I'm struggling to understand the exact circumstance where this would
What most folk want with a common emitter amplifier is for the DC collector voltage to be about half the supply voltage. So, when you have an AC signal present at the input, the amplifier's output (collector) doesn't clip the signal asymmetrically. Another way of phrasing it is ...
BOM optimization is a likely candidate. The value 2k is probably used elsewhere and adding another value would mean one more line on the BOM and one more slot in the pick-and-place machine.
Another BOM optimization could be availability. For example two 0603 resistors are easier to source than one 0805 (for power handling). You mention an eight resistor in ...
Each red LED drops about 1.5V. So three of them takes 4.5V, leaving only 0.5V for the resistors.
0.5 / (47+47+47) = 0.0035A or 3.5mA. That's rather low.
If you were to remove two of the 47 ohm resistors, you would get
0.5 / 47 = 0.011A or 11mA. That's brighter, but still less than most LEDs can handle.
I am guessing that you are looking at an SMT resistor marked "470" and that it is actually 47 ohms.
As to operation-
The TL431 draws as much current as required (up to 100mA+) to reduce the voltage at the sense input to 2.495V nominally. As long as it gets > 1mA to use itself it will function properly (400uA typically). If we assume the circuit ...
The emitter resistor serves mutiple purposes. It improves the linearity of the amplifier, raises the input impedance, and simplifies biasing. Let us look at each characteristic individually.
In the following I will refer to variation of the circuit with \$R_E\$ omitted as the 'grounded common-emitter amplifier'.
The gain of the grounded common-...
If either side of the LED is shorted to either supply it will probably survive with two resistors. With only one, it probably won't.
If the circuit has to be resistant to "knuckleheads" perhaps the 0.1 cent for the second resistor was deemed justified.
Edit: The comment from mkeith is also very much valid. If you keep the number of different parts ...
LEDs are nonlinear devices -- their current vs. voltage curve is exponential with voltage. In practical terms, it means that essentially no current flows at low voltages, and then the current very rapidly rises.
In the case of your LED, that "low voltage" is around 1.6V. If you look at the datasheet, figure 2, you'll see this.
In fact, that LED ...
Your battery, LED and voltmeter are all in parallel.
simulate this circuit – Schematic created using CircuitLab
Figure 1. You're not measuring the voltage "after" (or "before") the LED. You are measuring the voltage across the LED-resistor combination.
In this circuit it is the same as measuring the supply voltage.
To measure the ...
Look at the emitter circuit.
simulate this circuit – Schematic created using CircuitLab
As soon the base-emitter diode reaches its forward voltage, the bias runs away. You cannot completely solve this problem as you want to have current running into the base. But you don't want to have a diode characteristic at that place. You want it linear. That's ...
RE is needed to cope with the power supply voltage fluctuation, and it keeps the operating point stable.
For small signals, it will reduce the gain so a capacitor bypasses it for small signals. In other modes you have the same considerations.
If you look at the static load line you will see that a variation of VCE does not result in a large variation of IC, ...
Most likely, it's just the artist's idea of an "electronic component".
It may be justified by the site where you found the image classifying tin as a conflict mineral.
Tin is used widely for plating the terminals of many types of components, including resistors.
Tin is used for plating the resistor contact terminals.
Tin plating on the leads or pads of components makes them easier to solder.
Even though the substrate of the resistor is ceramic, the ends of the resistor have a metal coating.
In some special applications, the resistive material itself can also be made of tin or gold alloy.
As common solder types ...
The Congo, Nigeria, Uganda, South Sudan, and many other countries are a pit-full of rich resources and pitiful corruption breeds that profit from these. They rank pretty high on the corruption list but China has invaded many of these areas with massive loans for infrastructure that result in takeover of the assets created after failure to repay.
If you connect them in series, their forward voltages (Vf) add up.
Red LEDs have a Vf of about 2.0V at rated current. Three of them in series would yield Vf of 6V - more than your supply. So use two in series at most.
Also, you need only 1 dropping resistor. Knowing that 5mm (T-1-3/4) LEDs are usually rated for 20mA, we can set the current as follows:
R = (...
Typically, modern red 5mm LEDs drop about 1.7V-2.1V at 20mA, depending on the chemistry and size of the die.
That means that more than 2 of them in series will mean it's difficult to predict the current.
If you use individual resistors and connect them in parallel across the 5V, you'd want (for 15mA) about 220 ohms each and the current (using my Vf values) ...
Most of the answers forgot to state the most important thing:
You don't talk about the correct voltage supply for LEDs, you talk about supply current. That's the reason for having the LEDs in series instead of parallel!
For manufacturing reasons LEDs have a Vf (junction drop voltage) extremely variable. Also being them diodes have an exponential current ...
This won't work. You're asking a source of power (battery) that's designed to maybe supply 60 mW to supply 1 W.
End of story – you need a different power source, no circuit is able to create power (physics wouldn't allow that).
The answer to this question is just to look in the datasheet. You'll see that this cell is hopelessly wrong for what you want to do.
The Panasonic CR2032 states 225 mAh capacity and typical max. load current of 200 uA.
So it can't remotely supply the load current. Even if it could, it would be flat in 40 mins.
You must search for parts (battery) by their ...
Your battery is very, very dead. It has an internal resistance of around 500 ohms.
A fresh Panasonic alkaline battery (consumer grade, not industrial) has an internal resistance of less than 4\$\Omega\$.
Connect the black (COM) lead of your voltmeter to ground. Now, connect the red (+V) lead to each of the test points, TP7301 through TP7303, in turn. If the voltage you read is equal to BG_SUPPLY then the LED is open, if the voltage you read is zero then the LED is shorted. Since each test point corresponds to a single LED, it is clear which of the LEDs has ...
If I said that the voltage was \$10\cos(100t)\$ and the current was \$5\sin(100t)\$, would you be asking about the phase angle? I mean; the phase angle (when not mentioned) is implicitly assumed to be 0° so there wouldn't be any worry there if it wasn't mentioned. Are you OK with that?
So, if both voltage and current are shifted by 60° (or any phase angle), ...
There is 2A flowing counter-clockwise in your circuit, due to the constant current source. Starting at the circle there is a 4V voltage drop through the 2 \$\Omega\$ resistor. The more positive side is at the circle. Continuing counter-clockwise, the battery also has 4V across it. Again, the more positive side is the top, and the more negative on the bottom ...
How can I calculate an exact resistor value to my LEDs to get equal brightness?
You can only calculate this if you have bought LEDs that have been 'binned' (selected, sorted) to particular brightness bands.
Randomly bought LEDs can cover such a wide range of brightness that all you can do is adjust the resistors until they look matched.
The trick is not in matching resistors but in matching current, since even the same model of LED has differents Vf among parts and the resistors are computed over that.
This, of course, if you have the same model of LED. If these are different (like different colours) you need to look up the curves for the light emission. If you are lucky your leds come from ...
I would say to simplify that after long time the green current I is 0, so it is the voltage drop on the vertical resistor.
Knowing that in the second loop on the right also the current is 0 because we have at least 1 capacitor is series, the also the voltage drop across the horizontal resistor is 0.
The sum of voltages along the right loop is ...
The beauty of R3 is it lowers the output impedance by negative feedback. When you choose Ic around 1mA, Vbe becomes around 600 mV so the rise in the precision Zener @ rated current @ 5600 mV +/-x% matches Vbe to result in 5.00 output while biasing the Zener at the same time. As Iz >> Ib the circuit is almost independent of hFE.
You Icc calculation ...
9V in 10 ohms needs a current of 9V/10 ohms= 900mA.
9V in 100 ohms needs a current of 90mA.
A little 9V Name Brand alkaline battery cannot produce 900ma but can produce 90mA if it is brand new but for only a few minutes.
Your battery has no name brand so it probably is garbage.
Is there any reason why this circuit would need two 2k resistors instead of one 4k resistor?
I thought maybe it would be for power dissipation but unless I am making a stupid mistake, its only dissipating 1 milliwatt which even small SMD resistors could handle.
Subtract the voltage drop of two LEDs from 24V. To be conservative let's say 20V. 20V/4k\$\...
The resistors are split for better common-mode transient immunity (CMTI), i.e. for better separation of the two supply domains. The two sides are galvanically isolated, otherwise there would be no reason to use optodiodes (=isolators).
If there is any noise (more like disturbances caused by motors, or high current fast circuitry) on one of the supplies, then ...