7
The properties of aging on transparent substrate LEDs have been documented in research by many but are not handy to me the time of this writing. I have known this for about 10 yrs so Google or Microsoft Scholar ought to have some papers on this. I got mine from EOS by NDA.
Generally 10% of rated max current is needed to prevent some corrosion of the crystal ...
answered Jan 12 at 19:04
Tony Stewart Sunnyskyguy EE75
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5
This is a way of describing a positive or negative phase displacement.
(Image source)
A handy mnemonic is CIVIL
C IV Capacitors: Current leads voltage
VI L Inductors: Current lags voltage
5
Capacitors resist a change in voltage by consuming or sourcing current. So if you apply a voltage to a capacitor, you'll see that a lot of current flows in initially and then drops as the capacitor becomes charged to it's final voltage. Conversely since the voltage changes more slowly as the capacitor charges, the current will peak well before the voltage ...
5
What it means is that compared to a resistor where current and voltage are in-phase, when the load is a capacitor then the current peaks (and inverse peaks etc) come earlier than the voltage peaks (and inverse peaks etc)
It doesn't necessarily mean that the current in the capacitor somehow causes the voltage, just that the peakd and troughs come earlier.
...
5
You have a massive power MOSFET with a lot of capacitance. If you reduce the size of the MOSFET you can go faster, or change the type to a different technology than silicon.
A cheap one that should show a lot (like 5 or 10:1) improvement is the FDD1600N10ALZ.
They're cheap enough, but you can run some simulations to see what the predicted performance is. ...
4
If you're designing the actual silicon, you do this by making the two MOSFETs with different W/L ratios.
If you're designing with discretes, you can do it by including resistors between the FET source pins and ground, and adjusting the resistor values to give the desired current ratio.
4
Since the motor with the flywheel is being used as a generator
shouldn't the current be always negative?
No, it should always be positive (as your second circuit indicates).
So why does the current initially go negative? The unloaded motor has higher rpm due to not being loaded down with a flywheel, so it initially generates a higher voltage which ...
3
However, couldn't I use a single voltage amplifier which gives me the desired voltage to drive the load, why is there the need to have two separate amplifiers?
You can, but you might find that internally it uses two stages of amplification. If you read up on gain-bandwidth product you will find that this figure limits the maximum gain of an amplifier for a ...
3
Typical IR receiver needs a supply voltage between 2.5V ... 5.5V. Therefore you cannot use the 9V battery directly.
As for the circuit, you could try one of these circuits.
You can use any small-signal PNP you have in a TO-92 package.
And without any resistor, my IR receiver gives 15mA max of current through the LED at 5V supply. This is why I skip the ...
3
A static discharge is a “one way” circuit phenomenon and, is due to an unequal amount of surface charge on one body compared to the other. At the point of contact, the charge flow (aka current) concentrates for both parties and this means that the current density is higher and therefore the power dissipated in that small contact area is quite noticeable by ...
3
This sounds like a power problem. If you are using TTL chips, be aware that they are more power hungry than the equivalent CMOS chips.
It is not necessarily a battery problem. Breadboards are not known for being the best at power delivery, and there are certainly quality differences between breadboards. You're going to have power losses - voltage drops - ...
3
The way to obtain the curve is to realize that we can use the small signal model for the transistor. This is because the base-emitter voltage varies very slightly when the differential input is changed due to the emitter degeneration. To calculate how slightly is complicated because the tail voltage of the differential pair changes due to the constant ...
3
The specific equation for that curve is probably not so easy to mathematically reproduce because it's based on Gm which is a measured parameter of a real transistor. It is not a constant either. It varies with operating conditions so would actually be represented by some function which would probably be very large and messy since it's measured from the real ...
3
Any other improvement I could make?
There are resistors specifically designed to handle surges. In fact all will but, few have the information in their data sheet that allows you to make the right choice.
For instance, the Bourns CRS0805 range of surface mount resistors have a nominal power dissipation value of 0.25 watts but, the data sheet tells you ...
3
The voltage across an LED is about 2V and stays at just about 2V.
With a fixed power supply voltage it means that you have a fixed voltage across (and therefore a fixed current through) the series dropper resistor.
The resistor is acting as a constant current source and its current is shared between however many LEDs you have in parallel.
Lets say you ...
3
No.
When you have two devices in parallel their currents are independent. If i1 changes it has no effect on i2, as long as the power supply is able to keep the sum of both currents flowing.
2
If you compare the sine waves of voltage and current in an AC signal fed into a capacitor you will find that the waveform of the current is 90 degrees ahead of the wave form for the voltage.
Picture from wikipedia.
2
You can try something like this. R9 should be a 1W or 2W resistor. D6 prevents surge current from damaging the resistor. There are 4 op-amps in each LM324 so you would need two of them (or 3 LM358s). The 7805 creates a reference voltage which is divided down to the transition voltages for each LED. 20uA through the divider chain so steps of 40mV, ...
2
No, spinning motors are like capacitors.
When the power is turned on the current into the motor is high, but then the current drops as the motor speeds up, just like a capacitor.
When the supply is turned off, the motor acts as a generator and supplies current to the rest of the circuit, just like a capacitor.
The thing that causes the speed limit in ...
2
These switches have a much lower forward voltage than the 12v supply that I am powering the project with, ...
Switches don't have a forward voltage. The resistance of the switch will be very low when closed and the voltage drop across them will be very low.
... and so I need to find a way to step the voltage and current down to something the Arduino can ...
2
Try this:
Charge yourself up. Then hold a coin in your fingers. Have your "victim" do the same. Now zap the two coins together.
No shock!
Next, charge yourself up again, but use your coin to zap your victim (avoiding surface of eyeball, inside of nostril, use your imagination.)
You feel nothing. But they go "YOW!!"
See what's happening? You weren'...
2
Relation between current, resistance and voltage according to Ohm's law, Joules law of heat and P=IV
Answering about Ohms law:
Ohm's law has sometimes been stated as, "for a conductor in a given
state, the electromotive force is proportional to the current
produced." That is, that the resistance, the ratio of the applied
electromotive force (or voltage) to the current, "does not vary with
the current strength ." The qualifier "in a given state" ...
2
Whenever you have this type of question, usually the best approach is to compute the base-emitter voltage of the transistors. This is what you tried to do, judging from your comments.
The base-emitter voltage of the NPN transistor is \$V_\mathrm{BB}/2 - v_\mathrm{O}\$. The base-emitter voltage of the PNP transistor is \$-V_\mathrm{BB}/2 - v_\mathrm{O}\$. ...
2
Here's your schematic:
simulate this circuit – Schematic created using CircuitLab
(By the way, if you plan to use this site again you should learn to use the included schematic editor. It automatically numbers the parts for you and that provides us with an easier, more specific way to address ourselves to your schematic and questions. It's not ...
2
There are several parameters of interest;
Impedance of transmission lines and terminators
Signal currents
Short cct. currents to +12, -7V
Internal Switch Ron or RdsOn
Off switch capacitance.
The currents usually follow Ohm's Law if you have an accurate model for Ron and drive levels for PFET and NFET to short voltage for Hi and Lo.
The CANBUS is ...
answered Jan 10 at 18:01
Tony Stewart Sunnyskyguy EE75
89.1k2 gold badges33 silver badges132 bronze badges
2
why do we say "current drawn depends on the type of load connected
We actually say that the current drawn depends on the load AND the power supply voltage. In other words, what we should say matches ohms law.
Moreover does the above reasoning of using ohm's law hold valid for
any type of load? i.e. L load, C load, RL load or a combination of all 3)
...
2
If that's the case, why do we say "current drawn depends on the type of load connected"?
Usually we're dealing with a fixed / constant voltage supply. Most common examples would be domestic or industrial mains voltages or 12 V automotive systems. In both cases the voltage remains within certain tolerances so that means the other's have an inverse ...
2
Unlike other devices in ADS111x series, ADS1115 includes source MUX that selects which of the input channels will be used as inputs for differential amplifier.
Since you are measuring shunt voltage referenced to the ground, you have to select GND as one of those inputs. See "9.3.1 Multiplexer" diagram in the datasheet. This is done by setting bits 14:12 in ...
2
It seems to me like you could just use an emitter follower circuit. If you do this with the NPN transistor then the output voltage would just be Vout=Vd-0.6
If you need a current then just scale the output resistor accordingly.
If this answer isn't enough then you need to restate the problem in a more complete way.
2
There are a few problems with your layout , I can imagine from your resonance.
The 10:1 probe 10pF to 15pF causes Drain resonance with your wire jumper inductance from Vcc low ESR 5V Cap to Drain @ 10nH /cm. Solution add 50 OHm series near drain for a test pin
The inductance in the power to drain increases the resonant gain which can be reduced by using at ...
answered 2 days ago
Tony Stewart Sunnyskyguy EE75
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