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Calculate the R5 = (9v - 4x2v) / 20mA = 50ohm (use 47ohm) For Beta = 150 Calculate Q2_ib = 20mA / 150 = 133uA Calculate R2 = (9v - 0.7v) / 133uA = 62Kohm (use 47k) For R4 even 1Mohm should be enough, but it is not common to use such high values (except low power designs) due to noise affecting.

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I would put all 4 LEDS in parallel, each w their own resistor. Much more elegant and balanced. Also, draw your schematic with supplies inverted...much clearer. To turn on the PNP, you pull down the 1K with the NPN, and draw 8mA to GND. With a PNP gain of x100, it can support 8 x100mA going to LEDS...and you only need 100mA...so it turns ON hard..which is ...

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A small n-channel MOSFET such as a 2N7000, two resistors, and the LED. Plus a battery, waterproofing of the components (except for the sensing area), etc. Is this within your skill set? If so, I'll update this answer with a schematic.

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Yes - sure you can. You could detect the short with a logic-level circuit and control the power rail with a relay under feet. Simple short-circuit 'Sensor' builds from two wires-stubs or a dedicated PCB. Feed your 'Sensor' with V+ rail through a high value series resistor (10k). Connect the other end of the 'sensor' to an input pin of e.g. an Arduino. Pull ...

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Maybe you're in a bit over your head with analog design. The PD transimpedance amplifier is also going to be a challenge. You can buy an integrated pulse oximeter analog front end (AFE) from TI, for example. AFE4403. The IC contains an H-bridge driver and 8-bit independent constant-current settings for each of the two LEDs. The IC comes in a rather ...

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Spectrum of RGBA LED compared to low CRI (Color Rendition Index) 6500K LED (source): Just looking at the spectrum, it appears color rendition will be terrible. Color temperature gives information about how the light itself will look, but it only gives useful information about how objects will look under that light if the light source has a reasonably ...

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Hfe is an approximation for a transistor in the linear region, it doesn't apply to a BJT in saturation. You can use it to estimate a base resistance, but the characteristic curves will show you more detail. You should not rely on the saturation voltage of a transistor to limit the current in an LED. It can vary quite a bit over temperature and from unit to ...

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Inside each TLC5917 device, and in fact, most devices that use an external resistor like that, there is a constant current source that allows the device to detect the voltage based on the external resistance and then adjust the constant current used for powering the LEDs. When you link all of them together into one resistor, the current from all sources ...

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would it be more efficient to connect the MCU to the gate of a MOSFET and power the LED directly off the battery When you use a MOSFET and the LED is driven to be off, there will still be a leakage current through the MOSFET from drain to source. Go to your favourite MOSFET data sheet and look up the $I_{DSS}$ parameter; it might be 10 nA or it might be ...

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If your MCU is powered by a DC-DC it would be more efficient to use it directly, if it could actually swing enough I/O voltage for the LED (more about this below.) If it’s using an LDO, it doesn’t matter, the sum of IR drop power loss through the LDO and load resistor would be the same as going directly to Vbat with the load resistor. The problem in most ...

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If it’s like most motorcycles, the neutral switch has a single pin that connects to engine case ground (closes) when the selector is in neutral. It’s open otherwise. You can check this with an ohmmeter. Some motorcycles might use a 2-pin connector if the engine is in an anti-vibration mount (e.g., a newer Harley), but nevertheless would connect to ground via ...

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I see you've asked a few related questions about LED drivers, so I'll try and give a big picture answer. You put current through LEDs and photons are generated at some rate, roughly 1 photon per 1.5-3 electrons. To move that current you apply some voltage, with small changes in voltage leading to very large changes in current. Unfortunately, the specific ...

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How to Drive a Bi-Polar LED Having 2 Different Forward Voltages Given you are using an MCU with lots of I/O pins I'd suggest the following: simulate this circuit – Schematic created using CircuitLab Set I/O strength to 12mA set the required current using 1:3 upper and lower drivers Set pins to low output or as inputs when LEDs are off It's up to you ...

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tl, dr: you either need a buffer that has really low Rds(on) to ensure that your series resistor is the only thing (or at least, the dominant thing) setting your current, or you need to sense the current and regulate it. That's easy enough, as you'll see below. Also, to set the current in each direction you'll also need to steer the current so that each LED'...

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750umol *0.25m = 187.5 umol/s. Datasheet gives 2.72 umol/s/LED at the test current. 187.5umol/s / 2.72 umol/LED = 69 LEDs. Scale as appropriate if you use a different drive current. Since you're interested in an even illumination, consider how widely spaced they would be. If you evenly spaced them, then they're 6cm apart, or about double the height you want. ...

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There shouldn't be any great confusion here. How our eyes perceive the brightness of the LED as the average voltage? Our eyes don't detect variations in light intensity above about 30 to 50 Hz very well. If they did we would have been driven mad by AC powered lighting and TV/monitor flicker. It should make sense that if you blink a light at full power and ...

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Your schematic doesn't show the LEDs but I assume these are in series with R1 and R4 and that would be fine. The breadboard allows you to make few quick checks: Remove the capacitors. Now you have two independent (identical) circuits driving LEDs. The LEDs will be on (assuming a decent transistor gains). Remove resistors R2 and R3 and the respective LEDs ...

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Your method 2 is correct. You might find it useful to learn about load lines. These are useful when the device (the LED) doesn't behave in a linear fashion like a resistor. Figure 1. Various loadlines for a 5 V supply and various LED types. Image source: Loadline resistance graphic tool. Each loadline is drawn from 5 V, 0 mA to 0 V and the current ...

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You need to allow for the LED voltage drop, whether you have a simple LED, resistor and battery circuit, or the LED is controlled by an Arduino or other device. There is no "ideal" current for a LED. There will be an Absolute Maximum current, beyond which the LED may be damaged. LEDs will work happily at currents much below their rated maximum ...

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When calculating the current through an LED and a resistor in series, you need to take into account the voltage drop across the LED. So method 2 is the correct one.

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Microcontroller with enough I/Os can do that. If you want something more elegant and don't feel like writing code, you could program a Silego Greenpak to do this. The resulting circuit would be really tiny, and you could eliminate those diodes.

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If it is just for a Led effect, why not to use a transistor ring counter. With a little trick you can move backward also.

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If you have a look at the internal circuit of the CD4017 you can get a good idea of how to make one using other CMOS chips - if you have them. You can also see why it's a good idea to use the CD4017. Figure 1. Image source: Intersil. The obvious thing to do is use a micro-controller with at least six IO pins. If you have an Arduino or similar to hand that ...

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The LED is not connected to resistor, it is disconnected. On the other hand, it looks like there is a solder bridge which shorts the LED directly to buzzer pin. There might be other solder bridges as well at the op amp inputs, which may be the reason nothing works. It is also impossible to say if you have LEDs and the photodiode connected with correct ...

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Since the entire circuit is in the DC domain all you may need is a voltmeter to check voltages at the op amp terminals. What is the current required to drive the buzzer? Can the op amp drive it? Did you ohm-check the traces you separated on your universal PCB between pins 1-8, 2-7, 3-6 and 4-5 of the DIL package? (Did you remove all the copper there?)

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You seem to have used a 22Ω resistor for the LED instead of the 220Ω shown in your schematic, and the link between the LED and that resistor is missing (fortunately for that poor LED). The graph of figure 10 on page 10 of this LM358 datasheet, it can be seen that the output is unable to swing to the positive supply voltage, when sourcing current. If your ...

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You don't provide all the relevant information to give you absolute answers. The LED you have is a Phosphor conversion LED; it's actually a Blue LED with a phosphor disk on top and the phosphors are notorious for PCE droop (phosphor conversion efficiency droop). You have three effects going on: LED Vf is dropping due to thermal process (no matter how small) ...

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It's probably not the LED. It's probably the transistor that is powering the LED. To confirm, measure the voltage across the combination of the LED and the current limiting resistor. I think you will find that it decreases over time.

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@metacolin is generally right, but my first action would be to throw away the LED PSU and get a better one. The powerline ethernet works by injecting high-frequency currents into the power line. It takes care not to inject much, because for other devices this accounts for lowering the power quality. This is why it is easily affected by powerline noise. The ...

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The ground pin is tied to neutral at the breaker box, so it is perfectly possible for noise injected into the neutral wire to appear on the ground wire as well. Beyond that, most recent (not slow) powerline ethernet adapters use 2x2 MIMO, and utilize both line to ground and neutral to ground as the two MIMO pairs. If your adapter has 3 prongs, you can ...

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Your signal averages at 0VDC, it does not rectify and filter the input signal. Then the average brightness is with its input at 0VDC. You need a "peak detector" which rectifies the signal and filters the resulting DC voltage that represents the loudness of the input signal. Here is my peak detector:

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First thing, which I'm sure you already know: if you are concerned about lifetime, i.e. quality, "grab bag of LEDs off of Amazon" is not the place to look. Those might be very well for prototyping a project on a breadboard, but if you need them to be reliable, spend $0.25 at Digikey and get something from Cree, Lumex, Liteon, or any of the other ... 0 If the goal is only a smooth brightness level ramp-up through the full-range of possible brightness levels: The time step linearly decreasing as the brightness increases with a constant step, results in the brightness level exponentially increasing in time. Example: #define MAX_DUTY 255 #define MIN_DUTY 0 #define RAMP_DOWN 0 #define RAMP_UP 1 #define ... 2 Your LED has a maximum current rating: 700 mA. 3V just happens to (on average at room temperature) be the voltage at which this current flows. So, the moment you use a 1.5× higher voltage, you forced an exponential amount of additional current through your LED. That leads to a quadratic (of that exponentially increased current!!) increase in power converted ... 1 LEDs are driven with current, NOT with voltage. If its (LED's) threshold is 3V, then anything over 3V connected to it will act as short circuit with infinite (in practice, finite, but well over any spec) current. You always* therefore need a current limiting resistor. If your supply is 4.5V and LED drops 3V, then you will have 1.5V across the resistor. ... 0 LEDs are a goldrush right now, and as such there are a lot of "bottom of barrel" manufacturers in the game. People who buy mail-order or who buy cheap are particularly vulnerable to encountering this crud. Even a UL listing is no guarantee; UL certifies safety, not quality. Everything at CostCo and huge swaths of what I see at box stores are on ... 1 The LED light bulbs will flicker once every 30 minutes or so whereas a CFL does not flicker at all. HVAC, water heater, etc flipping on or off and injecting noise into the power lines. Your LED bulbs are extremely sensitive to it because they have cheap power supplies. Is there a reason why the LEDs would flicker when the CFL bulbs do not? Are they more ... 1 There is a good article here: https://www.ccohs.ca/oshanswers/ergonomics/lighting_flicker.html Fluorescent lights actually do flicker but the frequency will typically be high and the light dims a lot less at it’s minimum point than LEDs. LEDs tend to be instant on / instant off and run from DC. LED circuits can be designed to reduce flicker by adding ... 0 Is this the stair light? Also, question should clarify whether these are flickering when the power is ON, or OFF. Assuming OFF, it's because of stray power on the circuit, usually from electrostatic coupling from other cables (though degraded insulation is also a possibility). I've even seen a CFL flicker on a (UK, 230V) dual switched circuit (stair light) ... 0 My best guess is you've got a bad connection or other resistance in the wiring (maybe oxidation on switch contact etc) the comment to change bulb position(s) or feedback on trying this would help rule out the bulb itself. 2 Any kind of thing that has at least 2 discernible states can be used to represent binary, be it electrical, optical, or some other means. The question then becomes, how far, how fast, and how much cost? Let’s look at some optical options: if you have a flashlight you could send some kind of code using the switch - short pulse for ‘0’ and long pulse for ‘1’, ... 4 Could I use an LED to emit that binary code through light signals?. Of course, but if you want some device to receive it and decode it on the other side, it gets a bit more involved. If there is no outside interfering light source, for example in optical fiber, then you could simply drive a LED on/off with the bits, use a photodiode or phototransistor at ... 3 LEDs (all diodes) are non-linear devices. If you look at a voltage-current curve, you will note that the relationship is not a straight line as it is with a resistor: From 0V, as voltage increases, not much happens until the forward voltage (Vf) of the diode is reached. After that point, current increases exponentially. If there is too much current, the LED ... 13 LED D2 is green and it needs about about 2.0V to light up. LED D1 is red and it needs about 1.6V to light up. And when the transistor is turned on, it has a Vce drop of about 0.2V. Adding the voltage drops of the transistor and red LED D1, that is still less than what green LED D2 needs to work, so there is not enough voltage over green LED D2 to turn on. 0 Was there a specification for the voltage drop of the LEDs and the transistor curves/gain? You will need these to fully analyze the circuit. But in general, when the transistor is saturated, the voltage at D1 will be about 0.2V lower than D2 (the Vce at saturation). So unless the LEDs are different, I don't see how D2 would be off. 4 Ok I think it was silly to ask, The datasheet of the yellow LED says "with build-in IC for blinking" So this is it! 0 There are several reasons why it may not be working. Voltage levels. Termination. Timings for your transceiver The whole system. Let's address it one by one: Voltage levels. The LED WS2815 with datasheet is a 12V intelligent LED. The digital input considers $$V_{IH} = 0.7 VDD$$ and $$V_{IL} = 0.3 VDD$$. In your case $$V_{DD} = 12 V$$ $$V_{IH} = 8.4V$$$\$...

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