# 7410 and 7400 nand gate question

Discussion in 'General Electronics Chat' started by bdogpot, Nov 29, 2013.

1. ### bdogpot Thread Starter New Member

Nov 29, 2013
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My question is puzzling. I know how nand gates work so that not the question.

first off I supplied 5.04 v to each of the two chips. and connected each one to ground. what my problem is that if I measure voltage from ground to each of the input pins I get .304 v for the 7400 2 input nand gate chip, and I get 1.308 v on each input pin for the 7410 3 input nand gate chip. this is messing up the logic on the circuit because I am not getting zeros or low when I need them. I really need to know how I can fix this problem.

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Busted chips?

3. ### bdogpot Thread Starter New Member

Nov 29, 2013
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I fairly certain there not busted I just bought them and put them in

4. ### tubeguy Well-Known Member

Nov 3, 2012
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What is connected to the inputs ?

5. ### bdogpot Thread Starter New Member

Nov 29, 2013
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I have nothing connected to the inputs at moment. just power and ground. If I connect a switch to the inputs and lets say the switch is off so nothing goes to the input. So that would be low or '0,' and if I were to flip the switch it would be 5v so high or '1'. so at times I need there to be a low on the pin going in but if I read voltage over the pin it has 1.308 v, which makes the gate still think that it is high or 1.

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,971
616
It sounds like your pins were floating when you tested them. Kinda pointless exercise.

7. ### crutschow Expert

Mar 14, 2008
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"Nothing connected to the input" is not a TTL logic low. You have to sink current from the input of a TTL circuit to make it low. An open will thus appear as a logic "1" as you observed. So you need to connect the switch to ground for a logic "0" when the switch is closed, with a 10k pullup resistor to +5V to give a logic "1" when the switch is open.

8. ### bdogpot Thread Starter New Member

Nov 29, 2013
10
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I have tested them connected and un connected, im getting the same result so I wouldn't think that its point less. im just trying to figure out why voltage consistently shows up if no voltage should be going to the pin. it should read zero at the pin if the switch is not supplying voltage.

9. ### bdogpot Thread Starter New Member

Nov 29, 2013
10
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ok crutschow I will try that and see if alleviates my problem

10. ### GopherT AAC Fanatic!

Nov 23, 2012
7,262
5,978
No, it "shouldn't" not be reading a voltage from 0 to 5 volts. There is nothing in the datasheet stating what the floating voltage of an input shoud be because it is not intended to have floating voltages on the inputs.

The inputs should be connected to a logic high or a logic low, then the outputs will pick an output state (high or low). Without a solid high or low on the input, your output will do anything (high, low or oscillate). The oscillate state can vary with various duty cycles. A volt meter cannot see the oscillations which can be several megahertz or more.

11. ### MrChips Moderator

Oct 2, 2009
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Rule #1. Never leave an input with nothing connected.

With nothing connected to an input you can measure any voltage and the input can represent either logic LOW or logic HIGH or anything in between, whatever that may be. (Normally, an unconnected 7400 input is logic HIGH but don't make that assumption.)

You can pull a 7400 series input LOW by connecting a 100Ω resistor from the input to GND.

You can pull a 7400 series input HIGH by connecting a 1kΩ resistor from the input to Vcc.

12. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
I think the "unconnected 7400 input is HI" only applies to TTL families. If it is a CMOS family, all bets are off.

13. ### ScottWang Moderator

Aug 23, 2012
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When the 7400 and 7410 was floating then the voltage of input pin about 1.3V, Vin=0.65*3-0.65=1.3V

When the 74LS00 and 74LS10 was floating then the voltage of input pin about 0.3V, Vin=0.3V+0.3V-0.3V=0.3V

Using the 330Ω connecting to the input and another pin of 330Ω connecting to GND, and connecting the switch to the input pin and connecting another pin of switch to the +Vcc, this is one kind of TTL input arrangement when you using the switch to be the input signal, or using a 1K~33K resistor to pull high and the switch connecting to the input pin and GND.

The logic status when using the 330Ω resistor.
When the switch is open then the Vin will be pull low by the 330Ω.
When the switch is close then the Vin will be high by the +Vcc.

The logic status when using the 1K~33K resistor.
When the switch is open then the Vin will be pull high by the resistor.
When the switch is close then the Vin will be low by the switch.

14. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
Okay, so they mean "7400" as in the original 7400 TTL. Okay. Makes sense.

Whenever I see 7400, I tend to automatically read it as 74xx00 such that it is talking about the logic indicated by the '00 independent of the family that is indicated by the xx.

15. ### ScottWang Moderator

Aug 23, 2012
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I have some old TTL, they are 7400 series, I can't find the 7400 series in the Taiwan market, I think they are too old and draw heavy current, 74LS00 series still can buy from the market, but just a few people use them, maybe the students buy them for the school study.

This web still can buy the 7400 series.

16. ### ScottWang Moderator

Aug 23, 2012
5,478
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If that just connecting the input to GND is ok, if there is a switch connecting to the Vcc, if you do that then it will draw too much current as I = 5V/100Ω = 50mA when the switch is close and connecting to Vcc.

17. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
He's talking about pulling UNUSED inputs LO with 100Ω resistor, not inputs that you are going to have a switch to Vcc to pull it HI.

18. ### ScottWang Moderator

Aug 23, 2012
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I know that and I just mentioned that kind of the situation.

19. ### MrChips Moderator

Oct 2, 2009
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4,278
If you want to use a switch on the input of a 7400 TTL gate, use a 1kΩ pull-up resistor on the input. Connect the switch from the input to GND.

20. ### bdogpot Thread Starter New Member

Nov 29, 2013
10
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crustchow you were absolutely correct, thanks so much. I had been stumped for about a week on this project. I also want to thank everyone else for the valuable information. Its good to have stumbled across this sight with very knowledgeable individuals.