0
\$\begingroup\$

I'm learning about electronics and was studying the SN7404 hex inverter schematic when I saw this:

enter image description here

I naively expected just a single transistor (the one in the bottom right) and the input at the base of that transistor to give the inverted input at Output Y.

Yet there are 3 other transistors, a bunch of resistors and a couple of diodes in addition.

Can someone here help to explain what the purposes of these are and the reason for the resistor values?

Source datasheet here: http://www.ti.com/lit/ds/symlink/sn54ls04-sp.pdf

\$\endgroup\$
5
  • \$\begingroup\$ wiki.analog.com/university/courses/electronics/… \$\endgroup\$
    – dim
    Commented Jul 16, 2019 at 10:47
  • 1
    \$\begingroup\$ What you are describing would only be able to pull low and would not be able to give a low impedance output when high. \$\endgroup\$
    – Transistor
    Commented Jul 16, 2019 at 10:51
  • \$\begingroup\$ Also, here is a very good explanation on what happens within the phase-splitting/output driver stages: electronics.stackexchange.com/a/396739/107479 \$\endgroup\$
    – dim
    Commented Jul 16, 2019 at 10:54
  • \$\begingroup\$ Hi dim. Thanks for the link, that's a great resource which explains! :) \$\endgroup\$
    – Charles Nobbert
    Commented Jul 16, 2019 at 10:58
  • \$\begingroup\$ By the way, that resistive pullup on the first transistor's base is one cause of high timing jitter in TTL gates. In fact, LPSTTL uses 10X higher Rpullup. \$\endgroup\$
    – analogsystemsrf
    Commented Jul 16, 2019 at 11:13

1 Answer 1

Reset to default
1
\$\begingroup\$

This is a classic TTL gate.

To ensure that all TTL devices present the same load, the input is always a transistor with the emitter at the actual input (multi-input devices have multiple input emitters).

The next transistor is a drive stage and the output is a totem-pole stage which can both source (output high) and sink (output low) current.

The output current in the high and low states are different (and for a single fan out match the input current for TTL devices).

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for the answer. With the above overview, it gave me the right direction to look in. \$\endgroup\$
    – Charles Nobbert
    Commented Jul 16, 2019 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.