# Tag Info

0

There are a few possibilities come to my mind: Latched OLP (over load protection): Unlike the hiccup mode, the PSU shuts itself down on an overload or short circuit. And it requires a power cycle after removing the cause of the extreme situation. You may want to check the load side against any ahort circuit. The microcontroller (or microprocessor or ...

0

If this op amp is powered by a single voltage supply, connecting both inputs to a supply rail, as shown in your figure, is not a best practice. For reference, see Texas Instruments' TechNote SBOA204A How to Properly Configure Unused Operational Amplifiers which states at the top of page 1: Connecting the input pins together or to the supply rails can lead ...

0

It is as you suspect. An unused op-amp with inputs improperly connected. From How to Properly Configure Unused Operational Amplifiers Degradation Conditions Connecting the input pins together or to the supply rails can lead to device damage. If configured like this, the input stage can suffer permanent electrical overstress (EOS) damage, as shown ...

0

The point of this circuit is exactly as you've deduced; to stop the op amp's output from being unpredictable. The LM358 chip has multiple op amps on it, so if you don't need all of them it's good practice to tie all inputs of the unused amplifiers to some fixed voltage, so they don't just pick up noise and amplify it.

0

The LM358D is a dual operational amplifier, with two op-amp circuits per chip. If an odd number of op amps is desired, then one unit might be unused; in that case it should not left floating since floating inputs can pick up noise from the surroundings, causing undesired power dissipation or inducing issues with the other unit on the chip. Depending on the ...

0

The above circuit is used to measure the offset voltage in practical op-amps. Ideally, the op-amp is a perfect differential amplifier so the output of the above circuit should be zero. But in practice due to some non-linearities, the output of the op-amp is not zero when the input voltages are the same. This is called input offset voltage for the op-amp.

0

I'll somehow extend Cristobol's answer. #1. More hardware, less software If you can spend 21 I/Os (assuming you want to simulate key presses on all the keys) you can use the CD4066 IC. It is an analog bilateral switch and each package includes four of these. You can connect every key's two contacts to one of those bilateral switches and activate the switch ...

0

Connect an arduino GPIO directly to pin 6 of the 555, when configured as an input it will do nothing (allowing manual control via the button), but when configured as output (low or high) it will turn the 555 on or off. (then set it back to an input to allow the switch to work) if you need toggle action connect a different GPIO to pin 3 and write the ...

0

Yes, you could possibly measure the Voltage and apply it to the corresponding Button, but: It's easier than you think: The black circles on the inner side of the Buttons are made out of conductive material, wich close the circuit on the PCB. You can exploit this by using some sort of "external Button" to activate a Button. You can see four conducting ...

1

Those inputs are probably matrixed, since I don't see any obvious ground traces. Matrixing means that you drive one side of some switches low, and read the other side to see which one is pressed, then drive some others low and read them with the same input lines. Matrixing allows this particular keypad to be read with 9 I/Os instead of 20, but it means ...

2

When using a transistor as a switch, the "switching action" depends on the voltage between the base and emitter (or gate and source for a FET). But in your circuit the emitter voltage is a varying analog signal. This makes producing the correct base drive difficult. Enter the analog multiplexer, an IC designed specifically to switch a varying analog ...

1

The datasheet won't address what the transmission line should look like. It assumes you use a properly matched transmission line. As long as the impedance is correct along the entire length, it doesn't matter if the traces are straight, rounded, curved or zig-zagged. You may find it difficult to make a transmission line in a particular shape and still ...

0

It is convenient to break a loop where Rout=0 or Rin=infinity, as the analysis will become easier. So, as the output of the 'ideal' opamp has zero impedance, so it is easier to do the analysis by breaking the loop at output of opamp

3

Google is your first choice. Searching for FDS 6690A brought up this datasheet on the first page. The "F" would indicate that it was originally a Fairchild product, which ON semi acquired some time ago. FDS6690A Single N-Channel, Logic-Level, PowerTrench MOSFET

0

Your 7408-based solution sounds fine: just connect one input of all the AND gates to the common enable pin. Other than that, there's a lot of 74xx series octal buffers with three-state output and an enable pin. the Cheapest I found on digikey is the 74LCX541FT, which has active-low enable pins, but that's not really a problem: pull them up with a resistor ...

0

The crosstalk, even in a 50 ohm system, may be too high for a SPDT in one IC. I suggest you implement "T" switches: two SPST in series, with the center_node either "shorted to Ground" or floating.

1

It seems that you are asking about the power used by a fetal doppler device such as a baby heart monitor. It says in the instruction book that the power output is 20 mW/ square centimeter ... That tells you that the mechanical or sonic energy output by the device will be 20 mW/cm2. Therefore the total output power will be area × 20 mW where the ...

3

Your system is unstable because you have a pole in the right half plane, and the result is that the output grows unbounded. Any non-ideal op-amp has a finite DC gain $\ A_0$ and a finite bandwith (crossover frequency $\ \omega_c$ below infinity) such a non ideal op-amp can be approximated as a single pole low pass filter, with the following transfer ...

0

The "passive" in passive sign convention means that you are assigning positive voltage drops to passive components (i.e. loads). Negative voltage drops then occur across sources. Why you ask? Why not positive voltage drops across sources and negative voltage drops across loads (i.e. active sign convention which you will probably never see because no one ...

0

Create a resistive ladder with a zener shunt regulator. You can short selected resistors(BJT, MOSFET or analog switches) to change the effective voltage at the IADJ pin If you have a DAC in the MCU you can use it to change the voltage at the IADJ pin 2nd PWM signal to create a filtered analog Voltage using RC circuit and or an opamp as a buffer.

3

I think the problem is simply that the pulse is too short to charge the capacitor fully. I created a simulation using CD4000 series inverters at a time scale long enough that propagation delays are insignificant, and with an external resistor to ensure symmetrical output resistance. It showed the same effect as your circuit. With this short pulse length ...

3

You can actually buy a quad 2-input XOR gate IC, and use 2 of them to implement this. Part No: 74HCT86 Only about \$0.50 each on digikey.

0

Before detailing my answer, I want to stress the fact that we are dealing with large signal circuits, therefore it is no more possible to model the MOSFET as a simple voltage controlled current generator with the capacity of sourcing and sinking any amount of current: in this context, its behavior is more similar to a switch that sinks current from a ...

0

Maybe you're asking about Photoemission electron microscopy but this isn't generally used for manufacturing. It is used for (failure) analysis. How a semiconductor junction can produce photons is explained in the Wikipedia page.

0

A 74HC123 / 74HCT123 datasheet here draws 8 uA max at Vcc=5.5V at 25C. That's for the IC itself - current in the timing resistor will add to that. A lower current option is to build an oscillator from a single Schmitt trigger gate. Supply current is typically 0.02 uA and timing components add perhaps ~= 0.5 uA - less if care is taken. At 5V a CD40106 (...

1

Because the 123 has both positive and negative trigger inputs, it is possible to feed the output back to the input and create an oscillator. However, it is not as simple as it might seem because the timing capacitor needs time to discharge between cycles. The result is an oscillator circuit that is less accurate/precise/stable than a 555 circuit. Your ...

0

I've used those emissions in diagnosing high-voltage breakdown. Gotta turn off the lights, or use lid that closes.

0

If you're talking PCIe, then any PCIe NIC chip should work. Not sure if you can get quad port chips, or if you would need two dual port chips. But it should be doable. Switch chips are also an option, but bear in mind that a switch chip does not send all traffic through to the SoC, the SoC is simply another port on the switch. You can set up VLANs and the ...

Top 50 recent answers are included