What's the voltage drop across a forward biased silicon diode? My friend says .7, I say .6...?
82587
You guys are both right--good answers.  Junction temperature also can affect the voltage drop.
679488
simple use two resistors, 6 and 3k ohm , in series and take the volatage across the 3 ohm resistor, it will be 3V.....\n\nuse the laws,\n\nif you have a voltage source, and two resistors p and q in series with it, \n\nthe voltage across p is given by  Vx(p/(p+q)), here u know voltage across p(Vp)= 3V, or 4.7 V or whatever....\n\nso, Vp = V x p/(p+q)....\n\nu know the value Vp, and V so substitute them......\n\n3 = 9 x p/(p+q)\n\nso p + q = 3p, \n\nso "2p = q", take two resistors ( in kilo ohms, coz u cant use just ohms, coz they are very large) such that one is double the value of other.......\n\nand connect the circuit across the 'p' resistor......\n\nwith 4.7, u get,\n\n4.7 = 9 x p/(p+q).....solve ...and choose resistors accordingly
334912
The answer is a bit too long to go on this site.\n\nIf you would like to correspond by private message or by email I can help you.\n\n\nThere's one soulution here http://www.davidbridgen.com/5vdet.htm
1033484
Hi,\n\nHere is the diagram and description.\nhttp://www.kodkodkod.com/files/pin_diode.htm\n\nPlease let us know if you need more help!\nhttp://www.kodkodkod.com/index_files/submitquestion.htm\n\nKodKodKod Learning Consulting
948174
Of course. \nThe actual junction of an N type and P type is what prevents conduction when reverse biased.
185952
a diode blocks current from flowing in a certain direction. current will flow from the non striped side to the striped side
783633
YOU ARE ASKING A QUESTION THAT IS ANSWERED BY SOLVING THE EQUATION (E=IR) OR OHMS LAW. IF YOU GIVE ME TWO OF THE  UNKNOWNS THE EQUATION CAN BE SOLVED FOR A NUMERIC ANSWER.   LETS ASSUME THE VOLTAGE(E) IS 100VOLTS.  THEN THE CURRENT(IR1) WOULD BE (100VOLTS/110 OHMS) E1/R1 OR 0.909 AMPS.\nTHE SITE BELOW GIVE A LENGTHY DESCRIPTION OF THE SIMPLE CONCEPT AND JUST SEARCHING "OHMS LAW" WILL YIELD DOZENS MORE. YOU WILL FIND THAT THE VOLTAGE APPLIED TO EACH RESISTOR IS THE SAME AND THE AMPERES THROUGH EACH RESISTOR IS INVERSELY PROPORTIONAL TO THE RESISTANCE.
225576
TTL: 1 - 10k Ohms works, CMOS: 10 - 20 KOhms
700902
Ok, at the risk of being blunt, if you need to ask this, you might be better of leaving things well alone - or doing as the other person suggested and using a transormer to lower the voltage to begin with.\n\nHowever.........\n\nFirst of all, you need to plonk a diode in line really, LED's do not like AC so the diode will make things into "raggy DC", using a bridge rectifier would be better and would reduce the flickering, as would plonking a capacitor over the DC output - typically with AT LEAST a 380V rating if using a bridge and a 200V rating if using a single diode.\n\nStill with me?\n\nYou now need to work out the voltage drop over the LED's - this varies according to type - and mainly colour. About 1.8V for a typical RED LED and around 4v for a white one.\n\nYou then need to look at how many volts you have "left" after you have taken into account all of the LED's in series.\n\nEG:  You take 220V, use a single diode (110 v) and then pop a cap over to stop the flicker (around 160V)\n\nYou now have 30 red, 20 green and 10 blue LED's in series. Assume 1.8V drop on Red, 2.2V on Green and 4.1V on blue.\n54v  + 44v + 22v = 120v\nThis gives us 40v to "burn off" with a resistor.\n\nLet us assume you want to drive the LED's at 20mA\n\nR = V/I   therefore  R = 40/.02 = 2000 Ohms\n\nPower = I x I x R  (current sqaured 8 resistance) so the resistor will be dissipating 800mW - Allowing for plenty of play and wanting a warm rather than hot resistor, then 2w ideally.\n\n\nSo a 2K2 resistor - or 1K8 for a bit brighter would be your answer here (for the number of LEDS and type), rated at 2W or better.\n\nAs I said at the beginning, if ONE step of this doesn't make GOOD sense to you, PLEASE leave WELL alone.\n\n\n\nPLEASE DO NOT use this as a way to hook up the leds exactly as shown WITHOUT first checking their "forward voltage losses", I have used typical examples, different models can vary considerably.
942029
Use the sine wave as 2.8 peak to peak volts to a bridge rectifier. No clamping is possible without dc power source and capacitance.
254527
1.85 ohms. The first answer is correct
302538
Does the post code on the LED actually say "87" or does it say "8.7" with the decimal lit up as well?\n\nIf it's "8.7", then the system is hanging while it's checking the core voltage, not telling you to check it. POST codes tell you the instruction the system is attempting to carry out when it fails.\n\nIf it's "87", then it's hanging as it tries to assign IRQ's to PCI devices.\n\nRegardless, I'd start by ripping out every component other than the basics (CPU, Memory, Video card), as well as re-installing those components to make sure they're all seated properly. See if you can get it to POST at that point. If so, proceed to add individual components one at a time, attempting to POST each time you add another component. I'd also check to make sure your front panel connectors are installed correctly to the header - PWR, HDD Activity, Reset SW, etc.\n\nAs a side note, I've heard about some motherboards having issues specifically related to the main connector of the Power Supply. Is your power supply's main connector 24-pin, 20+4-pin, or 20-pin with a 20-to-24 pin converter? If it's the latter, that could be the problem. Some 24-pin mobos just won't work right with a 20-to-24 pin converter. It could also be a voltage issue with the power supply itself.
871859
Nominal voltage required for excitation of a circuit.\n\n- - - - - - -s-
888922
try this http://www.siliconfareast.com/grr.htm
800374
This newpaper article states what I have been saying for years (and my mother before me RE: transistor radios).\n\nhttp://www.hearinglossweb.com/discus/messages/11/807.html?SundayMarch1920060829am
291589
Use the node voltage method. Label each node with a voltage, then apply Kirchoff's current law at as many nodes as you need to generate enough equations to solve for any unknowns you have introduced.\n\nLet the node at the bottom of the circuit be ground, and then try to figure out the voltages at the other nodes relative to ground. Clearly the voltage at the node on the top will be 14V (I'll just call this V1) but no other voltages are apparent. So call the voltage at the node between R1, R3, and R4 Vx and the voltage at the node between R2, R3, and R5 Vy.\n\nNow that we have labeled all the node voltages, you can figure out every single current in the circuit in terms of V1, Vx, and Vy (using Ohms Law, the voltage drop across a resistor is equal to the resistance times the current in the direction of the voltage drop). For example, we know the current going down through R1 is (V1-Vx)/R1, the current going right through R3 is (Vx-Vy)/R3, etc. Now we just have to solve for the unknowns Vx and Vy.\n\nKirchoff's current law (KCL) says that the sum of the currents leaving a node is equal to zero. If we apply this law to the node with voltage Vx and the node with voltage Vy, we'll get a pair of equations that we can use to solve for Vx and Vy.\n\nApplying KCL at the node with voltage Vx should give you the following equation:\n\n(Vx-V1)/R1 + (Vx-Vy)/R3 + Vx/R4 = 0\n\nThe first term is the current going up through R1 (we already figured out the current going down through R1, but we want currents that are leaving node Vx so we had to flip the sign), the second term is the current going right through R3, and the final term is the current going down through R4.\n\nI'll leave it to you to figure out the equation for node Vy. Once you have that you'll have two equations and two unknowns and you can solve for everything. Once you have Vx and Vy, you can easily figure out a numerical value for any current in the circuit.\n\nBTW, the node-voltage method is pretty powerful and can solve most linear circuits. Check out the link below, it was written by an old EE prof of mine who is very thorough.
90104
If the transistor is stabilized in some circuitry, heat should not affect it too much. A simple amplifier using voltage divider biasing is very stable, whereas simply feeding a signal into the base with the B-E junction forward biased will be very unstable. In the first situation, the stability factor can be made to equal 5. In the second example the stability factor will approximate the beta of the transistor. (Lower S => more stable)