How to understand SN754410 datasheet and current?

[Munnu]

Hi everyone, could someone help me understand how to read the sn754410 datasheet (http://www.ti.com/lit/ds/symlink/sn754410.pdf) At the absolute maximum rating section (7.1) it mentions Ip (Peak output current) and Io (Continous output current). I want to understand if the power source that Vcc2 is connected and provides a current of 1A would that also mean that Io = 1A? Does the current at Vcc2 affect the peak output current and/or continuous output current? Should I even be concerned about the current at Vcc2 or as long as I supply a voltage in the 4.5~36V range my chip would supply a 1A current at its output pin? I want to better understand how the current value at the output pins are determined and the max current this IC could take.

 

[dlgoff]

Should I even be concerned about the current at Vcc2 or as long as I supply a voltage in the 4.5~36V range my chip would supply a 1A current at its output pin?

As long as you don't exceed 1 amp, yes correct. But that depends on what you are driving.

 

[Munnu]

I'm driving 5Ohm resistance hand-wound electromagnets at the output. I want to better understand what current I am getting at the output pins and how to figure that out.
You could correct my understanding in the next few lines. I was told that the input pins control the current at the output pins. To regulate the current to (0.7A) at the output (Y) pins, do a PWM at the input pins (A).

If this is true, how do you know what duty cycle would result in a 0.7A current (how is that calculated)?
Secondly, if I don't use PWM on any of the input pins and connect Vcc2 to a 12V DC power supply, what would the current be at my input (A) and output (Y) pins?

 

[jim hardy]

It's always advisable to find out what sort of circuit lies just inside the pins of your IC.
That way you can cross check your understanding of the rating against your understanding of circuits until it makes sense.

Here's the section of that TI datasheet that answers your question

that's known as "Totem Pole" output for obvious reasons...
it can either source(supply) or sink(accept) current depending on which transistor is switched on top or botttom..

Q: How does current get from Vcc2 to Output ?
A: Through that top transistor.
Q: What determines how much current will flow when that top transistor is switched on ?
A: Whatever is connected to output. Ohm's law.

I'd say from 'absolute maximum current' rating that top transistor can handle 1 amp continuous but 2 amps for only the briefest of instants.
Ditto for bottom transistor hence the dual + and - prefix to Io .

Observe from figure 1 of section 7.6,
at 1 amp there's about a 1.5 volt drop between Vcc2 and output
which means there's a watt and a half produced inside the IC, doubtless in that output transistor
and from section 7.3 junction to thermal impedance is 60 degC/w
so that watt+ will heat the device by perhaps as much as 90 degC
what if all 4 channels are at max Io ?
there's mention of internal thermal protection in section 7.1 note 3
so observe paragraph 12.1 and make a big fat trace for heat removal.

Spend some hours studying datasheets line by line until you understand every single entry.
Then you'll be able to speedread them.

old jim

 

[meBigGuy]

The current is not the spec you need to understand (other than not exceeding the maximum or getting too hot). You need to understand the output voltage at a given current and how it might vary from device to device.

Let's just talk about output high voltage for a bit.
You have a constant 50 ohm load.
From the data sheet, the minimum High-level output voltage, when driving 1A is Vcc2 - 2 (at free air temp range). It says nothing about the maximum. But, at a JUNCTION temp of 25C (essentially cold) it will vary between a min of VCC2-1.8 to a typical of VCC2-1.4. Still no mention of the max.

Now, if you are not driving a 1A, you need to go to the graphs to see how it varies (but those may be typical, so be careful).

You then do the same exercise for the output low voltage/current.

All the while you need to be thinking in terms of the output circuit design as described by Jim.

It is a puzzle to design a circuit that works reliably over voltage. temperature, and chip-to-chip process variations. The datasheet seldom gives you all the characterization data you really need, so you need to make some probabilistic assumptions and fully understand the tables and graphs.

Regarding PWM, you need to calculate the current at 100% on. Then, obviously, at 50% on you will average half. But the peaks will still be the same.

Once you get sick of reading the datasheet over and over, you are probably nearing enlightenment.

 

[jim hardy]

I'm driving 5Ohm resistance hand-wound electromagnets at the output.

I was told that the input pins control the current at the output pins. To regulate the current to (0.7A) at the output (Y) pins, do a PWM at the input pins (A).

If this is true, how do you know what duty cycle would result in a 0.7A current (how is that calculated)?

Start simple.
If 50 ohms is your DC resistance
then what continuous DC voltage would produce 0.7 amps ?Continuously on is 100% duty cycle..