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Flow of electric charge - typically movement of charge carriers, such as electrons. Measured in amperes (A).

3
votes
3answers
In this answer poster calculates a safety factor of 10 for the minimum base current in a transistor used as a switch. Seems like a lot to me, and for higher collector currents this may mean that a microcontroller won't be able to drive it anymore. Is a factor 2-3 not enough? …
asked Apr 28 '12 by Federico Russo
5
votes
1answer
I found this nomogram, which also appeared in an answer here I think. (larger version here) Is there a metric version of this?
asked Aug 16 '12 by Federico Russo
12
votes
4answers
Due to the base current, in a common emitter circuit the emitter current is a bit higher than the collector current: \$ I_e = \dfrac{1+\beta}{\beta} I_c \$ I was wondering how this is for a … phototransistor, like in an optocoupler. I would expect emitter and collector current to be equal because there is no base current. But a photodiode can create a current with zero volt across it, so it seems the photons can create electrons. So is \$I_e = I_c\$ for phototransistors? …
asked Feb 8 '13 by Federico Russo
4
votes
3answers
You can change a current mirror into a multiplier/divider by adding transistors on either side. Then the number of transistors defines the current ratio. The transistors have to be integrated on the …
asked May 10 '12 by Federico Russo
7
votes
6answers
As I understand it inrush current is the current when the contact closes. Resistance is not minimum yet, and still inrush current can be several times the nominal current, like 80A on a 10A relay … . How come that the inrush current doesn't weld the contacts? edit case in point: this relay can take 800A (!) inrush for 200\$\mu\$s …
asked Jul 9 '11 by Federico Russo
18
votes
2answers
The FDC855N comes in a 6-pin package, 4 of which are connected to the drain, and only 1 to the source. Why this difference? The source sees the same current as the drain, doesn't it? …
asked May 24 '12 by Federico Russo
5
votes
input current to your ADC is actually a resistor parallel to the 20 k\$\Omega\$. \$ \dfrac{115\mathrm{\,V} - V_{OUT}}{746.667 \, k \Omega} = \dfrac{V_{OUT}}{20 \, k \Omega} + 150\, \mu A \$ Or … have a low input impedance. However, checking the datasheet, Telaclavo is right, the 150\$\mu\$A refers to the reference input (page 3). On page 4 you can see the input leakage current for the analog …
answered Apr 20 '12 by Federico Russo