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0

I have seen Y used to label crystals and oscillators. This is how OrCAD labeled their library parts.


2

Just drag the output label over. It doesn't need to be touching the net at the label origin. Note: To remove those little origin markers, use the set Option.ShowTextOrigins 0; command. Also note that this does not disconnect the label from the net, they are still associated (renaming the net will change the label). This simply moves the origin to not be ...


0

You can solve this by placing in series an 'ideal diode' device such as ISL6146A. ISL6146A By using a FET as described you prevent voltage, down to 1V per the device datasheet, from reverse biasing anything before the external FET. However because of the body diode voltage would always pass through. If this is undesirable, a second FET in series with the ...


6

As Ignacio pointed out there is a parasitic body diode inside the regulator from the output to input. If VBUS has a very low source impedance then the C15 will get charged quickly through U5 and there is not much limiting the current. Or, anything connected to the VINLDO input could draw current from VBUS through U5 (for example, shorting to ground could ...


5

It is safe to apply a voltage to the output of a LDO regulator provided that it is not higher than the input voltage. Many LDO regulators have a parasitic diode from the output to the input that may become forward biased if the input is at a lower voltage than the output, thereby allowing a potentially destructive amount of current to flow.


1

This is a little old but could be helpful to other users. Create a project and then create your circuit. To reuse the circuit start a new project and you can import the old project in multiple times. Solves pick and place problem and without scripting.


1

This is an easy question if you have a solid knowledge in basics of circuits So These are steps that I will use to solve your problem, First find the equivalent resistance between L and M.And before that you want to ground a particular point of a circuit(this is a must because we want to find the voltage, simply we use that grounded point as the ...


0

R1 and R2 are NOT in series. Here is how you can do it: R1, R2 and R3 are in parallel. Calculate the parallel resistance of those three resistors so you can see them as one and call it R123. Now R123 and R4 are in series. They form a voltage divider. Now you should be able to calculate the volatge accross R123 (which is equal to the volatge across R1).


0

Ok, turns out I have to start the connection from the breadbord to the component - then Fritzing highlights the breadboard holes and treats them as a conductor. If I start from the component it tends to create a "wire" between components and disregard the breadboard matrix. Weird but can get used to it ;)


3

Schematics will be in the support documents for the particular development board. Every one I've seen has a full schematic (typically in an Appendix).


0

From the Help: Quartus II Help v15.0 > Creating Designs > Using the Block Editor > Viewing Blocks and Symbols On the Tools menu, click Options. In the Category list, select Block/Symbol Editor. In the Guideline spacing box, type or select the default guideline spacing for all new files. To snap to the grid, turn on Snap to grid. Click OK. It worked OK ...


2

It's a straightforward phase shift oscillator. Variable capacitors in 470,000 pF aren't widely available, so my suggestion would be to replace R422,R423 wit a 2-gang linear potentiometer, maybe 47k. And be prepared for the oscillation to stop at either end of the pot's range. Once the feasibility is established, restrict the range of operation by adding R ...


4

The BJT is a phase shift oscillator and uses the three caps marked C411 to C413 to generate 180 degrees phase shift to create sinewave oscillation. You could experiment with these to alter the frequency. This isn't the actual part of the circuit that modulates the audio but, the output should feed the modulator (the real heart of the vibrato circuit). If ...



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