We’re rewarding the question askers & reputations are being recalculated! Read more.
9

From the question below: A diode is put in parallel with a relay coil (with opposite polarity) to prevent damage to other components when the relay is turned off. This is called a flyback diode. See also How to choose a flyback diode for a relay?


6

The round symbol that contains several smaller circles is likely to be a DIN connector. There are several versions: - And, it looks like yours is the SV50. Is the circuit from a midi controller? Here's one that looks very similar in that the same pins appear to be used: -


5

Looks like a female TRS connector: (Image from here) So you would have left audio on the tip, right audio on the ring, and GBUF on the sleeve. (I'm assuming GBUF is the common ground for the 2 audio channels, but your schematic doesn't show it.)


4

The diode is known as freewheeling/flyback diode. Imagine that the relay is energised. A current will flow through it and will set up a magnetic field around it. When the relay is de-energised, this field will collapse as there is no more current to sustain that field. Michael Faraday found out that whenever a magnetic field associated with a coil changes, ...


3

Why are you showing the LM2596 "simple switcher" as a 3-terminal regulator? That's extremely misleading. In the real circuit, the regulator tries to keep its feedback pin at a constant voltage (1.235 V) with respect to ground. If you feed in more voltage from your DAC, then obviously, the regulator's output voltage must be reduced in order to meet that goal....


3

You need to add an Ethernet hub. They come in multiple flavors, with different channel counts, different interfaces, integrated PHYs, etc. You'll probably want one with 2 RMII interfaces and 1 PHY.


3

What does it mean to trim the reference? When the circuit is made there are all kinds of tolerances on all components. That means that the reference voltage will not be exactly what it is designed to be. You could use only "exact value" components but that could be costly (accurately measure each component) and or impossible (on a chip you do never have ...


3

The best way to solve this problem is to use a microcontroller that has interrupt/wake support on a group of its GPIO input pins. These would be programmed to generate the wakeup interrupt when any of the pins detects a state change. You connect this group of input pins to the MCU input side of the keypad matrix. Pullup resistors on these inputs keep all ...


3

Often, for board-level design one engineer will be responsible for an entire board. If it's complicated, or if it has a mix of critical digital and analog components, or if it's got an FPGA on it, more may get involved. But aside from that: Typically, yes, a designer will have topologies in their heads, waiting to be fit to a purpose. But we generally ...


2

How does 1 play out? I spend a portion of my time researching circuits. A good engineer is also a good researcher, they know how to find information. There are many sources, books, research papers, manufacturers app notes to look for circuits that have already been built. This gives one an idea of how to build circuits by understanding each component of ...


2

Basically you need to add up all the Vf (forward voltage) of each LED at your design current and subtract that from the supply voltage, then apply Ohm's law to figure out the total resistance required. You can then split that into as many resistors as you want, say if you want to use small SMT resistors and spread the heat dissipation out physically across ...


2

First, you don't need a resistor between all LEDs, one resistor is enough. To calculate it use the following rule: rule (voltage = current * resistance) V = I * R However, because of the forward voltage use (V - Vf) = I * R Thus (assuming 24 V), and 5 LEDs: (24 - 5 * 2.7) = 0.03 * R <=> R = 350 ohm If you don't have a 350 ohm resistor, use ...


2

Develop your requirements. This will be everything from the end functionality to how it's powered, to how every single component should function within the system. This gives you a set of parameters for your design as well as a rough idea of the needed modules. This is because you will be breaking down your idea into modules with which each can be designed ...


2

Designing something like this will take a while, and it's too much to really do through a site like this, but I can at least tell you how I would set about it: Break the task into chunks, and do each in turn, and Start with the important bits - in this case, it's going to be all about the interface. There's a whole lot of mechanical stuff to sort out too, ...


2

The simple answer is that you can't make a latch from just AND gates because it won't work. You can't do it with just OR gates either. Now, demonstrating that it doesn't work is a great homework question and I don't want to rob you of that learning experience. Make a truth table and figure out how the next output values will depend on the input values and ...


2

The best is to rewrite your transfer function in the proper low-entropy way, with a leading term. I have assumed the roots are all negative leading to left-half-plane pole and zero: \$G(s)=\frac{s+5447}{s+961.3}=G_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ where: \$G_0=\frac{5447}{961.3}=5.66\$ \$\omega_z=5447\$ rds/s or 867 Hz \$\omega_p=961.3\...


2

Have a look at the NL7SZ57/(SN)74LVC1G57, a "configurable multi-function gate": If you keep some input(s) constant, you get several possible logic gates from how the output reacts to the other inputs. Its datasheet says: The user can choose the logic functions AND, OR, NAND, NOR, XNOR, inverter and buffer. The NLS457 has four inputs. However, there are ...


2

With the addition of a simple filter, you have a classic CCM (continuous conduction mode) buck converter: simulate this circuit – Schematic created using CircuitLab So yes, the overall idea is quite viable, and there's tons of literature about this kind of converter all over the Web. Without the filter, it doesn't work — you never actually ...


2

Better approach: simulate this circuit – Schematic created using CircuitLab The control is opposite from yours — when the low, the gain is 1, and when high, the gain is 10 (note the modified resistor values). In this configuration, the MOSFET is perfectly happy to have both positive and negative voltages on its drain, as long as the signal ...


2

It's a pretty terrible way. The BJT has some offset and the base current will affect the output. A MOSFET would work better, but only for positive outputs, and the MOSFET has to have a low enough Vgs(on). I would suggest using an analog switch, but the details depend on the power supply rails and input logic level. In particular you should avoid running ...


2

You could try something like this: simulate this circuit – Schematic created using CircuitLab You'd need to operate the switch to allow current to flow to your load device. When the switch is open, no current will flow to the load, when the switch is closed, the PMOS will allow current flow to the load. Notice the high value for the pull-up resistor (...


2

Regarding positive/negative feedback, the answer is not so simple. Remember that each "negative" feedback will turn into positive feedback for rising frequencies due to unavoidable phase shift of the amplifier. More than that, positive feedback allows stable operation as long as the (positive) loop gain is below unity. Therefore, based on the classical ...


1

I'm not sure I understood corectly what you want, so if this is not the case, let me know so I can delete the answer. When you make the waveform window active, you can select in the menu Plot Settings > Select Steps and you'll get table-like entries for your steps.


1

First of all, never feel stupid because you don't know what a component is used for. Most components are actually pretty fairly specific. Looking through the speeds in that datasheet: it's unlikely this is modern technology. Feels like 1980's "low-voltage high speed logic", like 74LVCxxx. It's quite possible a high-volume customer came to ON Semi (or ...


1

It’s not clear to me how you are reading the thermocouples, but I suspect the cause of your disaster was a short between the thermocouple and the “hot” mains line causing current to flow through all the non-isolated analog bits back to the (grounded) computer motherboard. Insulation breakdown was caused by the heater/thermocouple assembly melting. That ...


1

The control topology of your chosen device is a strange one, and often difficult to analyze. It seems as the device operates far outside of the continuous conduction mode (CCM), and starts, what the data sheet calls, "Burst Mode operation". I guess it is some kind of pulse frequency modulation with constant on and off time until the output reaches some kind ...


1

IC2B forms an inverting integrator hence, when its output is summed with the input signal (at the junction with C10), DC offset is removed but, because the integrator will also only partially attenuate low frequencies, low frequency AC signals are also attenuated. As the input frequency gets higher, there is less attenuation. This is high pass filter action....


1

You appear to be using P-channel MOSFETs to do low-side switching which is an odd choice due to the more complex gate driving arrangement that would be required. On top of this you won't be able to turn off the load. You'd be better off with N-channel MOSFETs for switching loads to ground, the gate driving is very simple if you use logic level MOSFETs and ...


1

The standard formula for charging a capacitor via a resistor is this: - $$V_C = V_S(1-e^{\frac{-t}{RC}})$$ Where \$V_C\$ is the voltage on the capacitor, and \$V_S\$ is the supply voltage (5 volts in your circuit example). So, rearranging you get: - $$t = -RC\cdot \ln(1 - \frac{V_C}{V_S})$$ So the delay time is how long it takes the capacitor voltage ...


1

It is going to end up more complex than the circuit using two LM317- so the question might be more why you would use this circuit. The LM317 has a lot of (minimum) voltage drop (3V minimum recommended) and two of them in series have twice that, so a lot of power will be wasted in the regulators. A sense resistor of 0.1 ohm will drop only 100mV at 1A. You ...


Only top voted, non community-wiki answers of a minimum length are eligible