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Why does Q12 and Q8 conduct current from base to emitter despite the resistors.
- Thread starter Willon
- Start date
The 5v comes from the far left , and how can i limit the voltage without affecting the rest of the circuit
I see nothing on your diagram labelled IN1, IN2, Cin, OUT or COUT.
The diagram depicts two LEDSs, both RED.
Which one is that? Gold star for having reference designators in the schematic. Black dot for not using them in the text.
The #1 and #5 schematics are very hard to read. There are many lines going through the middle of components, so it is not clear at all where some connections are. Also, is the blue rectangle a signal loop that joins many components?
It looks like at least one signal goes through six base-emitter junctions and four resistors in series before getting to an LED that is grounded. That is way too many voltage drops for a 5 V circuit.
If possible, suppress or turn off the grid to improve readability.
Is there a reason you are not using logic gates for this?
ak
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By that transistor i meant Q7 and Q12(there has been a few chances so in the new schemathic its Q8 and Q12)
The blue line doesnt mean anything
I have attached a file thats hopefully easier to read
Im not using logic gates because it is a school project we need to use only transistors
If you redo the truth table in a more conventional manner, output condition patterns will be more clear. For example, across the top:
Cin 2in 1in | Out Cout
And columns with 000 at the bottom, increasing in a linear, binary sequence to 111 at the top.
ak
Cin 2in 1in | Out Cout
And columns with 000 at the bottom, increasing in a linear, binary sequence to 111 at the top.
ak
Last edited:
Consider this:
It looks like each pair of transistors functions as a logic gate; Q1 and Q2 form a NAND gate, Q7 and Q8 form an AND gate, etc. Look at your truth table. In a system with 16 possible outputs, there are only 4 that matter, two of those have the same input conditions, and the other two share two of three input conditions.. Do you really need 9 gates (or, more correctly, 18 gate inputs) to detect only two and one-half conditions?
ak
It looks like each pair of transistors functions as a logic gate; Q1 and Q2 form a NAND gate, Q7 and Q8 form an AND gate, etc. Look at your truth table. In a system with 16 possible outputs, there are only 4 that matter, two of those have the same input conditions, and the other two share two of three input conditions.. Do you really need 9 gates (or, more correctly, 18 gate inputs) to detect only two and one-half conditions?
ak
Last edited:
Attachments
If there is a better way to make a full adder with transistors i would definitely love to hear it
This is what the truth table currently looks like
Attachments
That is different from post #1. And - if #14 is supposed to be the table for a 1-bit binary full adder, it is incorrect.
ak
Last edited:
If it's supposed to be a full adder, it's still wrong.
