Output current for 74 series TTL

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Joe_J
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Greetings to all,

I am reading datasheet for a Schmitt trigger 74LS14 and don't understand the exact meaning of "output current".

ti.com/lit/ds/symlink/sn74ls14.pdf

It says:
Code:
IOH High-level output current MAX –0.4 mA
IOL Low-level  output current MAX 8 mA

Does not say anything about positive current for High-level output or negative current for Low-level output.


For example, which of the following currents are valid for High-level output? -10mA, -1mA, -0.1mA, +0.1mA, +1mA, +10mA
Which of the following are valid Low-level output? -10mA, -1mA, +1mA, +10mA

Thanks
 
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Look at the schematics at page 3. It has a totem pole arrangement for output. Thus for low output top transistor will be off and bottom will be on and the current will enter from the output lead. For high output opposite happens.
 
so the output transistor is sinking current NOT sourcing current ?

ie. the current is coming into the collector of the output transistor from whatever circuitry follows?


Dave
 
There are two transistors at the output stage. Only one of them is one at a particular time. For low output the bottom one is on (top one is off) and current goes in from the lead to the collector. For high output the top one is on (bottom one is off) and current goes out from emitter to the lead.
 
The TTL logic series totem-pole output is primarily current sinking logic, meaning that an output low is designed to sink (not source) current, while an output high is capable of providing (sourcing) very little current. In fact, an open input is treated as a high input. Positive 8 mA means sink 8 mA, and negative 4 mA means source 0.4 mA. Unused inputs should be tied high (to Vcc) with a resistor.
 
they key to remembering this is to look at the input circuit.

This stuff is intended to drive not milliametrers but other 7400 chips.

So look at the input and notice there's a 6k resistor pulling that base "high" to turn on the input transistor. (or a 20k for LS series)
To turn that transistor off, whatever drives the chip must sink the (around a)milliamp that comes out that input pin through 6K. Look at parameter IIlow
To turn it on, whatever drives that chip need only refuse to sink that milliamp.

So 7400 output needs to sink better than it sources. Hence the difference between high and low output capabilities.
Normal fanout is 10, ie one output can drive ten inputs
so the output drive is adequate to sink 16ma when low (see line VOL on page 3)
which gives more than required to sink ten inputs.

When studying these IC's it is necessary to maintain awareness of what's just inside the pins .
You should study that schematic as was suggested to understand how they achieved that asymmetric output drive capability.
The ubiquitous 555 timer chip makes a great logic interface, it'll drive most anything. Tie TRIG and THRESH togethetr and you have a schmitt trigger inverter, works great for a comparator to interface with analog.