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Most likely there is an ESD protection diode connected between the input pin and the VDD net on the chip, in such a way that it is normally reverse biased (A schematic showing the configuration is given in Peter Smith's answer). The idea is that when there is a positive ESD event, current will flow into the lower-impedance VDD net where it will do less ...
18
This is due to the input protection diodes.
A typical input looks like this (CMOS inverter shown):
simulate this circuit – Schematic created using CircuitLab
The diodes in newer parts are schottky devices. These diodes are for short, low energy transient events and cannot handle much current (a few mA generally).
11
The 0.3V drop comes from the Schottky clamping diodes used to protect the pins of the chip. These diodes typically connect between each pin and the two power rails. If they are forward biased by more than 0.3V, arbitrarily large currents can flow.
The diodes are designed to absorb transient currents produced by ESD, which represent limited amounts of energy ...
9
To be honest, you're asking 'How do I understand a massive subject quickly so I can design a project' and the answer is 'you can't', as you probably gathered. Your best path is to contact an experienced electronics engineer who can work with you on what you're trying to do.
As a start, search/look for demonstration boards that may be similar to your ...
6
Actually, the Schottky clamping diodes and the VDD + 0.3V are both present for the same root cause and that is SCR Latch-up. The design of all CMOS ICs actually creates a pair of BJT transistors intrinsically. It simply results from out the p-type and n-type silicon substrates are laid out. This picture from VLSI Universe shows it well:
https://1.bp....
5
Most likely one company came up with the design first.
Then other companies reverse-engineered the design to sell copycat parts.
Some customers might even insist that a second source of supply be available for such parts (linear regulators, jelly bean transistors, basic logic gates, etc) so it wouldn't help the original vendor to try to stop the copy-cat ...
5
There are two things:
The signal level
What the signal means, ie assertion
The signal level is either digital Low or High
The signal meaning is attached to either Low or High, so we say the signal is asserted low or the signal is asserted high. Usually a bar or a slash indicates a low signal assertion level.
In the case above the reset is asserted low, ...
4
what makes a "standard" transistor like one you may find in an electronics store in a bag
You mean discrete semiconductors like transistors and diodes.
Most modern discrete semiconductors are indeed made in the same way, or at least a very similar way, as integrated circuits are made.
Both are made on silicon wafers using photolithography. Most structures ...
4
Summary: It's an incremental process, not a "big bang", to go from college theory work to a substantial project. There is no substitute for experience.
After reading the whole question, I don't think your real question is the title of "Where and how do I search for ICs?". Instead I think your real question is more like - how can you choose components for a ...
3
It means the signal is inverted (like a NOT gate). Let's take this 555 timer below as an example
Picture can be found here... Not my picture (and excuse the massive compression for this picture, hence the ugly pixels)
Say that a signal that goes to this pin is a 1 or HIGH. Since Pin 4 is active low, it will end up being a 0 or LOW for this pin. The ...
3
There's no reason to drive the GPIO pins high at all. Just configure them as open-drain — drive the pin low when you want to activate the corresponding "button", and tristate it otherwise.
As long as the fob voltage is not higher than the MCU voltage, it will be fine.
And running the fob at 3.3 V should be fine, too.
2
Well, obviously, a triple 3-input NAND gate (74LS10), with various unneeded inputs tied high, will meet the stated requirements.
2
You could make a custom charger BUT I'd strongly recommend against it.
The energy difference between 4.2 & 4.4V will be small and you will get useful whole of life capacity improvement by using 4.2V end of charge.
You can test this by
CAREFULLY manually charging to 4.4V with a power supply set to 4.4V and current limited to Imax,
then letting the ...
2
But what about the bond wires? Is it possible that they desoldered/detached?
The bond wires generally aren't soldered to the chip or to the package. They're pressure welded.
If the bond wires are embedded in the plastic package material (I believe they will be for most QFN package types), and you heated the plastic beyond its glass transition temperature, ...
1
you can melt the microwires attaching the wafer to the package
Sure but if that is happening, you're using the chip in the wrong way.
The bondingwires shouldn't get that hot. Solve it by using more in parallel (to more package pins) and/or use thicker bondwires.
When this happens, can the melted microwires cause shorts between the pins of the chip
No, ...
1
A darlington pair with its base connected to the handle or a MOSFET with its gate connected to the handle (and 15MΩ to Vdd) will do the trick.
If you want the MOSFET variant in a more modern and more reliable fashion, use a capacitive touch switch IC as e.g. the IQS228 or a pure software solution on a µC. This method has the advantage it also works through ...
1
You should do some research into touch sensors and touch sensor circuits. There are even some microcontroller manufacturers that are offering touch sensing application notes and some even include the touch sensor circuitry right on the MCU.
You want to concentrate on capacitive touch sensor technology as it does not depend upon a direct current path ...
1
Low level "IC designers", that is those who are not designing MCUs etc are those using things like FPGAs or designing ASICs (Application Specific IC). FPGAs with millions of gates are now available and even amateurs can afford to use them. So if you want to create your own data flow computer, go for it.
1
With some caveats, it is actually possible to test if the bond wire is connected if you remove it from the PCB. First of all take a known good IC. Then use a volt meter in diode mode. Connect the black probe to IC GND and the red wire to one of the pins you want to test. On the known good IC note the forward voltage shown in diode mode. You may also want to ...
1
It is very normal to add some series resistance with your decoupling to damp the response. You should simulate this WITH your package inductance and determine how much series resistance you need. The exact amount of peaking you can tolerate will depend very much on your circuit. Key parameters to look out for are the affect of decoupling network peaking on ...
1
I can't tell for certain, but it looks like you have IN-HI connected to the +5V line (red wire) which matches up to the schematic, and the COMM pin (32) is going to the ground side of the supply. The COMM on these chips floats below the positive supply, and tying it down to the negative side will put the amplifier out of its common mode range. What that will ...
1
The schematic from the datasheet shows those two pins simply connected together. And this is a plastic housing. So it doesn't much matter.
Usually when the part has a metal case, one or more of the unused pins is connected to the enclosure of the part, so in that case connecting them to ground will provide some shielding and prevent some radiation as @...
1
Active LOW means that a 0 V level is considered to be a logic 1.
For instance, consider a logic input tied high using a pullup resistor and pulled to ground through a pushbutton switch.
Whenever the switch is not pressed, the input is at the pullup voltage, 5 V for example.
When the switch is pressed, the input is pulled to ground.
That input can be ...
1
I believe I found the problem - one of the example circuits has a pullup resistor between OUT and V+ when working with a transistor. I added a 4.09k and now it seems to be working.
EDIT: The output was still not quite right, so I added a second 4.09k pulldown resistor between the PWM line and ground. This completely fixed it, giving a perfect square wave.
1
Some DRAM can be affected in that way, because storing a charge on a capacitor is a linear operation, and the effects of perturbations accumulate.
But the ferroelectric material used in FRAM is bistable, and requires a certain minimum energy to change state. If any individual perturbation isn't enough to change the state, then the effects of multiple ...
1
It won't work at all in this case. The output of the opamp goes high, and the diodes turn on when the output goes high. The diodes are in an "or" configuration allowing either of the comparators to pull the 4049 high (towards 5V).
If the diodes were reversed, very little current would flow and the circuit would not function as intended and the input of the ...
1
First off, that is a very specialized field with probably not that many jobs globally. It's almost as specialized as, say, writing video chip graphics drivers.
However, there are related fields which are very active at present, most notably in 2D physics eg graphene, molybdenum disulfide, topological conductors semiconductors insulators etc with an eye to ...
1
Si forms an insulating layer of SiO2 by thermal oxidation. This layers is a native oxide and the thickness is controlled by diffusion. I.e. self-limiting oxidation...
The thickness thermally oxidized layers follows the Deal-Grove model. Deal and Grove developed this model while working at Fairchild semiconductor. Their work in thermal oxidation of Si was ...
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