Tek scope. You should be able to get similar results with the Rigol, this is an older scope with a modest 2Gs/s capture frequency. My circuit is just a standard 10:1 probe connected across a 6mm tact switch with a 1K pullup to +5V supply. Not all the captures were this messy, some were pretty ideal looking. Pushing it ...


7

If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.


6

Assuming that the pull-down resistor is a reasonable value (1k - 10k), the very next thing that I would check is to see if there is a filter active on that channel. I wouldn't be looking for signal averaging - this is a single-event occurrence and the trace shows that single event. But it is entirely possible that there is a very-low frequency low-pass ...


6

One way to achieve your progressive LED lightup as you turn the rotary switch is to use a current sink on the common of the switch and then wire the LEDs across the selector switch terminals as shown below. The constant current sink shown is a low cost way to get a 20mA sink for the LEDs so that there is no brightness variation as the number of lit LEDs ...


5

Figure 1. The guys down at photo-forensics found this. There are several factors: You have a nice new clean switch that bounces very little. Your scope is loading the circuit and the 15 pF is enough to help. This is unlikely, though, with what appears to be a resistor with a value in the hundreds of ohms. (The colour rendition of your photo is poor.) ...


4

Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions. Sunyskyguy has a clever solution if you have a high voltage available.


3

Simple momentary push buttons can be made dirt cheap in quantity. Push-push switches, definitely not. The cost ratio between a momentary switch plus a bit of electronics*, vs. a mechanical push-push switch, is huge. As a bonus, a momentary plus electronics is cheaper to make reliable. So on a cheap consumer product, the on/off switch is almost certainly ...


3

Relays are still used and with good reason. In your case: The relay separates your logic circuit from your high-current and higher voltage pump circuit. Any faults on your pump circuit will not affect your logic circuit. A faulty relay can often be a plug-in replacement. A transistor is cheaper but requires that the common line must be shared between both ...


3

That is not how you connect a motor to a switch. You connect everything in series like so: simulate this circuit – Schematic created using CircuitLab Such that opening the switch stops current from being able to flow into through the motor. In your circuit, when the switch is closed the switch is much much lower resistance than the motor so most of ...


3

If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes. (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol) simulate this circuit – Schematic created using CircuitLab


3

It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine ...


2

It's pretty easy to get the wrong pin diagram using a random datasheet for the generic "2N2222 part. There was a time, when these were not consistent and it seems even now, there are different datasheets There were many reasons for this, which may be shared by esteemed colleagues.


2

The switch is not perfect, there is some voltage drop across is so power must be dissipated of load current x voltage drop. That's in addition to the base or gate power which is usually not very significant in a well-designed circuit. For a triac the loss is usually of the order of 1W per ampere of load current (implying a voltage drop at typical currents ...


2

Look at the curves on page 3 of your MOSFET's data sheet: You are not fully turning on your MOSFETs. This means the drain to source resistance is higher than you think. You need to look at more than just the Vgs(th). If you don't see the phrase "logic level" on the first page of the data sheet, you should scrutinize the conduction curves before trying to ...


2

You wrote that, "I am total rookie in electronics, so I need it to be as simple as possible." Since none of my betters (I'm just a hobbyist without even so much as one day's class of DC electronics training) has yet bothered to provide something I'll try to follow your guiding words and provide something simple and easy and relatively available. simulate ...


2

It's ok to use an NPN in the 'middle' of the circuit. If you were to use mosfets, there could be problems with Vgs if used in the 'middle', the gate needs to be higher than the source or the mosfet will not turn on. An NPN transistor will function like a mechanical switch, but will have a voltage drop across it similar to a diode. The NPN transistor most ...


2

Oldfart and Mattman944 give very similar answers involving complex diode networks. If brightness variation is acceptable a simple diode ladder is sufficient. Red LED's typically have 2V voltage drop and diodes typically have 0.6V voltage drop, so the combined effect of the diode voltage drops in a ladder is can be significant. With a 9V battery and the ...


2

Yet another example of trying to use an N-channel MOSFET as a high-side switch. It simply doesn't work. Use P-channel MOSFETs and pull the gates to ground to turn them on.


1

The circuit below will provide DC bias for inputs to the switch, and remove DC bias on the outputs. simulate this circuit – Schematic created using CircuitLab "in" is any input to the switch. "out" is any output from the switch. The input bias circuit has a corner frequency below 1 Hz. It will pass frequencies above that. The output bias will ...


1

As you don't wish to use a relay, you could possibly use a CD4066, CMOS Quad Bilateral Switch (datasheet): Although as suggested in Finbarr's comment, as you might want to use a double pole switch, you would want to use all four switches within the IC package. See image below taken from How to Build a 4066 Quad Bilateral Switch Circuit Just modify the ...


1

Use a 50mA PNP with proper bias or consider a LOW side NPN switch on Cathode of LED. with 5mA base current. at 1.1V in order to drive 50mA or so. (below) simulate this circuit – Schematic created using CircuitLab This is a level shifting NPN-PNP pair switch assuming a CMOS 1.1V driver Here I assumed that the base voltage was 3V and when closed the ...


1

How about using a MCB + latching motor switch? Switches look like this: MCB sits in between of power supply and the rest of the setup. Latch switch sits in parallel of the resistor: To turn the motor ON, you lift the MCB and then press the latch switch after a second or two (depending upon the time it takes for charging). It provides additional protection ...


1

The 555 is a great idea, but it's worth pointing out that an arduino nano could also do the job. The power switch likely jumps a pin pulled to 5v to ground. In such case, a 5v powered arduino could do the job. It may seem like overkill, but it's a software driven solution that requires no soldering - just an arduino nano with pins soldered (around five bucks ...


1

The title says you are asking about a triac but most of your post is about transistor heating. Because the collector and emitter will just act as a closed switch and no power will be lost there right? No. Try measuring the colletor-emitter voltage when the transistor is turned on. You will find that it is probably about 0.2 V if you have adequate base ...


1

simulate this circuit – Schematic created using CircuitLab Note this is assuming ideal parts. Reality check: every part has resistance. Battery (ESR) , wire {mOhms/meter} Motor DCR coil resistance and switch. Each have current limits Current in loop I = V/R on startup then reduces and depends on mechanical load and RPM.


1

That unit uses a two-wire connection so it relies on being able to pass a little current through the load while "off". The load resistance needs to be low enough that enough current can pass without a large voltage drop across the load and without turning on the load. They have chosen 20 W. Options are adding in a parallel load, using a > 20 W lamp or, ...


1

You need a high pass filter on the input of the 555. The resistor/capacitor values here are fairly arbitrary, but they should work. You should also implement Randy's points 4 and 5 to generally improve your circuit. If you don't implement a high pass filter, the output of your 555 will remain activated if you hold down the button longer that 500ms, until ...


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