New answers tagged

0

I'm going with dumb low-tech solutions: A "superbright" 5mm LED at 1mA will be quite visible in interior lighting. Add a 270k resistor in series, or better two 150k resistors to spread the voltage. The resistors will dissipate about 1/4 Watt. That's tolerable. Stick an analog voltmeter on your 270VDC line... Use your 28VDC supply to light it with LEDs. ...


1

Since you mention that you have a separate 28V rail available, you could use an opto-isolator with the HV rail as input, as shown in the schematic below. simulate this circuit – Schematic created using CircuitLab The opto-isolator must be chosen to be capable of the 28V output level. Please make sure that you size the resistors to achieve the ...


0

If you can have an independent power supply for the the indicator LED then you can use a Hall effect sensor to sense the high current flow and switch on the LED.


0

An indicator LED does not draw a lot of current -- maybe 30 mA or so. If you use a simple active regulator, like an emitter follower with a voltage reference, you will have to draw that 30 mA from the 300V source, so the circuit will have to dissipate about 9W. That's a lot for an LED, but not a lot compared to your heater, so maybe that will do for you. ...


2

Can I connect 1-3 tp4056 module in series ? No. They will not share the voltage equally. The one that draws the least current will get the most voltage, then it will blow up and probably become a short, putting all the voltage across the other one which will then also blow up. Don't do it! I thought to use l7805 or lm317 but 1A is seemed too much for ...


1

You can use opto-isolators to control low-voltage DC from high-voltage DC. Wikipedia says you can use these with massive voltage differences: Commercially available opto-isolators withstand input-to-output voltages up to 10 kV. This would require you to have another power rail at a lower voltage, though. Perhaps using a voltage regulator attached to the ...


2

If you were going to do what you proposed in Figure 2, then why bother with the divider at all? Just use a resistor connected to +V with the appropriate value to produce the desired bias base current. But this is BIASING, not just sending a lone current through the base-emitter. Biasing implies you might end up having another signal ride on top of (...


1

This is not directly connected to your question but it will surely help. I suggest you to put an opamp as buffer after voltage divider output. And RC network to filter some noises caused by voltage divider resistances and opamp.


4

You will need to connect the "bottom" of your voltage divider to ground, but do not make any additional connections between ground and your batteries. So, you will need to use four voltage dividers. The first will measure the voltage from cell 1, the second will measure the total voltage of cell 1 + cell 2, and so on. In your software, measure the voltage ...


1

I solved the righthand side (below) picture. Well, in order to calculate \$\text{R}_\text{th}\$ we get: $$\text{R}_\text{th}=\frac{1}{\frac{1}{1000}+\frac{1}{1200}+\frac{1}{4700}}=\frac{282000}{577}\approx488.735\space\Omega\tag1$$ And, the short current is: $$\text{I}_\text{th}=\frac{5}{1000}+\frac{6}{1200}=\frac{1}{100}=0.01\space\text{A}\tag2$$ So: $...


1

You have to use a logic level shifter IC. You can use a dedicated IC which can translate the signal from one level to the other. you can also use the below circuit if there is only one signal to convert from one voltage level to the other. The MOSFET can be logic N type MOSFET. Image and many other solutions from this link: https://next-hack.com/index.php/...


0

Chances are high that in addition to feeding 6V into a pin that's not even 5V tolerant, you're trying to run a 74HC part with woefully inadequate pin voltages. The easiest solution is the proper voltage divider, run your sensor at 5V, and use a 74HCT165 (note the 'T' in the part designation). The 74HCT line is made to run from 5V and accept TTL input ...


1

While the datasheet is not clear on what the absolute maximum ratings for the IO's are, I am certain that it's 3.6V. Usually "+0.3V" means that the inputs are protected with a diode that turns on at Vdd+0.3V. So this means don't tie Qh directly to the port of the ESP32, it could burn out the diode. Source: https://www.espressif.com/sites/default/files/...


2

Your circuit can be solved using the suoerposition principle. First set Vbatt equal to 0 and calculate the effect of Vm which is simply a voltage divider consisting of R5 and the parallel combination of R6 and Ro. Then set Vm equal to 0 and calculate the effect of Vbatt which is simply a voltage divider consisting of R6 and the parallel combination of R5 ...


1

a resistor based voltage divider requires some current to flow, it doesn't have to flow to ground, and you can adjust what the resistances are to change how much total current is drawn, but for what your describing, it would likely end up to ground also add to this that your arduino's ADC generally cannot handle large value resistances before its readings ...


2

Yes, you always need a closed loop to measure a voltage. This means, you have to connect the ground of your arduino with the ground of the voltage source (your DUT, device under test). You could also use a differential probe, if you can not connect both ground signals, but this is way more complex and you still need two inputs (+ and -) to connect your DUT. ...


4

Figure 1. Typical oscilloscope probe. Problem is oscilloscope only needs one input to measure voltage ... No, your hunch is correct. All voltage readings require a reference point so a ground reference is required as well. This is typically the ground of the circuit under examination. A typical probe uses a coaxial or screened cable with the probe tip ...


2

Strictly speaking, your circuits are not voltage dividers. The notion of a voltage divider, where the voltage at a point between two resistors is determined by the ratio of their resistances, works only if the resistors are in series. When elements are in series the current that flows through them is exactly the same. In practice, you can use a voltage ...


1

Here's a way you could isolate the C64's +5V from the ESP8266, and still be able to 'listen' to the joystick switches. simulate this circuit – Schematic created using CircuitLab When the joystick switch or ESP output pulls to Ground, Q1 is turned on through R1. When the joystick switch is off The transistor's Emitter is pulled up to ~2.7V by R1, ...


2

On the C64, all pins (up, down, left, right, button) are pulled high internally by resistors and will show 5V. When a classic joystick is moved, it will mechanically connect the corresponding pin to GND, forcing it to 0V. To emulate this from a microcontroller you can use either: An NPN transistor with emitter to ground, collector to joystick pin, base to ...


1

Make a "FET level shifter" from discrete components. All you need is 2 resistors and the FET, for each data pin. It limits the voltage to 3.3V microcontroller and it can still pull the 5V side down.


Top 50 recent answers are included