This is related to , I want to find the values for the two resistors so that the led will get 100mA.    Vcc will be 5V and I have either 2N2222 or 2N3904 transistors. Let's say I use the 2N3904.  They mention the equation: hFE(VCC-VBE)/RB. (Is the full equation hFE(VCC-VBE)/RB = ICC?)  I tried to plug in the values: Vcc = 5, Hfe = 30 (from the sheet), VBE = 40*10^-3R since the arduino's pwm output is 40mA and from ohm's law, ICC = 100mA (the desired current for the led).  so we get: 30*(5-40*10^-3R)/R = 100*10^-3 => R = 3.3/(100*10^-3/30 + 80*10^-3) = 39.6.  This is not even close to 1000 like in the answer, so I'm doing something wrong here.   So how are you supposed to find the B line resistor value? Also, how do you find the C line resistor value?  This is the led I will use btw:   "Arduino" is Arduino UNO for now (Wemos D1 mini later).  I don't think it's a duplicate since my circuit is a little different and what I ask for is different. I also have no idea what is that ...
50914
I have 3 12VDC/40A automotive relays () that I want to use with my Arduino. Based on the tutorial I am following () I need transistor, resistor, and diode. I am not an electrical engineer therefore I am unsure about the parts and calculations I made. For the start, relay coil resistance is 90+-10% Ohm per datasheet. So I proceed by calculating the current flow.  Voltage=Resistance*Current Current=Voltage/Resistance Current=12V/90 Ohm Current = 133mA  For the transistor I can get 2N3904 or 2N4401. At this point I have to calculate resistance for the base of transistor. In tutorial its as following  hfe = Ic / Ib Ib = Ic / hfe Ib = 0.03 A / 75 Ib = 0.0004 A =&gt; 0.4 mA R1 = U / Ib R1 = 5V / 0.0004 A R1 = 12500 Ohm  2N3904 datasheet states the H(fe) is 30-300 when lc = 100mA (mine is 130mA) and Vce = 1V. At this point I have no clue what is going on, thus I need help. Edit: Here is what I ended up with. RLY1 in the picture is 12VDC/40A ()
3603
This is my circuit. I need to find the power dissipated by Vlamp. So the lamp turns on if the Vlamp is more than 2V. When the lamp is on it has an internal resistance of 10 Ohms, and when it is off it acts like an open circuit.  How would I find the power dissipated? I know how to do it for Voltage sources and Resistors, but not elements like these.  One more thing, I already figured out that: Battery has open circuit voltage of 4V R1 = 5 Ohms R2 = 5 Ohms k = 0.2A/V When lamp on and Vlamp = 2V, Rbat = 10 Ohms   .
73319
For the circuit shown below what is the time it takes for Vc to rise from 0 V to 5 V in (second). Assume Vc(0+) = 0 V    The answer is \$ t = 2\ln{2}\$, but I am unsure why. What I have so far:  $$ V(t) = [V(\infty) + V(0)]e^{-t/T} \qquad \text{where $\;T = RC$}$$ R = 2e3 and C = 1e-3 , so T = 2  $$ V(t) = [V(\infty) + V(0)]e^{-t/2}$$  at t= infinite, the capacitor acts like a broken circuit so V(inf) = 0 but the problem also states that V(0) = 0 so that would set whole thing to 0   Am I doing something wrong, or am I even on the right track here?
33560
I need to work out what resistor I need for a series circuit containing 3 LEDs of 2 different types, and a 9v battery.  I can do the maths, but I need to check I'm doing it right as I'm trying to learn this from what I can google!  LED1 VF = 3.2V @ IF = 20mA LED2 VF = 2.2V @ IF = 20mA  Both LEDs have a Forward Current of 20mA at these Forward Voltages.  The circuit will contain 2x LED1 and 1x LED2.  So, using R = (VS - VF) / IF, do I calculate the total VF as 3.2V + 3.2V + 2.2V, therefore VF = 8.6V?  Using the figures I have, I get:  (9 - 8.6) / (20 / 1000) = 20 ohm, so I'd need a 20Ω resistor?  Also, can anyone advise what type of resistor would be best?
74403
If I had a 12V battery and wanted to use an LED that used 2V and 20mA, then I need a resistor that will drop 10V and provide enough resistance to produce a current of 20mA. Therefore I would use R=V/I = R = 10/0.020 = 500 Ohms.  Therefore I would need a 500 Ohm resistor. I believe this is correct so far?  My confusion comes from when we then look at the circuit as a whole. The circuit has a current of 0.020A and a resistance of 500 Ohms, which when looking at V=IR equals 10V (the resistor), not the 12V that it should be there for the whole circuit?  Where am I going wrong? How do I make V = IR work regarding the whole circuit so that V = 12?  I'm pretty new to this sort of thing and no doubt I've probably made a daft mistake but if someone could just help clear up my confusion it would be appreciated.  Thanks.
69024
I'm quite new to electronics and I'm having a hard time understanding the "pull-up resistor" principle. I have read a lot of articles about that, and I think I've got it but I'm not 100% sure so I have a question. In , after the first image, it says:     When the momentary button is pressed it connects the I/O pin to Vcc and the microcontroller would register the input as a high.   But I don't get it. Where is VCC? From what I see, there is no power source on this schema, just a microcontroller wired to a button that are both wired to ground so how can there be any voltage at all in this circuit?
55201
I have trouble with finding out resistance of a resistor. It has 5 bands but somehow I cant get the code right.
113583
The image of the circuit is shown below and it required to find \$V_0\$,  My first attempt at solving this problem is by changing the current source into a voltage source with 1-V and 2k\$\Omega\$ resistance. The fact that the inverting and non-inverting terminals aren't grounded make this problem look difficult.To the point which, I don't know how to proceed with this question or where to start. I would appreciate any help. Following some thought and another schematic from a hint suggested by Alfred, I produced a schematic representing our work.  And  my solution for the problem is below, Using node equation at nodes A and B we have, $$\frac{V_A-V_B}{1k}=-0.5 \text{mA}$$ $$\frac{V_B-(2+V_A)}{1k}=-0.5-x$$ where x is the current that is sent in the output of the op-amp. Using KCL, at the bottom node near the current source we see that the same current that goes through the op-amp also goes through the \$2\text{k}\Omega\$ resistor. Hence, we have, $$\frac{-V_A}{2k}=x$$ Replacing this in the second equation, $$\frac{V_B-(2+V_A)}{1k}=-0.5+\frac{V_A}{2k}$$ And and solving the equations yields \$V_A=-2 \text{-V}\$ and \$V_B=-1.5 \text{-V}\$
84471
I need to implement the following circuit in LTSpice but I don't know if my schematic is ok. Also, I don't know what  values for I should use for Cinf and R2.  The following constraints are imposed:   VCC= 10V VCM = 1.5V Vgis a sine wave with a frequency of 1 kHz. Its amplitude will be chosen so as to ensure minimal distortion of the output signal (i.e. v0 should also be a sine wave) R1= 15kΩ.    I need to choose a value for R2 and Cinf in order to have a (theoretical)gain value Av=1500.   How can I compute the theoretical expressions of the DC collector currents of bipolar transistors Q1–Q8, considering VBE≈ 0.6V and the theoretical expression of the small-signal voltage gain Av= vo/vg?   The first picture is the schematic and the second is the LTSpice implementation.  What I found is that:   ICq5=ICq6=ICq7=ICq8=(Vcc - Vbe)/R2 ICq1=ICq2 = [ICq7 / 2 ] * [Beta/(Beta+1)] ICq4=iCq2-current of Q5 ICq3=ICq1-current of Q3-current of Q4  Beta in my case is 182.1 because I'm using a BC847A transistor. A=Vo/Vg = Gm_q2 * Beta_q5 * R1Gm_q2=ICq2 / Vt that is approx 40*ICq2   How can I find the value of R2? This is what I have now, an sine like wave for Vg
82109
I am trying to drive  with an  microcontroller.  Both devices are listed to work with I2C clock frequencies up to 400 kHz. The two devices are the only devices on the I2C bus.  However, working out the calculations for the pull-up resistor bounds gives some rather odd values.  Calculating the I2C minimum pullup resistor value:  \begin{equation} R_{min} = \frac{Vcc - 0.4}{3mA} = 966.7 \Omega \end{equation}  Looking at the uC datasheets, on page 92 lists the max pin input capacitance of 10pF.  However for the LCD, it has on page 8 something called the Capacitive load represent by each bus line labeled as Cb, and is listed at a max value of 400pF. I'm assuming I should just add this value to the 10pF uC input capacitance, but this seems really high and the calculations are wonky.  For example, when I try computing the maximum pullup resistor value for a 400kHz clock:  \begin{equation} R_{max} = \frac{300ns}{10pF + 400pF} = 731.7 \Omega \end{equation}  Am I misinterpreting the LCD datasheet? Obviously the max allowable pullup resistor value cannot be smaller than the min allowable value.  likewise, if I assume a maximum net bus capacitance of 400pF, I get:  \begin{equation} R_{max} = \frac{300ns}{400pF} = 750 \Omega \end{equation}  still under the maximum allowable value.
118768
I know how to calculate resistance in parallel and series connected resistors, and how to reduce most circuits to these, but I missed my lecture on cases where you can't just reduce the circuit to series and parallel and now I'd like to catch up...         &ndash; Schematic created using   How to calculate total resistance between A and B in such case? R5 rains over my parade because I don't know a'priori which direction current will travel through it, and I don't know how to transform this into something I'd know - I could try to calculate potentials at points p1 and p2, to get the current that flows through R5, except that current modifies these potentials...  How do I solve a circuit like this?
112857
I learned in high school the basic rules of how to compute resistance between two points, given a network of resistors.  You try to rearrange the circuit so into parallel parts and series parts, then compute the resistance of each piece individually using those rules, until you're left with just one number, and you're done.  Maybe I explained that poorly, so consider the following schematic:         &ndash; Schematic created using   Sorry it's so large, that seems to be the default.  Anyway, it's easy to compute the resistance from A to D (regardless of the values of the all the resistors)- A-B-D, A-D, and A-C-D are all series connections (easy to compute) then you combine them with the parallel formula and you're set.  But!  How do you compute the resistance from B to C?  There seem like lots of possible ways to think about it, and they all generate different answers.  I've left the specific numbers off the diagram because this isn't homework and I don't care about the answer specifically, I just want to understand the principles involved and how to solve the problem.  Any help, or links to especially readable guides, would be very much appreciated.
86676
I am a software developer trying to understand how computers work at a lower level.  I have managed to create a few simple circuits, one which lights up an LED.  I have noticed that all the instructions advise you to connect the components in a certain way.  Please have a look at the circuit .  The LED is connected to column E and column F.  Why can't you just connect the LED to G7 and H7 and the black cable to F7? Why does the LED plug in across the gap?  I seem to remember something about this from school.  I have checked my GCSE textbook and also two textbooks which I am working through.  I have also Googled it but I cannot find an answer and hence the reason for this very basic question.
61037
After always being interested in learning more about electronics, I figured LED-strips would be a great start.  I'd like to power a 4 meter WS2812B-LED-strip (60 LEDs/m). These run on 5V and draw about 60mA per LED. That's 14.5A, or 75W. However, I've yet to find any power supplies for these figures, only imported industrial ones - which feels a bit scary with my limited knowledge.  I've understand that powering it through the Arduino is a bad idea, but what's the simplest and cheapest solution for powering it? I do have a bunch of old supplies for laptops and whatnot lying around.
75689
I am attaching a motor to an ATmega32 chip. I want to build an H-bridge but before doing that, I want to control the motor using the chip. I bought few NPN Transistors from RadioShack. I have 5V running though the breadboard. I attached Vcc to the Collector, Base to the Pin, and Emitter to the motor. And other wire of motor is hooked up to Ground. Unfortunately, the power coming out of the transistor is too low, the motor is barely moving (As opposed to when I hook up my motor straight to 5V). What is wrong? Do I need a different transistor or is my setup wrong? If it is the transistor, can you please recommend the right one, I am not too good at those.
122783
I'm making a voltmeter for the college senior project, and the problem is that voltage divider should equally split 17V into two 9V(nominal value of the batteries. As of now, an actual voltage is 8.5V and eventually will go down as the batteries die, so don't freak out too much by my round errors.) sources, but as soon as I connect Arduino board to the circuit, voltage( across the resistor it's connected to) drops to 2-3 Volts. Which is apparently enough to power the micro controller, but not enough to light the display up. (it's also enough to use the other resistor as a space heater with 14 Volts across it).    Please feel free to look at the schematic of the device provided above. (either models, or specs of the components are written on the schematic)  Now I'll explain the design thoroughly: The idea is to use piezoelectric element (thin laminated  flexible piezoelectric beam), place it into a fluid flow (air) and rectify the outgoing AC signal with a full wave bridge. The next step is to amplify the signal, because it has very low magnitude and arduino analog pin can't register it without amplification.   Amplifier needs at least +-7V supply, so it's provided to it from the two 9V batteries, that are connected to a voltage divider. Therefore, the virtual ground of the circuit is in between the two batteries.   At the same time I want the arduino board (Actually it's not arduino, it's arduino compatible board dubbed Pro Micro 5V) to be powered from the same two batteries.  The whole thing works fine from USB, but from the batteries it works as it pleases!  Here are the design constraints of the device:   The entire device shall hang somewhere in a remote location, so it has to be powered from an autonomous power supply (two 9V batteries in my solution). The device should be as light, and as small, as possible. (solutions like adding another battery are not desirable). Arduino board have to have 5V on it to light the display.   Additionally I'd like to learn why micro controller has variable input resistance and what it depends on.  I've tried to figure out what is the resistance across the arduino's ground and RAW by assuming that the resistor of the voltage divider and the arduino are connected in parralel. In three different setups I've got the arduino's resistance to be equal to 221.93, 527.73 and 743.4 Ohm.  As I understand, the board has some kind of reducer at the voltage input that prevents the board from burning at supplied voltages above 5V and below 12V, but why it drops the supplied voltage from 8V to 2-3V?
33467
I have a simple circuit:    The max rating (current) of the LED is 30mA.  How can I work out what the resistance of the resistor needs to be? By using ohm's law I found it to be \$3V/0.03A = 100 \Omega\$. However using software called , and trial and error I got the minimum possible resistance to be 36&Omega;. However, if I use a 35&Omega; resistor, then the LED breaks. Is the software wrong, or (more likely) am I doing something wrong?