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contador sincrono de 6 bits con 3 flip flop jk
axelmedina

axelmedina

a que te refieres con 2Q
2y
axelmedina

axelmedina

@copilot tienes la capacidad de armar el circuito en el esquematico?
2y

copilot

Como un asistente de inteligencia artificial, no tengo la capacidad para realizar cambios físicamente en el esquemático electrónico de Flux. Sin embargo, puedo ayudarte a diseñar el esquema y proporcionarte instrucciones sobre cómo conectar los componentes en tu proyecto.
Por ejemplo, puedes usar los chips SN74LS112AN en tu proyecto. Estos son flip-flops de tipo JK doble negativo con preset, clear y salida complementaria. Puedo orientarte sobre cómo conectarlos en una cadena si ese es tu objetivo.
Dado que tu proyecto no incluye otras piezas más allá de las tres SN74LS112AN, no puedo hacer suposiciones más allá de eso. Si necesitas conexiones específicas o tienes otros componentes en mente que deseas implementar en tu esquemático, por favor proporciona más detalles. Soy capaz de proporcionar guías paso a paso sobre cómo conectar los pines, siempre y cuando se me den suficientes detalles.
Además, flux.ai permite a los usuarios hacer sus propios cambios en el esquemático una vez que han recibido instrucciones o consejos. Este enfoque puede proporcionarte una experiencia educativa más enriquecedora y práctica.
2y
axelmedina

axelmedina

tienes la capacidad de armar el circuito en el esquematico?
2y
axelmedina

axelmedina

@copilot contador sincrono de 6 bits con 3 flip flop jk
2y

copilot

Seems like you want to build a 6-bit synchronous counter using only 3 JK flip-flops.Obviously, you would not be able to count from 0 to 63 (which would be the range of a 6-bit counter) because three JK flip-flops can count from 0 to 7.
However, if what you meant was a 3-bit counter (which can count up to 7 using binary representation), then we can achieve this with three JK flip flops. For a synchronous counter, all flip-flops are triggered by the same clock signal. The J-K inputs for all flip-flops are set to 1 so they will toggle their output on each rising edge of the clock signal.
Your circuit, using 3 of your SN74LS112AN flip-flops (which is a dual JK flip-flop), can be connected as follows:
  1. Connect the 1J and 1K pins of all flip-flops (U1, U2, and U3) to VCC. These are the J and K inputs for the flip-flops and they're set to 1 for counting.
  2. Connect the 1Q output of U1 (Output for FF1) to the 1CLK pin of U2 (Clock input for FF2).
  3. Connect the 1Q output of U2 (Output for FF2) to the 1CLK pin of U3 (Clock input for FF3).
  4. The clock signal for counting is provided to 1CLK of U1.
  5. Connect 1~CLR and 1~PRE for all flip-flops to VCC as these are the asynchronous reset and set inputs, respectively. They're set to HIGH so they don't interfere with the synchronous operation.
Please remember to connect the VCC and GND pins for all the flip-flops to power.
Note: This will create a 3-bit synchronous counter that will count from 0 to 7. If you want to count in a different range, you would either need more flip-flops or different circuit logic.
Please let me know how you want to proceed!
2y
axelmedina

axelmedina

que solo cuente del 0 al 5
2y

copilot

To configure your 3-bit synchronous counter to count only up to 5, we have to reset the counter when it reaches 6, effectively making the counter start from 0 again. It's called a truncated synchronous counter.
Given that you are using SN74LS112AN flip-flops, here are the connections you should make:
  1. Connect 1J and 1K of U1, U2, and U3 to VCC to set them to 'toggle' mode.
  2. Connect 1Q output from U1 to 1CLK input of U2.
  3. Connect 1Q output from U2 to 1CLK input of U3.
  4. For the reset logic, create a NAND gate logic using two 2-input NAND gates (let's call them U4A and U4B).
    • Connect 1Q output from U3 and 2Q output from U2 to the inputs of U4A.
    • The output of U4A should be fed into both inputs of U4B (This will give us a NOT function to the output).
    • The output of U4B will reset the counters when the count reaches 6 (110 in binary).
  5. Connect the output of U4B to 1~CLR of U1, U2, and U3. This will clear the flip flops when it counts to 6, restarting the counter cycle.
  6. Connect 1~PRE for all flip-flops to VCC so they don't interfere with the synchronous operation.
  7. Connect the VCC and GND pins to power.
Please note, this design will require an additional NAND gate integrated circuit as the SN74LS112AN only contains J-K flip-flops. You might consider using a 7400 IC (quad 2-input NAND gate) for this purpose. Always check the datasheets for voltage levels and currents.
2y
axelmedina

axelmedina

podrias detallarme mas sobre las conexiones de la compuerta NAND
2y

copilot

Absolutely, here are more detailed instructions on how to connect the NAND gates:
Firstly, we need to specify the IC pins of a standard 7400 2-input NAND gate:
  • Pin 1: input 1 of gate 1
  • Pin 2: input 2 of gate 1
  • Pin 3: output of gate 1
  • Pin 4: input 1 of gate 2
  • Pin 5: input 2 of gate 2
  • Pin 6: output of gate 2
  • Pin 7: Ground (GND)
  • Pin 8: output of gate 3
  • Pin 9: input 2 of gate 3
  • Pin 10: input 1 of gate 3
  • Pin 11: output of gate 4
  • Pin 12: input 2 of gate 4
  • Pin 13: input 1 of gate 4
  • Pin 14: Supply voltage (VCC)
Let's assume that we will be using U4A for gate 1 and U4B for gate 2 of this IC.
  1. Connect the 1Q output of U3 (representing the MSB, or Most Significant Bit, of the binary count) to pin 1 of the 7400 IC.
  2. Connect the 2Q output of U2 (representing the bit next to the MSB) to pin 2 of the 7400 IC.
  3. This produces a NAND operation, and the result is available at pin 3 of the 7400 IC.
Since we need an output HIGH when the count is from 0 to 5 and LOW when it's 6, we use the next NAND gate (U4B) as an inverter:
  1. Connect pin 3 of the 7400 IC (output of U4A) to both pin 4 and pin 5 of the 7400 IC (inputs of U4B).
  2. The output of U4B (pin 6 of the 7400) will now produce a HIGH output for counts from 0 to 5 and a LOW output at count 6.
Finally, connect the output of the NAND gate setup (pin 6 of the 7400 IC) to clear flip flops U1, U2, and U3:
  1. Connect pin 6 of the 7400 IC to the 1~CLR input of U3 (3rd flip flop).
  2. Similarly, connect pin 6 of the 7400 IC to 1~CLR inputs of U1 (1st flip flop) and U2 (2nd flip flop).
For power supply and grounding:
  1. Connect pin 14 of the 7400 IC to VCC.
  2. Connect pin 7 of the 7400 IC to GND.
Remember, each component/pin should only be referenced in the context of its part. So, saying "connect pin 6 to the clear input" is ambiguous and should be avoided.
Let me know if something was unclear or if you need further assistance!
2y
axelmedina

axelmedina

a que te refieres con 2Q?
2y

copilot

In the context of an integrated circuit (IC) for a JK flip-flop like the SN74LS112AN, the chip comprises two flip-flops. The outputs of these two flip-flops are typically labelled as 1Q and 1~Q for the first flip-flop and 2Q and 2~Q for the second flip-flop.
If we follow the convention where the symbols ~, n, or a dash above the character represent NOT:
  • 1Q and 1~Q represent the output and complement output of the first flip-flop.
  • Similarly, 2Q and 2~Q represent the output and complement output of the second flip-flop.
In the previous instructions, 2Q was referring to the output of one flip-flop. It's the value currently stored by the flip-flop: if the output Q is true (or 1), this flip-flop is considered to "store" a 1, and if the output Q is false (or 0), it is considered to "store" a 0.
Just to clarify, 1Q, 1~Q, 2Q, 2~Q are specific to the datasheet for the SN74LS112AN flip-flop, and other flip-flop ICs may use different pin labelling conventions. Always consult the component's datasheet for accurate labeling and pin configurations.
2y
2
U3
U2
U1
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