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I have decoder IC 74138 and I found almost all decoder make selected output low (0) and I want it high (1) and I don't want to connect NOT gate to all of the outputs (it is illogical).

What should I do to make output high? Or is there any way to use this one without buying a new non-inverting decoder? If there is not what type of decoder should I buy (IC number)?

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  • \$\begingroup\$ What is the decoder connected to? If something like a LED or transistor you can probably just change the circuit configuration. \$\endgroup\$ – PeterJ May 21 '14 at 8:26
  • \$\begingroup\$ i want to connect it to 7 segment. \$\endgroup\$ – user3649714 May 21 '14 at 8:30
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    \$\begingroup\$ If you used a common anode 7 segment display (if that's possible) you could connect the anode to Vcc via a resistor and then the segments being low would turn them on. \$\endgroup\$ – PeterJ May 21 '14 at 8:34
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It is natural for TTL to use active low signaling. This is because of its polarity asymmetrical nature. Low output drivers are typically stronger than high output drivers. Also unlike CMOS, TTL input switching voltages are much close to 0V than Vcc. Furthemore, you ca directly tie logical input to 0V, but not to Vcc, you need a resistor there. This cause open-low outputs are much more common than open-high ones (I cannot remember single chip with them). While it is better in many cases driven wire correspond to active state (e. g. for Wired-OR and hot plug) it is better to have low state as active.

So it is natural when designing TTL circuits to have active-low signalling.

If you still want design active high circuits you should look toward CMOS (4000 series chips): 4028, 4514, 4555 are decoders with active-high outputs. Also there 'migrated' chips like 74HC4028 which are (TTL-compatible) copies of corresponding CMOS circuits.

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TTL (and therefore 74S, 74L, 74L LSTTL and 74HC CMOS) decoder outputs are typically active-low. The reason (which is the question in your title) is not completely obvious to me, but were I to speculate I might point out that it is somewhat arbitrary for normal push-pull outputs, and may be related to the way the ancients would "wire-or" open-collector outputs (eg. 7405 and 7438), implying an active-low.

If you're driving a 7-segment display, you probably want a BCD-7-segment decoder such as a 7447 or variant, or something from another family. The outputs of this chip are indeed active-low, so you'd connect the anode of a common-anode display to the supply voltage (+5 typically) and each output to a segment (through an individual current-limiting resistor, of course). Each "low" output results in the respective LED turning "on".

If you happen to have a common-cathode display, all is not lost, you can use a part such as 74HC4511, which has active-high outputs. I recommend this part, as it's likely to be around for quite a while longer.

You could also use the 74LS48 (link with 7447 above), which has active high outputs and internal 2K (nominal) resistors to directly drive a common-cathode display with no external resistors (brightness may vary too much to be acceptable between adjacent digits- it's very loosely guaranteed, so this is a bit risky, and the part may be hard to find).

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what should i do to make output high?

Use a 74HCT238. This is a pin-for-pin substitute for the 74138 and has the active high outputs you want. It may be possible to find surplus sources of the 74LS238, if you're really keen on sticking with "real" TTL, but frankly I doubt it's worth it. Your call, though.

As to your title question, the answer is that the original TTL family and its bipolar offspring (7400, 74LS, 74S, 74AS, 74ALS, etc), all used totem pole outputs - see page 2 of http://ee.hawaii.edu/~sasaki/EE260/Labs/Datasheets/7400.pdf

In this configuration, the lower NPN transistor can sink more current than the upper transistor can source, so active low is the preferred output. Furthermore, the heart of the input is a multi-emmiter NPN, and this works faster for low inputs than high.

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