New answers tagged

1

I came here looking for the solution to this problem myself, and then realised the answer. With a normal binary comparator, all bits have a positive assigned value, bit 0 is 1, bit 1 is 2, etc. So, with 5 bits, bit 4 would be 16. Dealing with signed numbers, using the 2's complement, the last bit is just the negative value of whatever it should be, in this ...


1

The supply voltage is in the formula for calculating the op-amp output voltage, so it is not left out. But as the NTC and the fixed resistor use the op-amp supply and ground for the voltage divider, all it is needed to know is that the divider output will be within the supply voltages. The R1 value is calculated with a geometric mean instead of arithmetic ...


3

It means the components are drawn in the schematic and there are places reserved for them on the board, but during board manufacture these components are not assembled on the board. So they are not needed for basic functionality but if someone wants to try out a feature such as the RTC clock crystal it can be soldered there for experimenting and prototyping ...


2

It is quite common to have components which are not populated on a PCB during manufacturing. They may be omitted for several reasons: The design does not require the component but it could be added later for various reasons including noise immunity, stability, performance, tuning, etc. This is common in prototyping and early product versions. The schematic ...


0

I think \$R_C\$ is in the wrong place. You have accidentally created a first stage with gain in the tens of thousands, or whatever the open loop gain of the opamp is! Also 1: If you consider the push-pull stage to be a voltage amplifier with unity gain, then U2B with feeback via the potential divider Rf & Ra produce a total gain of: $$ \begin{aligned} \...


0

Not an expert here, but diodes D1 and D2 do not look like they are in the right place. You want something like this: Picture is from https://www.electronics-tutorials.ws/amplifier/amp_6.html. Take them out and run your simulation. If everything else is OK, you should see a sine wave on your output with a bit of discontinuity at the crossover regions.


0

Rc should be in series with C1 as in the first circuit diagram. As you have it in the simulation the first stage is acting as a differentiator. Insert a 2.2uF dc blocking capacitor in between Ra and ground. Add small (1 Ohm) emitter resistors in series with each of the emitters of Q1 and Q2 to take up the voltage difference between the double diode drop and ...


0

You do compare the input sine voltage VS. output pulsed voltage of discrete values, of course nothing useful could be . L You have to put a LC lowpass filter on the output stage and take the feedback signal only after the LP filter.


0

The VCC voltage of 5V that biases Q1 and Q2 might be too low compared to the signal that drives the bases of Q1 and Q2 themselves. Try VCC = 15 V.


1

At 30 mA you can just power the buzzer from the 555's output (assuming the 555 is running from a suitable voltage)


1

You could still have thermal issues with derating the Zener in an enclosure above 25'C, unless designed properly such as using 2 in parallel. Always derate power by 50% when you want it to run at 50% of the max temperature rise as MTBF drops 50% for every 10'C rise. If you operate more than 5V it will draw proportionally more current and fail sooner from ...


1

Based on buzzer's datasheet max current and voltage you can find the maximum Power needed. P= V * I= 8(V) * 0.03(A)= 0.24 Watt. I suggest you to simply do this schematic: For 24V as Vcc you have 5v on the buzzer. So 24-5= 19V. If you have a resistor with value 1k you will get 19mA current flowing to your buzzer. If you limit the current then your buzzer ...


5

In this case, a "strap" refers to a low-impedance connection between the substrate and a power rail, in order to provide a stable substrate voltage at the body of a MOSFET. If this impedance is too high, then a parasitic PNPN structure formed between the power rails can act as a parasitic thyristor under certain conditions, leading to a short ...


2

You could feed each Thermistor output into an input of an Analog 2:1 Multiplexer, and also into a Comparator. Then use the comparator output as the Select signal for the Analog 2:1 Multiplexer. Which thermistor goes into the positive terminal of the Comparator and which goes into the negative intput of the Comparator depends on whether the thermistors are ...


0

The CD4053 has three independently controllable SPDT analog switches, with a global inhibit. One chip can handle all three modes for inputs 1, 2, and 4. Input 3 never is inhibited, so that would use one channel of a second CD4053, with the inhibit input grounded. https://www.ti.com/lit/ds/symlink/cd4053b.pdf?ts=1634601279843&ref_url=https%253A%252F%...


0

There is a lot to say about each component changes I made to improve this design. But Rather than explain how your circuit works and why it overstresses an LED and why the frequency control of the pot is suboptimal, allow me to show a better solution with slight changes in values to reduce base drive currents and LED output powers from 10W pulses to 50 mW ...


1

Let me tell you how I did it. I used an analog input port with a 1 megaohm resistor between +5V and the port. With zero load, it gives a pretty good high signal - over 1000. Then, I connected the port to a stainless steel probe, in a stainless container (this is for an automatic distiller), container is grounded. When the distilled water hits it, there is a ...


2

In the circuit below, focus on C1 and R1 to understand the differentiator behavior: Assuming \$V_z\$ always zero, \$\frac{dV}{dt}\$ at the capacitor is \$\frac{1V}{10ms}\$, so: \$I_c = C \frac{dv}{dt} = 100 nF * \frac{1V}{10ms} = 10 \mu A\$ Considering this current goes only through R1: \$V_{out} = -10\mu A * 100k\Omega = -1V\$ R2 and C2, with such small ...


0

For manufacturability, don't put those vias between pins 3/4. FWIW, to keep the power clean I would put a series L or ferrite in series to create an analog supply for the op amps. You could also have divided grounds by splitting the plane. In Altium this can be done with polygons or by drawing lines on the negative plane layer if you used that. The ground ...


1

Comments on the updated design: You want to minimize the length of the traces connecting the photodiode to the opamp to the feedback network. Move the opamp closer to the photodiode, even directly over it so that you can eliminate the boxed trace: Since you are putting components on both sides, consider putting C1/R1 on the backside of the board, directly ...


0

In this low frequency range, it is not usual to work with the 50 Ohm system. This is rather used in the RF range, in which you want to achieve maximum power transfer. So you have both 50 Ohm output resistors and 50 Ohm input resistors (the actual value of the resistors can be different if they are all the same, e.g. in the 75 Ohm system). In your ...


1

You are mixing up with different things, protective earth as "ground" required for protection against electrical shock in high voltage system. in low voltage ground is used for protecting system from EMI. do not use for that ground from receptacle.separate grounding system is used in labs, etc. in your case the system will work without ground. just ...


0

The current-carrying parts of a domestic light switch must not be grounded - there should be no connection between those parts and the mounting bracket.


1

You have misread the datasheet. The ADC input does not have input impedance of 3500 ohms to ground. Your simulation has 3500 ohms to ground so a series resistor of 100k will form a resistor divider network and thus explains your simulation having low voltages at the ADC input. You can fix the simulation by just removing the 3500 ohm resistor.


1

R13 when pulled low draws 10 mA with 23V across it or 1/4W at burning hot temps . Change R13 to 10x R11. R18 turns off LED until R19 is pulled high.


1

I figured it out. I had it grounded out because on the seven segment pins 3 and 8 are connected. I did not know this and connected pin 3 to vcc and pin 8 to ground. After cutting the connection of pin 3 to vcc the pcb worked as needed!


2

When you are using a BJT as a switch (and not just as an analog amplifier), you will usually have a base current more than the minimum needed. That's not really driving it hard. For the 2N2222 in particular, the max base current is 200mA. See https://html.alldatasheet.com/html-pdf/15067/PHILIPS/2N2222/745/3/2N2222.html That would be absolute max. I wouldnt ...


4

It's intentional, this is a known design for an off-line switcher with buck topology. The internal supply of the chip charges the capacitor on the VDD pin with the voltage differential between the drain and GND (source) pin when the internal MOSFET switch is off. The capacitor will then continue to power the circuit long enough when the MOSFET is switched on ...


0

First, if the PE has to be disconnected, then also the power has to be disconnected at the same time. In order to find if your car has a leaky connection from the battery to the chassis you have to measure that before connecting the power and PE, first. simulate this circuit – Schematic created using CircuitLab Eval board tester from TI


2

4 layers is overkill, but these days it is pretty cheap and it saves time, so no problem with that. how can I send signals to the middle 2 layers in a 4 stack without the use of vias? When you use a standard thru via, it goes through all the layers. It will link all the layers where you connect a track to the via, and it will also connect to power/ground ...


1

This exercise is really good because it covers a lot of material in a seemingly simple question. Most of the work you did is good (except getting the sign of e wrong), but you missed a key aspect, as you'll see. Forgive me for redrawing the circuit with some extra labels, because it's going to make things a lot easier to refer to later. simulate this ...


1

You can try using a conductive adhesive (e.g. silver paint) I have tried it myself. Insert your wire in and after that, paint it from both sides at both entrances on the cloth and let it dry. I suggest: Silver paint 503 is a flexible, high-temperature conductive material designed for a wide variety of uses, and adheres to most substrates. Source: Electron ...


0

It's largely a matter of consistency. Note the top half of the schematic. It uses an AND/OR structure. The bottom half has a row of ANDs, mostly connected to XOR gates. In the case of the Cn+4, since the left-hand input comes from one of the AND gates, they continued the use of OR gates, even though it required negating the inputs.


0

I guess 249Ω resistors using for low pass filter. You can look 10.2.1.2.7 Input and Reference Low-Pass Filters section on ADS1248 datasheet.


1

simulate this circuit – Schematic created using CircuitLab That would be more correct. You may not bias all nodes with 680 Ohm, only at the both ends, where also the termination resistor is connected. If you want a 120 Ohm characteristics impedance termination, you should calculate the equivalent of all resistors, including the bias resistors. EDIT: ...


1

The basic difference between a MAX485 and MAX487 is the driver slew rate. There should be no difference to the number of nodes you can run with either of them. I've done hundreds of installations with MAX485/487 with around 16 nodes over 1km. You need to ensure the 0V is connected between all nodes. Also your biasing resistors of 680Ohm are way too low. With ...


3

Forget about the cable capacitance. Drive it source terminated through a \$100\Omega\$ resistor as I suggested in the other question. The cable looks like another resistor (about \$100\Omega\$) during the transient. At the far open-circuit end 10m down the cable, you will see a clean edge, just 50ns later. If the cable was 20m long, you would still get a ...


0

For long-distance and high frequency signal transmission, you can try the following two methods: Use shielded coaxial cable. Use subsystem, the control signal transmission by differential method, CAN, RS-485 eg. The method 2 is recommend.


2

Well, they are quite overpowered for the job but a mosfet push/pull driver would do the work. Your voltage would be at least 5V to make them work correctly. A pair of consideration: 10m of twisted pair for only 1pF seems a little low to me. Are you sure it isn't more like 1nF ? The AVR GPIO is rated for 20mA, 40mA is the absolute maximum Since you are ...


2

While the IC allows you to use differential input signals, it doesn't mean you have to. It's perfectly valid to just tie one side of a differential input to ground and drive the other side with a single-ended signal. Be careful with your current return paths, though, you don't want to have high power circuitry attached to the same ground as your sensitive ...


3

As these are cascadable, and the datasheet claims that they can drive 10m to the next one, why don't you simply put one pixel in your controller and turn it off. Use it to drive the 10m to the next one. If you want to make your own driver, then there are two things to watch out for: driving the capacitance of the cable to give you a reasonable risetime ...


2

You could have tested this idea out using a SPICE simulator. Just for fun, I ran this in LTspice (free and full featured SPICE simulator). R15 needs to be much lower in value due to capacitance from the FET and cable. Four runs were made with the cable capacitance at 1pF, 50pF, 500pF, and 5nF. At 500pF the signal is having a hard time getting above 2.7V and ...


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