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Above is a TTL totem pole output NAND gate. There is a 120 ohm pull up resistor there. Since it is called pull up, can we say that the HI output will be connected to a very high input impedance? If so what can it be as an example? Aren't NAND gates in an IC are connected to each other by being ones output is other's input? I mean in the figure the output will be the input of another gate right? If so HIGH will not be 5V since there will not be input impedance unless it is not connected to a very high resistance.

Where is this NAND gate's input coming from and where is the output going to? If the output is 5V isnt it high for a new TTL gate input and if is not 5V why do we call the resistor pull up in there?

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  • \$\begingroup\$ the OUTPUT is low impedance, the INPUT is high impedance. If one drives another, then it results in nice low current/power. \$\endgroup\$ – KyranF Oct 24 '14 at 22:04
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    \$\begingroup\$ Also the 120 Ohm resistor is not a pull up, it's a series current limiting resistor for probably a very low size/power rating transistor for the "push" stage of the totem pole driver. \$\endgroup\$ – KyranF Oct 24 '14 at 22:05
  • \$\begingroup\$ Imagine what would happen if there was no resistor there? The output stage of the totem pole would be almost a short circuit (other than the limiting factor of the DC Current gain of the BJT.. but whatever) and things would blow up if the output was shorted to ground for example. In this example, the NAND gate can "source" ~41mA. \$\endgroup\$ – KyranF Oct 24 '14 at 22:08
  • \$\begingroup\$ lets call this transistor above T1 and say we feed the output in this figure to another transistor T2's input. what will be the high impedance you are mentioning? the T2's emitter inputs without any extra resistor? and what will be the low output impedance of T1? 120 ohm of the T1? could u expound on that a bit? \$\endgroup\$ – user16307 Oct 24 '14 at 22:10
  • \$\begingroup\$ The input to this circuit is high impedance because Q1 is acting in high resistance, to let through the InA and InB signals. If you chain these devices input to output etc, the 120 ohm value will not even matter \$\endgroup\$ – KyranF Oct 25 '14 at 10:17
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The primary purpose of the 120\$\Omega\$ resistor is to reduce the current spikes when the output switches (when Qo and Qp are both on simultaneously for a brief moment). See, for example, here. It's a component part of the (active) pullup circuit, but it's not a 'pullup resistor'.

Totem pole outputs like this one use an active pullup, which is Qp, Rc and Rcp. When Qs is 'off', the base of Qp is pulled to Vcc by Rc, so the effective pullup resistance is limited by the collector resistance Rcp - so it's about 120\$\Omega\$, meaning that for a 50pF load, the time constant is about 6ns. Without the collector resistor it would behave more like a few ohms (1.6K divided by the current gain of Qp).

TTL inputs are defined as 'low' if they are less than or equal to 800mV and 'high' if they are greater than or equal to 2V. TTL outputs will be 400mV or less when low (and sinking 16mA or less), and 2.4V or more when high (and sourcing 400uA or less). That drive capability guarantees each TTL output the capability to drive 10 TTL inputs with a guaranteed noise margin of 400mV or greater.

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  • \$\begingroup\$ i dont understand how to connect this gate's output to another NAND gate as an input. do we need a resistor. her my question repeated form the above comment: lets call this transistor above T1 and say we feed the output in this figure to another transistor T2's input. what will be the high impedance you are mentioning? the T2's emitter inputs without any extra resistor? and what will be the low output impedance of T1? 120 ohm of the T1? could u expound on that a bit? \$\endgroup\$ – user16307 Oct 25 '14 at 0:12
  • \$\begingroup\$ @user16307: TTL logic outputs can be connected directly to TTL logic inputs with no additional components. The same statement applies within any logic family, but only sometimes between families. \$\endgroup\$ – Peter Bennett Oct 25 '14 at 0:41
  • \$\begingroup\$ but then u re saying ttl is 2.4V between eachother, but in the figure high is around 5v. i do not understand where i dont get. besides one of my questions nobody answered that is what is the input resistance? is that a resistor or in this case the input emitters are called input resistance? \$\endgroup\$ – user16307 Oct 25 '14 at 1:12
  • \$\begingroup\$ According to the TI TTL databook, a TTL input will accept anything ovcer 2.0 volts as a high, and anything below 0.8 volts as a low. A TTL high output will be typically 3.4 volts, while a low output will be less than 0.4 volts. It is not right to speak of a TTL input having a resistance. It is an emitter of an NPN transistor, and, when taken Low will source 0.4 mA for the 74LS series, and 1.6 mA for plain 74xx. The input High current will be under 100 uA for all varieties. \$\endgroup\$ – Peter Bennett Oct 25 '14 at 1:51
  • \$\begingroup\$ 2.4V is the guaranteed minimum voltage the output will give under load (driving 10 standard inputs high). Typical is more like 3.4V (maybe 3.6V open-circuit), and 5V (or even a bit more) is an acceptable input. Anything more than 2.0 is guaranteed to be interpreted as a '1'. A similar thing is true for the '0'. Anything less than 800mV is guaranteed to be interpreted as a 0, and the output will be no higher than 400mV even when driving 10 standard inputs low. Typically the output under load will be 200mV (even less open-circuit). \$\endgroup\$ – Spehro Pefhany Oct 25 '14 at 8:54

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