New answers tagged schematics
I don't understand what is happening on this diagram. Referring to the diagram in OP. This MOSFET-TRANSISTOR duo is used to enable and disable the power supply of the module, when MCU_CTRL is high Q101 is enable, which in turn enables the Q102 by pulling the GATE of the PMOS Q102 to GND, and enable the power for module on VBAT pin. When MCU_CTRL is ...
Most modems need up to 2A peak current. I have not checked your specific one but most LDOs will struggle to supply 2A bursts. That's why most designs use DC-DC supplies.
The real world application varies upon the application and the type of the signal. If linearity is not a concern, then the above circuit diagram can be improved to remove the 47(ohm) resistor and the circuit will act as a digital circuit, with almost no sound when there is no light. On the other hand if linearity is a concern for analog applications, then ...
That is a really crappy circuit since it depends on the gain of the transistor being just right. Real transistor gains vary widely, even within the same production lot. Competently designed circuits work from the minimum guaranteed gain of the transistor to at least 10 time that, preferably to infinite gain. I would start with the transistor stage setting ...
This is known as an emitter follower. There is no voltage gain in the circuit, this acts as a impedance buffer i.e. it can drive heavier loads that the CDS cell could on it's own. Light on the CDS cell generates current which goes into the base of the transistor, this current is multiplied by the transistor (this gain is known as Hfe). This current when ...
It is a simple non-inverting transistor amplifier. The resistor to ground is what allows it to be non-inverting. When the transistor is off (the CDS cell not conducting), the Headphone output at the emitter of the Transistor is pulled to ground level through the 47Ω resistor. When the transistor is on, the headphone output gets pulled up to 3V, with the ...
After searching much longer than I expected to search for this, I have come to the conclusion that this does not exist. You could easily reverse engineer it by using the reference design schematics, specifically the Sample Application Circuit diagram and the table of description of external components diagram. You may need to infer some names to put along ...
The crystal capacitors are called "load" capacitors, and are required as part of the oscillator circuit. The value of them depends on the specifications of the crystal being used. The 100nF capacitors on the Vcc and AVcc pins are called "decoupling" capacitors, and one is required on every power pin. They keep the power stable - without them the chip will ...
This will minimize "pops" on plug-in, and will keep the amp quiet when nothing's plugged into it:
Usually the tip is the right channel, the ring is the left channel, and the sleeve is ground. Therefore on your diagram, 1 is the (-) input, 2 is the (+) if you want to use the left channel, 5 is the (+) if you want to use the right channel. Pins 3 and 4 are connected to 2 and 5 when there is nothing connected and disconnected when something is plugged in, ...
If you want both the left and right channels to provide audio you need to make a little mixer. Two resistors (around 10KΩ) connected to the left and right inputs (pins 2 and 5), then joined together, will make a simple mixer. Then you link that to the LM386's IN + through a capacitor (around 10µF is usually OK). Pin 1 links direct to the IN - You ignore ...
1.This is being fed a 48V AC source Right on. 2.These leads are a source and a ground? Not quite. There is no specific source/ground attribution. Unless the source they get connected to is shown as grounded, they can be assumed to be floating. 3.I'm not sure what that is That is some sort of input spike limiter, such as a transzorb. 4.Two alternating ...
Yes. No, they're just wires going to the source. It's a MOV (Metal Oxide Varistor) to absorb surges (through-hole types look like ceramic disk capacitors). Markings on the symbol for the AC input of a bridge rectifier. A bridge rectifier (four diodes). More markings on the symbol for the DC output of the bridge rectifier.
It's a bridge rectifier - it consists of 4 diodes and always does its best to route the most positive part of the AC power input to the + output and likewise on the negative side. I wouldn't assume the power in is "source" and "ground" - it's more likely to be a floating (groundless) output from a transformer secondary.
Without the pull-down resistor the net would be floating and you would measure random states (H or L) on it. Even if you would always see a L on the pin during development that might suddenly show up as a H due to external circumstances (i.e. noise) and cause you big headaches.
You need something to pull the voltage down to GND when the jumper is removed. A pull down resistor, as in your second example, is the usual way. Some microcontrollers have a built-in pull-down resistor, but it consumes more power because it requires a FET to switch it in. Your second example with the resistor is probably the best option, and the only one ...
Yes, it is possible in Altium Designer. You can design your own custom parts and build your circuit with them. All of the circuit elements (even the IC and capcitors) in the following picture are designed by me. You can learn about creating custom component libraries in Altium Designer from here.
LTspice has the option of changing colors and fonts to an extent. LTspice IV
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