Help with 7400 TTL (Current Limits)

hp1729

Joined Nov 23, 2015
2,304
Excuse me Sir, but I'm not that smart with 7400 chips yet. As easy as the NAND gate logic by itself might seem.

Why are we looking at what happens at the output as we vary the resistance, simulating a varying input current going low? What does that even mean? So basically, you lowering the voltage at VCC on the right, while pin 3 stays at 0V.


It's confusing that you didn't label VCC & VCC as VCC(1), VCC(2).

Does it really matter if you raise or lower the resistance on the resistor? The voltage on Pin 3 shouldn't change, if there's 4.95v on VCC on the left, or both inputs of the NAND gate. It's output (pin 3) should be 0V. Because you turned it into a NOT gate by joining pins 1 & 2 together.

Where is it labeled "output voltage" on your circuit? That's pin 3 right? Because VCC on the right is is sourcing current & voltage.

So basically, if pin 3 is 0V, then the VCC on the right is actually sourcing current to the NAND, and the NAND is sinking current. Correct?



I don't know.. Can we..? Why can we exceed what the data sheet says for I guess your talking about current limits of sinking current to the 7400.




Yup. As with any device, or electronic component. Yes, I didn't buy the chip yet. I don't understand the circuit, and how it's even hooked up, so it's a good idea to talk theory, and actually understand how the chip works in the first place, before giving Mouser or JAMECO any money, that I'm just going to burn up anyways, if I don't hook the Quad NAND chip up right to start.

The bottom line is that the chip doesn't seem that difficult to hook up, I simply don't understand how something can source or sink a negative current, i.e. -0.4mA. Which is TINY, 400 uA. I could just make a NAND gate, and be done with it, but that could be very bulky. I could then, later down the road, use SMD 7400, and make a project even tinier.
Excuse me Sir, but I'm not that smart with 7400 chips yet. As easy as the NAND gate logic by itself might seem.

Why are we looking at what happens at the output as we vary the resistance, simulating a varying input current going low? What does that even mean? So basically, you lowering the voltage at VCC on the right, while pin 3 stays at 0V.


It's confusing that you didn't label VCC & VCC as VCC(1), VCC(2).

Does it really matter if you raise or lower the resistance on the resistor? The voltage on Pin 3 shouldn't change, if there's 4.95v on VCC on the left, or both inputs of the NAND gate. It's output (pin 3) should be 0V. Because you turned it into a NOT gate by joining pins 1 & 2 together.

Where is it labeled "output voltage" on your circuit? That's pin 3 right? Because VCC on the right is is sourcing current & voltage.

So basically, if pin 3 is 0V, then the VCC on the right is actually sourcing current to the NAND, and the NAND is sinking current. Correct?



I don't know.. Can we..? Why can we exceed what the data sheet says for I guess your talking about current limits of sinking current to the 7400.




Yup. As with any device, or electronic component. Yes, I didn't buy the chip yet. I don't understand the circuit, and how it's even hooked up, so it's a good idea to talk theory, and actually understand how the chip works in the first place, before giving Mouser or JAMECO any money, that I'm just going to burn up anyways, if I don't hook the Quad NAND chip up right to start.

The bottom line is that the chip doesn't seem that difficult to hook up, I simply don't understand how something can source or sink a negative current, i.e. -0.4mA. Which is TINY, 400 uA. I could just make a NAND gate, and be done with it, but that could be very bulky. I could then, later down the road, use SMD 7400, and make a project even tinier.
The point of the exercise is to realize what the data sheet means when it is talking about output voltage. As you draw current through the output the voltage changes. In order to stay within recognizable high and low limits the current must be within a certain range. We are not limited to that range if we are driving other than logic. We can drive an LED without a resistor using the current limiting built into the high side driver. It is not limited to 400 uA. Some TTL gates can drive up to 80 mA safely on the low side driver, or more if you don't have to stay to a legit low logic level. See a data sheet for a 7439 for an example..
 

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Guest3123

Joined Oct 28, 2014
404
The point of the exercise is to realize what the data sheet means when it is talking about output voltage. As you draw current through the output the voltage changes. In order to stay within recognizable high and low limits the current must be within a certain range. We are not limited to that range if we are driving other than logic. We can drive an LED without a resistor using the current limiting built into the high side driver. It is not limited to 400 uA. Some TTL gates can drive up to 80 mA safely on the low side driver, or more if you don't have to stay to a legit low logic level. See a data sheet for a 7439 for an example..
Ok, but it's sinking current, that allows it that high or current. That means the output of the NAND is sinking current that allows 80mA, or 16mA. If it's sourcing current IOH, then it's limited to 0.4mA, which is fine by me, I'll just use a high value resistor on the gate and source on the n-Channel MOSFET and drive whatever I feel like driving.
 

Thread Starter

Guest3123

Joined Oct 28, 2014
404
If the output is connected to a grounded resistor then it will source (negative) current when the output is high and have zero current when the output is low.
If the output is connected to a resistor that goes to V+ (Vcc) then the output will sink (positive) current when the output is low and have zero current when the output is high.
Why..?
Because Electrons are attracted to protons, and protons are attracted to electrons.
Because Protons flow from source to sink, and electrons flow from sink to source.

Atoms are made of extremely tiny particles called protons, neutrons, and electrons. Protons and neutrons are in the center of the atom, making up the nucleus. The charge on the proton and electron are exactly the same size but opposite. Since opposite charges attract, protons and electrons attract each other.

It's not rocket science, it's elementary.
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AnalogKid

Joined Aug 1, 2013
10,987
Because Protons flow from source to sink,
No, they don't. Not ever.

Without getting into high energy particle accelerators, only electrons move. If you can't accept that, most of electrodynamics will not make sense. Depending on the circuit or application or context, there are many terms to describe electron flow - in, out, positive, negative, up, down, through, between, around, etc. Sometimes, because of the way terms developed over time, some terms appear to contradict each other (as in post #5). You can struggle against these apparent contradictions and quirks, or just learn them and move on to more interesting things. Remember that all - *all* - labels are arbitrary. We brand the charge polarity of an electron as "negative", but it just as easily could have been called "left".

ak
 
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