I'm aware of the previous post a few years back. Sink and source currents of series 74LS TTL.

I need help understanding the current limits of the 7400, I don't understand what Mr. Chips was saying.

For 7400, 7404
IOH is -0.4mA or (-400uA)
IOL is 16mA

IOH, and IOL means Output high current, Output low current. Correct?
I'm planning on using the chip with either an n-Channel mosfet or a 2N3904, to drive whatever I feel like driving.

Question #1. I don't understand how high output current can be negative.
Question #2. How the heck can low output be that high?
Question #3. That's the current limit on pins 3, 6, 8, & 11 on the 7400, right?

According to the datasheet of the 2n3904, when the collector is at maximum current (100mA), the HFE is 30.

Some useful simple NPN transistor equations. Ib = Ic / HFE, Ic = Ib*HFE, HFE = Ic/Ib
Solve Base current : 0.1A Ic / 30 HFE = 0.0033A Ib.
Solve Collector current : 0.0033A * 30 = 0.1A
Solve HFE : 0.1A Ic / 0.0033 Ib = 30.30 HFE

Ok, let me be honest, I really don't care about using an npn transistor to drive things, I'm more interested in using an n-Channel mosfet. I can lower the mosfet gate to 0.1mA, and still have it work. Because I've tried it. Also the n-channel mosfet can also drive very big things, as high as 120vac, using a SSR. So If I want to use an n-Channel mosfet, I'll use something like a 15k on the gate, and only draw 0.3mA from the output of the 7400.

So the output pin(s) of the 7400, is 5 Volts? That's what I'm understanding on the datasheet for the 7400.
Typ (Note 2), and note 2 says : All typicals are at VCC = 5V, TA = 25°C (77°F)

@blocco a spirale : Before you start another thread, please read this : https://en.wikipedia.org/wiki/Concision

I tried.

 

Positive output current means the chip is sinking current and negative output current means the chip is sourcing current.
From the 7400 datasheet:

With the output high and sourcing 0.4mA the output voltage may be only 2.4V. A MOSFET has a very, very, high input resistance so the gate current is essentially zero, so you can rely on getting 2.4V on the gate and there are MOSFETs that will work at that gate voltage.

 

I'm aware of the previous post a few years back. Sink and source currents of series 74LS TTL.

I need help understanding the current limits of the 7400, I don't understand what Mr. Chips was saying.

For 7400, 7404
IOH is -0.4mA or (-400uA)
IOL is 16mA

IOH, and IOL means Output high current, Output low current. Correct?
I'm planning on using the chip with either an n-Channel mosfet or a 2N3904, to drive whatever I feel like driving.

Question #1. I don't understand how high output current can be negative.
Question #2. How the heck can low output be that high?
Question #3. That's the current limit on pins 3, 6, 8, & 11 on the 7400, right?

According to the datasheet of the 2n3904, when the collector is at maximum current (100mA), the HFE is 30.

Some useful simple NPN transistor equations. Ib = Ic / HFE, Ic = Ib*HFE, HFE = Ic/Ib
Solve Base current : 0.1A Ic / 30 HFE = 0.0033A Ib.
Solve Collector current : 0.0033A * 30 = 0.1A
Solve HFE : 0.1A Ic / 0.0033 Ib = 30.30 HFE

Ok, let me be honest, I really don't care about using an npn transistor to drive things, I'm more interested in using an n-Channel mosfet. I can lower the mosfet gate to 0.1mA, and still have it work. Because I've tried it. Also the n-channel mosfet can also drive very big things, as high as 120vac, using a SSR. So If I want to use an n-Channel mosfet, I'll use something like a 15k on the gate, and only draw 0.3mA from the output of the 7400.

So the output pin(s) of the 7400, is 5 Volts? That's what I'm understanding on the datasheet for the 7400.
Typ (Note 2), and note 2 says : All typicals are at VCC = 5V, TA = 25°C (77°F)

@blocco a spirale : Before you start another thread, please read this : https://en.wikipedia.org/wiki/Concision

I tried.

That high output current limit is the limit that still allows the output to be a valid logical high. If you are driving just the transistor and no other logic you can get a lot more current out of the high side driver. If you are using a family that has high side current limiting you can drive an LED from ground with no resistor, but it isn't considered a "good design practice".
I use the concept often but my designs don't go anywhere but my work bench.

 

Positive output current means the chip is sinking current and negative output current means the chip is sourcing current.
From the 7400 datasheet:
View attachment 113421
With the output high and sourcing 0.4mA the output voltage may be only 2.4V. A MOSFET has a very, very, high input resistance so the gate current is essentially zero, so you can rely on getting 2.4V on the gate and there are MOSFETs that will work at that gate voltage.

So the datasheet lies. Basically. The 7400 can source +0.4mA. The 7400 can sink 16mA.

Like these? I love drawing circuit in InkScape btw. If anyone since the time I've been on here knows that by now. Takes hours sometimes, but if I learn something from it, and so will others when they read my threads, that's VERY GOOD. Because I put detail in my threads, with the excellent help from knowledgeable people here on AAC. I also watched a couple videos on Sourcing and Sinking. Which I'll list below the images. Also here's the datasheet for the MOSFET I received from mouser a couple months ago. Also, yes, I'm aware that the LED in the circuit is INCORRECT. Meaning, the 7400 CAN NOT source or sink 20mA of current. The drawings are for demonstration purposes only to illustrate my current understanding on SINKING & SOURCING Current with the 7400 IC.

So here's my current understanding in picture format.


and


Here's some videos I've watched to help understand.

VID #01 : Sinking & Sourcing (LED Example)
VID #02 : LOGIC Sinking & Sourcing

Let me know please, if I missed something, or didn't get what is being talked about.

I also read a little about this post here on AAC. : Why HIGH Level Output Current is negative.

 

So the datasheet lies. Basically. The 7400 can source +0.4mA. The 7400 can sink 16mA.

No, it doesn't. Leaping from "I don't understand..." to "the datasheet lies" is a sophomoric mistake. Consider for a moment that the 50 year old datasheet reviewed by *millions* of designers is correct, and that you, as you said, don't understand.

A TTL output can both source and sink current, and from the point of view of the pin they both cannot be considered the positive direction of flow because they are opposites of each other. By convention, one direction is called positive and one is called negative. These conventions are important because they affect the algebraic sum of the Vcc and GND pin currents.

16 mA isn't a particularly high current value. AC series CMOS can source and sink 20 mA, later logic families can sink 64 mA, and even old opamps are rated for 20 mA continuous and 50 mA short circuit. The 16 mA comes from the characteristics of the original TTL input stage (also documented on your datasheet). You have to sink -1.6 mA out of the input pin to ground to cause the gate to change state, so a fanout capability of 10 means an output must be able to sink 16 mA.

If you want to question something, wonder about why current out of a TTL output pin is considered negative, while current out of a power supply terminal is considered positive.

ak

 

Hello,

The attached PDF might be an interesting read for you.

Bertus

 

No, it doesn't. Leaping from "I don't understand..." to "the datasheet lies" is a sophomoric mistake. Consider for a moment that the 50 year old datasheet reviewed by *millions* of designers is correct, and that you, as you said, don't understand.

A TTL output can both source and sink current, and from the point of view of the pin they both cannot be considered the positive direction of flow because they are opposites of each other. By convention, one direction is called positive and one is called negative. These conventions are important because they affect the algebraic sum of the Vcc and GND pin currents.

16 mA isn't a particularly high current value. AC series CMOS can source and sink 20 mA, later logic families can sink 64 mA, and even old opamps are rated for 20 mA continuous and 50 mA short circuit. The 16 mA comes from the characteristics of the original TTL input stage (also documented on your datasheet). You have to sink -1.6 mA out of the input pin to ground to cause the gate to change state, so a fanout capability of 10 means an output must be able to sink 16 mA.

If you want to question something, wonder about why current out of a TTL output pin is considered negative, while current out of a power supply terminal is considered positive.

ak

Is this better...? PIN 8 is Sinking Current now. It's allowed to sink +16mA. (Looks like the same thing talked about in this video).

 

No, it doesn't. Leaping from "I don't understand..." to "the datasheet lies" is a sophomoric mistake. Consider for a moment that the 50 year old datasheet reviewed by *millions* of designers is correct, and that you, as you said, don't understand.

A TTL output can both source and sink current, and from the point of view of the pin they both cannot be considered the positive direction of flow because they are opposites of each other. By convention, one direction is called positive and one is called negative. These conventions are important because they affect the algebraic sum of the Vcc and GND pin currents.

16 mA isn't a particularly high current value. AC series CMOS can source and sink 20 mA, later logic families can sink 64 mA, and even old opamps are rated for 20 mA continuous and 50 mA short circuit. The 16 mA comes from the characteristics of the original TTL input stage (also documented on your datasheet). You have to sink -1.6 mA out of the input pin to ground to cause the gate to change state, so a fanout capability of 10 means an output must be able to sink 16 mA.

If you want to question something, wonder about why current out of a TTL output pin is considered negative, while current out of a power supply terminal is considered positive.

ak

Why current out of a TTL output pin is considered negative, while current out of a power supply terminal is considered positive? I guess that's the question I'm supposed to ask. So please, pretty please, do explain it to me, because I'm clueless.

All joking aside.

 

Consider the circuit below. The battery is 1.5V. If the output of the inverter is a logic '1' then its output voltage will be 2.4V and current will flow from the output into the battery, charging the battery. If the inverter output is logic '0' then the output voltage will be close to zero and current will flow from the battery into the inverter output, discharging the battery.

It seems only sensible to describe the direction of these two currents as opposites.

Way back in the mists of time it was thought that current flowed from what was called the positive to what was called the negative. Later it was discovered that an electric current was the movement of electrons and that the electrons moved from the negative to the positive. This has led to a lot of confusion about the direction of current flow. We speak of a current going from negative to positive but describe the same current as a 'conventional' current flowing from positive to negative.

The convention for describing the current into or out of a chip pin is that sinking is considered positive and sourcing is considered negative. This is a fairly arbitrary choice (though there are good reasons for making this choice) and it is an agreed convention throughout electronics so it is best to learn and accept that convention.

 

The Spice simulator convention is that all currents into a pin (including power supply pins) are positive and all currents out are negative.
Thus a negative current flowing out of the positive end of a supply will become a positive current flowing into the pin of an IC, keeping all the polarities consistent.

The one place it makes a certain sense is if you have the current going into a resistive node.
For that a positive current (into the node) generates a positive voltage.

 

7400 input and drive stuff. Can you drive an LED with a high 7400 output? Yes, it draws about 5 mA from the high side. Can you drive an LED with a 7400 low output? Yes, but it draws about 60 mA. There are a few 74xx devices that are capable of this, 7439 (DM8881) being one of these devices.

Attached here is a pdf file )from Excel) that shows what the output will be under different loads to ground or VCC. The resistors simulate an equivalent I in Low and I in High from some other driven circuit.

OOPS
The pdf file doesn't come through for some reason.


This applies only to 74xx devices, not 74LSxx or other 74(whatever) type devices.

 

Consider the circuit below. The battery is 1.5V. If the output of the inverter is a logic '1' then its output voltage will be 2.4V and current will flow from the output into the battery, charging the battery. If the inverter output is logic '0' then the output voltage will be close to zero and current will flow from the battery into the inverter output, discharging the battery.

It seems only sensible to describe the direction of these two currents as opposites.

Way back in the mists of time it was thought that current flowed from what was called the positive to what was called the negative. Later it was discovered that an electric current was the movement of electrons and that the electrons moved from the negative to the positive. This has led to a lot of confusion about the direction of current flow. We speak of a current going from negative to positive but describe the same current as a 'conventional' current flowing from positive to negative.

The convention for describing the current into or out of a chip pin is that sinking is considered positive and sourcing is considered negative. This is a fairly arbitrary choice (though there are good reasons for making this choice) and it is an agreed convention throughout electronics so it is best to learn and accept that convention.


View attachment 113497

Alright, I think I understand. Like these circuits I made?

 

The Spice simulator convention is that all currents into a pin (including power supply pins) are positive and all currents out are negative.
Thus a negative current flowing out of the positive end of a supply will become a positive current flowing into the pin of an IC, keeping all the polarities consistent.

The one place it makes a certain sense is if you have the current going into a resistive node.
For that a positive current (into the node) generates a positive voltage.

Can you draw a circuit showing what you've talked about. I think that would be best for me, seeing I'm more of a visual type learner. Yes, you explained it, but it's still a little difficult to understand. If you have a piece of paper, or maybe a circuit simulator with actual current flow, that would be really nice. Thanks.

 

View attachment 113503

7400 input and drive stuff. Can you drive an LED with a high 7400 output? Yes, it draws about 5 mA from the high side. Can you drive an LED with a 7400 low output? Yes, but it draws about 60 mA. There are a few 74xx devices that are capable of this, 7439 (DM8881) being one of these devices.

Attached here is a pdf file )from Excel) that shows what the output will be under different loads to ground or VCC. The resistors simulate an equivalent I in Low and I in High from some other driven circuit.

OOPS
The pdf file doesn't come through for some reason.


This applies only to 74xx devices, not 74LSxx or other 74(whatever) type devices.

Oops is correct. I've tried looking at the circuits you have in your image, but when I go to click it, it doesn't increase in size, the words in the image are blurry, and unreadable. When all else fails, I usually use YouTube, or ImgUr.com to upload my HQ images so people can look at my work.

Also, I'm more of a visual type learner, so if you could draw up examples instead of using charts that would be nice.
I don't understand the chart, please draw a circuit as an example either on a piece of paper, InkScape, or in a circuit simulator. Thanks.

 

I´d say please stop using the falacy simulataor aka the Falstad... Ok I haven´t used it for a while, but as far as I know it is only good for switches and lightbulbs, and anything more comlex or using proper models just doesn´t match the reality. But I´d be glad if someone proved me wrong.

 

Can you draw a circuit showing what you've talked about. I think that would be best for me, seeing I'm more of a visual type learner. Yes, you explained it, but it's still a little difficult to understand. If you have a piece of paper, or maybe a circuit simulator with actual current flow, that would be really nice. Thanks.

I will redraw things. Give me a few minutes.

 

I will redraw things. Give me a few minutes.

Thank you. Please take your time, no need to rush.

 

Oops is correct. I've tried looking at the circuits you have in your image, but when I go to click it, it doesn't increase in size, the words in the image are blurry, and unreadable. When all else fails, I usually use YouTube, or ImgUr.com to upload my HQ images so people can look at my work.

Also, I'm more of a visual type learner, so if you could draw up examples instead of using charts that would be nice.
I don't understand the chart, please draw a circuit as an example either on a piece of paper, InkScape, or in a circuit simulator. Thanks.

Looking at what happens at the output as we vary the resistance, simulating a varying input current going low. The chart starts with a high resistance. Something similar to what a CMOS load would be like. We lower the resistance and see what the output voltage does. We can exceed what the data sheet says by a LOT. The spec on the data sheet states the allowed limit that still allows the output to stay within the legit limit. We can go well beyond that if we are not driving other logic.
Somewhere in the higher current range the chip will be destroyed but the chip in the exercise survived short exposures to very high currents. No chip was hurt in the making of this exercise.

 

Looking at what happens at the output as we vary the resistance, simulating a varying input current going low.

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.

We lower the resistance and see what the output voltage does.

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?

We can exceed what the data sheet says by a LOT. The spec on the data sheet states the allowed limit that still allows the output to stay within the legit limit. We can go well beyond that if we are not driving other logic.

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.

Somewhere in the higher current range the chip will be destroyed but the chip in the exercise survived short exposures to very high currents. No chip was hurt in the making of this exercise.

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.

 

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.