New answers tagged

2

Connect them all. The reason why multiple pins of the same potential are brought out of the die is for two reasons. 1) this reduces the resistance. When you do that you reduce the voltage bounce that occurs from current pulses that are formed upon switching. So you get a cleaner voltage source on the internal power rail. 2) it reduces the inductance of ...


1

Generally, a device with multiple supply/ground pins intends for those pins to be externally connected to the appropriate supply/ground. It is particularly important to connect the ground pins together using wide traces or polygons/planes to avoid ground loops. The datasheet, section 1.1, for the 328p indicates three specific power connection types: VCC, ...


4

Connect all of them and more importantly apply very localized decoupling capacitors on them all.


1

A simple and direct answer to your main question, is that this is one way of providing the differential voltages that the op-amps need (+ & - 5v). By floating the ground (to +5v) the single 10v source can supply + & - 5v! You should now be able to understand that the output of op-amp I, is being used. It creates a virtual ground (or reference ...


2

(TL,DR: see paragraph 5) The meter module requires the voltage on its GND pin to be between its V+ and V- supply pins. It converts and displays the voltage between its IN+ pin and GND. The 7810 regulates the input to 10V between nodes labelled +5V and -5V. R2 and R3 provide a mid-point voltage with Thevenin impedance of 2.5 kOhm (= 10K // 10K) in ...


26

While I agree with @pipe and in fact upvoted his answer, an additional nuanced reply is that a ground is more than "just" a reference. What I mean by this is a ground is not just a voltage, but something that can source & sink current and stay at the same potential. The ground created by that op-amp can both source and sink current and remain roughly ...


34

The first OP-amp is actually creating the circuit ground. The 7810 creates a stable 10 volt, which is then divided by the voltage divider R2 and R3, filtered by C3 to make a stable 5 volt level relative to the most negative level. The OP-amp then buffers this, and the rest of the circuit uses its output as the reference ground. Remember that ground in a ...


7

A1 and A2 are potentiometers. (This diagram uses the European notation.) This diagram could have been drawn in a general-purpose (not specialized for electrical engineering) diagramming tool such as MS Visio. It was not drawn in Altium, or Eagle, or OrCAD. update: why would there be potentiometers in such a circuit ??? :( Potentiometers are there ...


4

You can absolutely do this. Here's the process: Select all of the net labels you want to change Go to the bottom right of your screen and click the "SCH" tab Open SCH Inspector At the bottom of the window that appears, it will say how many objects are selected. Make sure this is correct. Within the SCH Inspector, you can change any attribute you want, and ...


0

The device measures the current flowing between the screw terminals, which are isolated from everything else. the pins are to power the hall effect sensor only, however, there's no reason you couldn't have power going through the screw terminals and then going to anything you want, those screw terminals are isolated from the pins so measuring itself is fine. ...


2

Some Altium parts (mostly logic chips) have the supply and ground automatically connected to a supply net (you can spot these parts by their lack of visible power pins), I've never really understood the advantages, try putting a net label or something on a capacitor with the same names as the hidden nets and connect them to your existing power and ground ...


2

What kind of this electrical schematic? It's a line diagram. It seems to show a mix of individual wires (such as the 24 V DC) and cables such as those going to be box on the bottom left. I have to re-draw this schematic by Office Visio ... Not a great idea. If you only have one drawing it may be OK. If you have a lot of work get an electrical CAD ...


1

Your circuit has a few problems, the first being that since Q1 is PNP, making S1 won't turn it ON, it'll just charge up C1 which will keep Q1 OFF, forever, since C1 has no discharge path. The same problem exists for Q2 in that a positive rectified and smoothed PWM signal never turn it ON either. Then there's the catch diode, which should be across the ...


1

Go to symbol page Right click on pin Select properties In 'Visible' dropdown select 'pin' Click OK and save


1

The resistors on the cable side are supposed to terminate each pair being treated as one "wire". The purpose is to reduce EMI and they do not functionally affect the actual signals. The impedance is not so well defined. With 20 years of experience, the magnetics nowadays with built-in common mode chokes are effective such that those terminations do not make ...


0

A couple of points in addition to the ones posted above. The first answer is fairly heroic but there's one thing I don't agree with. Pin order in schematic symbol. Why to reorder pins It makes aesthetically more pleasing schematic that may be easier to interpret depending on how the pins are laid out. Why not to reorder pins It's asking for trouble, ...


1

This is tricky, but one common solution in EDA packages is net ties. These are essentially "dummy" components that act to do what you want. They tie the nets together to allow you to name them differently, and then you create some kind of PCB component to represent the copper connection. That is also one of the reasons I don't like doing net ties -- creating ...


1

Without seeing a schematic, this is difficult to answer. I would suggest the first thing to check is that the port is ACTUALLY connected. This warning is in place for people foolish enough to use a grid of 1 unit haphazardly, as you run the risk of ending the track just shy of what you are planning to connect to. This results in a schematic that looks like ...


0

You actually have separate GND net and chassis net. Ground being the inverse pyramid as you probably know. Usually how this works is that chassis is the metal enclosure as the name implies. This may or may not be connected directly to the external ground from PSU and/or mains PE. The ground on the other hand is usually is your zero-volt point for whatever ...


1

Based on the picture in the question you have a PCB stripboard such as following So I would layout the components such that all the pin connected to ground use one strip. Also Circuit 1) and Circuit 2) are both the same with respect to GND. I suggest that you look at fritzing. They have tools that will help you layout the board on CAD before you start ...


1

To understand circuits you need to know that there are no absolute voltages. All voltages are properly called potential differences, i.e. the difference in potential between two points. Ground is a conventional name for the point from which all other voltages are measured. (Confusingly, it's also a name for a point on the circuit which is connected to the ...


1

Op-Amps come in two forms. Single Rail/Supply, and Dual Rail/Supply. Your op-amp, the NE5532 accepts +15V to -15V, and can be used in a single or dual supply mode: Single Rail are typically Positive Voltage to Ground reference, while Dual Rail op-amps typically have Positive to Negative. Single Rail are more common in newer, DC based designs, while ...


0

Vcc- is only for the Op-Amp and it is consider to be the negative supply voltage. Op-Amps usually have dual power supply like +/-5V, +/-10V, so their output can vary between these voltages. The reason behind this is that the Op-Amps should operate in both polarities of the input signal, given a 2Vpp sine wave. Without a negative voltage reference the ...


0

The minus sign inside a circle is a dedicated negative voltage for the op-amps, usually -3v to -18v, often labeled (Vee). Normally it is the opposite of the positive supply pin (Vcc).The signal ground (usually 3 horizontal bars) is the true ground for the circuitry. It may tie to earth ground at a screw somewhere on the board. If the board had a mix of ...


0

In Design Entry HDL, go to: Tools ==> Options ==> Grid, Set the grids to "Show...", "Dots" and multiple set to "1". Also make sure you have View ==> Grid, checked. Grid may not be visible when you are zoomed out because it would be to fine and cover the whole drawing. You need to zoom in and it will turn visible.


8

In short Parts inside component A Component may have more than one parts, for ex: LM324 has 4 op-amps inside. With the part option, you select which one to use. Altium will create netlist for the pins with the part you have selected. Part locked Now, when you annotate the schematic, Altium not only changes the component names with ? (for ex U?), but ...


-1

Expanding on Viral - Embedded hardware's answer. The locked function also makes the component non-selectable which means a single click would not select the component. This feature comes in handy when you have overlapping components.


-1

Locked function when enabled, it means components value,name,annotate cannot be modified. This function is given to prevent the accidentally changing of the any of the components in schematic.


1

Typically for Xbox related power supplies, you need to tie PS_ON to the 5V standby rail. You could do directly, but I would use a 1kΩ resistor just in case. R-Sense normally indicates a Remote Sensing option for a power regulator. It would normally be connected at the far end of the power supply's connector, to compensate for any voltage droop from the ...


0

Noise on the analog reference means noise on the adc converted signal. So, for a good adc conversion (ENOB = effective number of bits), the noise on the analog reference is crucial. I am pretty sure the bead is to reduce noise. It is a common method.


0

There is a picture of the finished vero board in the link. The 2 x 1,5kV capacitors can be clearly seen - 2 small blue devices. Obviously not 1000uF.


5

A bit of Googling finds this page, which has the following correction: Capacitors C3 and C4 Note that there was a typo for the values of capacitors C3 and C4. The correct values for C3 and C4 are .01uf at 1.5KV DC or higher, ceramic disc.


2

M is possibly an abbreviation of micro-micro farad. A micro-micro farad would be a pico farad. See here for a discussion of micro-micro farads. This answer is possibly the correct explanation of the marked values for C3 and C4, but is probably irrelevant, as RJR has found the correct values for the schematic in a later article that mentions that the ...


0

It is definitely a ridiculous value. A 1000µF capacitor, rated for 1500v, would be very large, very heavy, and very dangerous. A 1000MF of the same rating? As big as a house, or more. As the others have hinted to, the small size of the component dictates that it must be much, much smaller in capacitance. 1000pF sounds reasonable. It could be a translation, ...


1

As per comments, I'm posting an alternative solution that may better suit your needs. If you make the base of the piece and the square in the chessboard a closed system, you ~might~ be able to identify parts by how much light they reflect back. For this I believe you need 4 things: 1 Chessboard and piece to be a closed system (outside light may mess with ...


3

For now you have two buses with same name, that is how Altium sees it, and it makes sense. ADC[0..2] on both sheets. You can not make Output for bus containing 3 signals, from two different sheets. Your A2 sheet contains only one signal and A3 two signals, so solution is to make Bus entries at top level and connect those signals with outputs from sheets A2 ...



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