circuity Posted February 25, 2006 Report Share Posted February 25, 2006 hi,I have simulated this circuit in Multisim and it works... but when i build it did work.. I have several question about this circuit.1) do i need to add any other component for the real circuit (hardware)2) i already put together the circuit using SN74LS21N (AND gate), SN745LS04N (NOT gate), SN745LS32N (OR gate) .. but there is no output.3) the output (pin 8 and 6) is high when only the Vcc and Gnd of the SN74LS21N is connected... the same happens with the OR ic.4) do i need a different supply for the Vcc? or can i use the same 5V supply for the input and Vcc...Thanx Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 25, 2006 Report Share Posted February 25, 2006 Hi Circuity,Your inputs are always high. Nothing in your circuit makes an input a logic low!Learn about the inputs of TTL logic ICs that are at a logic high unless something pulls them to a low. Quote Link to comment Share on other sites More sharing options...
circuity Posted February 26, 2006 Author Report Share Posted February 26, 2006 thanx audioguru for the fast reply.but i dont understand what you are trying to say....does it mean that i need to add other components..what should i add...?i dont have any clues..im a newbie in electronics.... ???thanx.... Quote Link to comment Share on other sites More sharing options...
Guest Kasamiko Posted February 26, 2006 Report Share Posted February 26, 2006 He means to put some PULLDOWN resistors at the other side of the switches to put that side near at ground potential whenever the switches are open..a 10k resistors is enough I think..rhonn ;) ;) Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 26, 2006 Report Share Posted February 26, 2006 A resistor can be used to "pull-down" the input of a TTL logic gate. Each input when a few are in parallel needs 0.4mA at about 0.4V for a 74LSxx input, which is 1k ohm. An ordinary 74xx input needs 1.6mA at about 0.4V which is 250 ohms. Quote Link to comment Share on other sites More sharing options...
circuity Posted February 26, 2006 Author Report Share Posted February 26, 2006 do u guys mean like this?so i need a 1k resistor.....okthank u very much u guys... Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 26, 2006 Report Share Posted February 26, 2006 Most of your 10k resistors have two 74LS inputs, each with up to 0.4mA. Therefore the voltage across the 10k resistor is 2 x 0.4mA x 10k= way too high!If you use a 1k resistor then two inputs with a 1k resistor will be at 2 x 0.4mA x 1k= 0.8V which is the threshold voltage of a logic low and might not work.Each logic input needs a 1k resistor, so use 470 ohms where you have two logic inputs connected together. Quote Link to comment Share on other sites More sharing options...
circuity Posted February 27, 2006 Author Report Share Posted February 27, 2006 is this circuit ok? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 27, 2006 Report Share Posted February 27, 2006 Hi Circuity,Now that you have learned about the input currents of 74LS logic gates, you need to learn about the output currents.The output is guaranteed to be a logic high of 2.7V or more when driving a load resistance to ground of 6.8k or more. The max recommended output high current is only 0.4mA!Therefore the outputs probably won't drive the 1k resistors and certainly won't drive 200mA into your lightbulbs. Quote Link to comment Share on other sites More sharing options...
circuity Posted February 28, 2006 Author Report Share Posted February 28, 2006 is this circuit ok?im also thinking about adding a relay at the output (where the lamp is).. relay with rating should i use...?wht else should i do to make this circuit work...?sorry... im not tht good in logic circuits... ???thanx Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 28, 2006 Report Share Posted February 28, 2006 Hi Circuity,Logic gates don't need pull-down resistors at their outputs. The output of a 74LSxx TTL gate is guaranteed to drive an 8mA load from a positive source down to 0.4V.Your circuit still has very high current lightbulbs. A 74LSxx TTL output can drive the base of a transistor through a current-limiting resistor, but not a lightbulb nor a relay. Quote Link to comment Share on other sites More sharing options...
circuity Posted February 28, 2006 Author Report Share Posted February 28, 2006 does this mean that the resistors that i added are correct?can i use BC547 and connect the base BC547 to the output of the circuit with 5.6k resistor in series? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 28, 2006 Report Share Posted February 28, 2006 does this mean that the resistors that i added are correct?No. The resistors at the outputs of the gates are not required since they don't do anything.can i use BC547 and connect the base BC547 to the output of the circuit with 5.6k resistor in series?Yes, but the BC547 won't have enough base current to drive a lightbulb or a relay. Quote Link to comment Share on other sites More sharing options...
circuity Posted February 28, 2006 Author Report Share Posted February 28, 2006 do i need to remove all the 6.8k and 20k resistors or only the 20k resistors?what resistor value should i use then?....or do i need to find another type of transistor...?... how about BC107? Quote Link to comment Share on other sites More sharing options...
Staigen Posted February 28, 2006 Report Share Posted February 28, 2006 Hi circuityTry to get a copy of the book "TTL COOKBOOK" by Don Lancaster, and read it, specially the chapter about intefaceing to real world. :)I belive amason.com have it, or at least a used one.You can also try to borrow it from your local library.If you found it as an E-book, please let us share it!//Staigen Quote Link to comment Share on other sites More sharing options...
circuity Posted February 28, 2006 Author Report Share Posted February 28, 2006 i couldnt find the book with the info i needed..... the one with the info has not been returned to the library...and i need this circuit as soon as possible...that's y i asked here...sorry guys....i really need ur help.... Quote Link to comment Share on other sites More sharing options...
circuity Posted March 2, 2006 Author Report Share Posted March 2, 2006 so do i need to remove all the 6.8k and 20k resistors or only the 20k resistors?for the BASE of BC547 what resistor value should i use then?....or do i need to find another type of transistor...?... how about BC107? ???thanx Quote Link to comment Share on other sites More sharing options...
audioguru Posted March 2, 2006 Report Share Posted March 2, 2006 so do i need to remove all the 6.8k and 20k resistors or only the 20k resistors?I repeat: "The resistors at the outputs of the gates are not required since they don't do anything."for the BASE of BC547 what resistor value should i use then?....or do i need to find another type of transistor...?... how about BC107?A BC107 has the same chip as a BC547 but is in a metal case. They don't have enough current gain and aren't powerful enough to drive your high-power light bulbs. A BC547 could use a 6.8k resistor in series with its base from a TTL output and could drive a more powerful transistor to drive your lightbulb. Quote Link to comment Share on other sites More sharing options...
stevedabear Posted March 2, 2006 Report Share Posted March 2, 2006 A darlington pair perhaps ? Quote Link to comment Share on other sites More sharing options...
audioguru Posted March 2, 2006 Report Share Posted March 2, 2006 A darlington pair has a voltage drop which is a high percentage of this 5V system. Quote Link to comment Share on other sites More sharing options...
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