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I'm trying to make 1 bit RAM using four transistor ( 2 NOR Gates ).

I built the 2 copies of the following circuit:

enter image description here

I connected a LED to the output "F". Also, I connected Another LED to the output of the second circuit ( because I build two Identical copies ) to test them before I continue.

The problem is that when I connect any Input of both circuits to HIGH, The two LEDs turns off rather than one led. And that means the outputs of the two circuits are somehow connected together. In other words, when the base of any transistor is HIGH, the transistor works as a short circuit ( wire ) for the 2 LEDs so that they both turns off.

I Thought I was mistaken and completed the circuit as shown : enter image description here

But the result is that the LEDs are always off.

What can I do to separate between the two outputs Q and Q' ?

Does a floating input mean it is LOW ?

Note : I used only R1 and R2, and I Didn't use R , R3 and R4. I connect the bases of the transistors directly to the positive terminal To make them HIGH.

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    \$\begingroup\$ Connect resistors in series to the LEDs... and do not omit the resistors R, R3 and R4 (try by yourself to understand why). \$\endgroup\$ – Circuit fantasist Jan 24 '15 at 6:23
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    \$\begingroup\$ Another hint is to insert LEDs in series to the collector resistors R1. Then they will light when the output is low. \$\endgroup\$ – Circuit fantasist Jan 24 '15 at 6:36
  • \$\begingroup\$ Thank you very much @Circuitfantasist I connected R, R3 and R4 and it works :) But I can't Imagine the secret yet.. Would you tell me, please ? I will appreciate your answer as a best answer... actually you should be appreciated more than that .. thank you agian :) \$\endgroup\$ – Michael George Jan 24 '15 at 7:12
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    \$\begingroup\$ If you omit R3 and R4, and connect T2 and T4 bases directly to the opposite output, the outputs cannot rise above 0.7 volts, as that is the maximum voltage you can have across a Base-Emitter junction. \$\endgroup\$ – Peter Bennett Jan 24 '15 at 16:52
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Well, @Michael George... I can't refuse such a tempting promise:)

You can build the most elementary memory cell only by two inverters (transistor switches). I do this every year with my students at the beginning of the laboratory exercise about latches in the laboratory of digital electronics (you can see how in this movie and also in this Wikibooks story written by my students). Here is the building "scenario".

  1. To make a latch (flip-flop, memory cell, 1-bit RAM...), you need only a non-inverting amplifier - just connect its output to the input... and it will begin "memorizing"... keeping by itself at the "top" (+Vcc) or the "bottom" (ground). BTW this clever idea was proposed in 18th century by Baron Münchhausen... so that he can be considered as the inventor of the most elementary RAM cell:)
  2. But, for some reasons, in electronics there are only inverting amplifying stages - a common-emitter amplifier, in your case. So, to make a non-inverting stage, you connect (by the yellow wire, in the picture below) in succession (cascaded) two inverting stages, and then connect (by the black wire) the output of this non-inverting amplifier to its input. Actually the two stages are connected in a loop but for some reasons (maybe to hide the idea:) they draw this configuration as two cross-coupled stages. You should connect the base resistors R3, R4 to extend the small (0.7 V) input range of the the bipolar transistor up to Vcc (5 V). If you don't connect them, the forward-biased base-emitter junction (like a diode) will fix the collector voltage of the other transistor at +0.7 V, and your LEDs will never light (if connected to ground) or will ever light (if connected to Vcc).
  3. You can drive this latch in various ways (as you can see in the movie):
    • The most correct of them is to connect the one end of the (red) wire to the ground and to touch the base of the turned-on transistor with the other end, or to touch the collector (the body, if it is made metallic) of the turned-off transistor (this is the role of your additional transistors T1 and T3, they act as the red wire).
    • An incorrect way is to connect the red wire to Vcc (as shown in the picture), and to touch the collector of the turn-on transistor... but it is interesting that it still survives:). It is extremely interesting to see why since exactly in this way the memory cells of today's modern RAMs are controlled...
    • And the most incorrect way is to touch the red wire directly to the base of the cut-off transistor (since you apply +5 V across the base-emitter junction that tries to fix it at 0.7 V). The result - more likely, either the transistor or the power supply will be damaged; that is why (to limit the current) you have to connect the base resistors R. But in the circuit shown in the movie, the transistors still survived when my students touched their bases by +5 V... it is interesting to see why...

Building a transistor latch

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