I  want to activate the 6v relay using the output from an and gate whose output is around 5 v.how can i do it?
822904
I use an NPN transistor to turn my relays on/off. See pic...\nhttp://img47.imageshack.us/img47/9813/untitled1ut4.png\nR1=(Voutmax)/(Ioutmax)\n\nThe other option is to use open-collector logic. They use a pulup resistor to what ever voltage you want. This gets sketchy because logic cant handel much current: not to mention the current needed for the relay. See pic...\nhttp://img47.imageshack.us/img47/3778/untitled2vf8.png\nI would use the transistor methoid. Its less trouble then the open-collector methoid. If you need more current through the relay: just decrease the bias resistance.\n\nEDIT: Left_Field is right about the diod. Slipped my mind.
975955
generally pc work on 18 volts dc not 12 . they actually us e less than 12 volts . but for prop sake and etc . if you can tell the kind of pc you us e i can give you a digram of what to do . dc to dc is not a solution for your problem
48905
There's a great article explaining thyristor operation and showing basic circuit diagrams here: http://www.ibiblio.org/obp/electricCircuits/Semi/SEMI_7.html\n\nThere was also a very good reference book called "SCR Manual Including Triacs and Other Thyristors" published by GE from the early 1960's into the late 1970's.  It's a highly prized source for many engineers, and you -may- be able to find one for sale, if you're lucky. (No, you can't have mine :-P )\n\nDo a Yahoo! search for "thyristor circuit diagram" and you'll get thousands of results. :)
518938
Your question is very vague.\n\nGive more description as what you want. or contacta  hardware engineer.
736596
It depends on the current.  You want to drop 6 V across the resistor to get the voltage to the load to 6 V.  The voltage drop across the resistor is V = R*I.  The resistor value you need is R = V/I.  Once you know the resistance, you need to determine its power rating.  The power dissipated in the resistor is P = V*I, where V is the voltage across it and I is the current through it.  Or, you can use P = (I^2)*R.  You should choose a power rating at least 1.5 times the power you've calculated so it doesn't run too hot.
527732
no, it is not.\n5 volts and 500ma are the max. that doesn't mean you can't hack it up though (i've been looking how too. lots of documenation.) might want to give this a shot\nhttp://www.usb.org/developers/docs/hs_usb_pdg_r1_0.pdf
623671
There's both an explanation as well as a diagram\nof a 2 input AND gate built just from transistors at:\n\nhttp://tams-www.informatik.uni-hamburg.de/applets/cmos/cmosdemo.html
434630
dont forget the formula I= R/U\nsort them and u will get the answer\n\n Current: 625mA (milli amperes)\nResistance: 19.2 ohms
808764
Besides buffering the output to provide a higher drive level.  Another factor is that in reset the 8051 outputs are at a high level.  So your circuit needs to be designed to have the relays on or off as desired in reset or startup.
755501
you need to regulate 12 v down to 5v.\n\nYou could use a common 555 timer, and adjust the delay with a potentiometer for the Resistor in the RC network.
240898
This is stuff I have not worked with in 20 years. LOL. I can not draw this for you here, you can figure it maybe from what I tell you. \n(1) You want a circuit that has say 10V steered to four outputs depending on the control inputs. We will say the control inputs are 0 & 5 volts we are steering 10 volts. You are looking at a circuit with 7 lines. 1 input, 2 control and 4 outputs. The voltage to your controls has 4 possibilites.\n0&0 control volts input 10 volts to output line #1\n0&5 control volts input 10 volts to output line #2\n5&0 control volts input 10 volts to output line #3\n5&5 control volts input 10 volts to output line #4\nThat about all I can say here. May use AND, OR, NOR Gates??\n(2) Is even more complicated and I can not even begin to tell this in words. Hope this helps, I doubt it will, but, I tried.
508767
It sounds to me like you have an Interruptible Power Supply (UPS) where you have a primary AC power supply and a battery backup. It seems from your question that you are looking to automatically switch to the battery backup if the AC fails. Of course you must rectify the DC from the battery to AC of the proper current and voltage. Assuming this is the case, you can wire the load through an AC relay. Connect the AC power source to the relay coil in parallel to the load. Hot on one side, Neutral on the other.  Wire the primary AC hot wire to load  through the Normally Open contacts. Wire the DC rectified AC Hot wire through the Normally Closed contacts. When you have good AC power the relay coil will energize, closing the Normally Open contacts (and opening the Normally Closed contacts). This will energize the load with primary AC power. If you loose primary AC power, the relay will de energize. This will close the Normally Closed  contacts(and open the Normally Open contacts)  and port DC converted AC to the load. \n\nYou will probally have to come to a common point after the relay contacts such that power takes the path of either the AC or DC but goes through to the load via the same path after the relay.\nEnsure that you have compadible neutrals between the AC and the DC they will probally be common. Only the hot wires go throughthe relay. \nChoose a relay that will be able to handle the current and voltage you desire. Ie you need enough power to energize the relay coil and you do not want to melt the contacts by giving them too much current. You can obtain them as mentioned above frpm granger.com there real cheep. Order the relay and base. \nEnsure the load works well with both power sources.
143811
Aboslutely, provided the fans are for 12V service.  Leave the secondary center tap unconnected, and put each end of the secondary to the AC input of the rectifier.  The + goes to the red lead of the fan; the - goes to the black lead (traditionally speaking of course.) Now the supply will be very ripply, with a 120 Hz waveform which could be more than a volt.  If you put a 470uF 35V capacitor across the plus and minus of the rectifier, you will smooth out the ripple, but increase the voltage above 12 volts.  Make sure the fans can take the increased voltage.  Transformers are rated at the RMS output at the 3 amps in your case.  The peak voltage will be around 1.4*12.6, but in practice I have never seen it go above 15 V with a load.  Therefore at 1 amp, if you put a cap on the supply, it may produce excessive voltage.  Check it with a voltmeter-- you may luck out.\n\nI have been in a similar situation.  Get a 5 watt resistor at maybe 1 to 3 ohms.  Insert it in series between the capactitor positive and the fans wired in parallel.  When running, use a voltmeter on the fan circuit.  Increase the resistor till you get 12V or thereabouts.  The resistor will get hot, so leave the leads long and connect them to a terminal strip or block if possible, and mount it in the airstream.\n\nCheck out the LM7812, however, at 1 Amp.  Google it, or check out jameco.com for the part.  Put a 470uF at 35 V at each end of the regulator to keep things very smooth.  Your fan will run perfectly with the voltage regulator.  Be sure to mount the tab on a heatsink to keep it cool since you are running it at full capacity.
700354
Since you have no contact info, I have no other choice...\n\nIn reference to http://answers.yahoo.com/question/index?qid=20061115060801AAEHWQA&r=w&pa=FZptHWf.BGRX3OFMhzdTUCYEW59bygAIQoCa4DY8zIlrypTkPQ--&paid=answered#NbUvWju8VzmdRHfXRHQU\n\nYou suggested to the asker to wire his sub to 1 ohm when it clearly states in his question "500 watts X 1 @ 4 ohms". You do realize you told him to blow his amp up by wiring it that way.
104828
You need two blocks in your circuit. \n\nFirst you get a thermocouple signal conditioner. There are two types analog output and digital output. If you want to get fancy use a microprocessor use a digital, otherwise go with the analog.\n\nNext you take the output signal and turn something on. \nIf the output is analog and you want to go simple you need a comparator which you add a resistor network with a trimpot to adjust the setpoint and the other input is the temp signal which then turns the load on and off.  For both types the outputs may need to drive a relay to control more wattage load since you are probably using it to control the temp.\n\nIf you are interfacing it with a microcontroller you can use a chip like this:\nhttp://pdfserv.maxim-ic.com/en/ds/MAX6674.pdf\nIt will just give you a digital temp and you can trigger the load with another output and even add a PID loop to control the temp in a smoother nicer way.\n\nIf you want the analog version try something like:\nhttp://pdfserv.maxim-ic.com/en/ds/MAX1464.pdf
341482
Ok break it down:\n1) A power supply converts AC power (110 in the states) to some other required voltage. For this (lab) project I'd presume a DC regulated voltage from, say 1.5 to 15V something like that.\n\n2) Digital and programmable means you'd control the output voltage with some sort of digital signal like up/down buttons, a keypad, or a PC hookup.\n\n3) You are supposed to use an 8051 to accomplish this.\n\nSo there are several ways to do this. One way would be to have the 8051 accept a command to set the voltage and produce a PWM signal. Let's say the PWM signal is a 5V output switched from 0 to 100%. You can use an RC filter to produce 0V to 5V from this signal. If your instructor is nice, you could argue that this meets the criteria -- especially if you use it to drive a pass transistor to develop some serious current (but your voltage is now 0V to 4.3V because of the BE drop).\n\nBut I'd instead use an op amp with gain X3 (plus a little extra) so that I got from 0 to 15.7V and then use the pass transistor. Another idea would be to use the voltage to drive a simple current source which would program a LM317 voltage regulator.\n\nWeb searches on things like LM317, PWM, and "pass transistor" are your friends here.
488998
it should not exceed 5 mSev :) ..