# SN754410 output supply current - heating

I am experiencing excessive heating problem with SN754410NE (SN754410). I know that I can use heat sink, but I want to be sure that I am not doing something wrong.

I am using SN754410NE to drive a bipolar stepper motor. My circuit is similar to schematic below. The current rating of the stepper motor is about 0.6 A and since continuous output current is given as 1.1A in the data sheet, I don't expect excessive heat.

However, the other issue is output supply current is given as 70mA in the data sheet. So, for example, if I energize one solenoid of motor output current of '1out' pin will be 0.6 A (which is below rating) however '2out' pin supply current will be also 0.6 A (which is above rating). Am I correct? Can this cause excessive heat?

(source: g9toengineering.com)

• What kind of heat sink are you using? Commented Mar 16, 2014 at 17:44
• I don't use any now. It heats up to around 60-70 C in 10 secs I guess. I just don't know if it is normal or not... Commented Mar 16, 2014 at 17:47
• The chip can handle up to 2W of dissipation, but only if it has a proper heatsink. Commented Mar 16, 2014 at 19:01

According to the data sheet, at 1A, the typical voltage drop per half bridge will be 2.6V. Assuming you have two half bridges energized 100% of the time, the total power dissipation at 0.6A (assuming the same voltage drop, which is a bit pessimistic, but they are 'typical' values, not guarantees) is 3.1W.

That doesn't even include the 20 to 70mA of supply current, it's just the output drop times the current. At 24V/5V, the additional power dissipation due to the supply current (half high, half low) is 0.27W max (from 5V) and 0.64W max (from 24V) for a total of 4W.

That's clearly in excess of the absolute maximum power dissipation rating of 2W at 25°C free-air $T_A$, so you're way over what that chip is capable of safely handling. If you want it to work with $T_A$ = 50°C you need to keep power dissipation under 1.7W, as per "Note 2".

Always carefully read the fine print in datasheets. When they say "1A capability per driver" that often means that you can only use one driver at once, and maybe not continuously, and often not at the maximum temperature.

• Thank you Spehro, It is very enlightening. I would like to ask you to give a little information about what is output supply current (20 to 70 mA). BTW, what does continuous total power dissipation means? Does it mean that I can keep the chip @25C with 2W power dissipation or do not exceed 2W? and how did you get 1.7W:) Commented Mar 16, 2014 at 21:51
• OK I get it =) 2075 - 25x16.6 = 1660 mW So as the temperature increases power dissipation decreases. So, I should not exceed 2W @25C otherwise It is getting worse and worse... Thanks again... Commented Mar 16, 2014 at 22:40

The 754410 is not a very good motor or stepper motor drive circuit because it will generate a lot of heat. Look at Voh and Vol on page 4. They say, in effect the following. If you are taking half an amp from a high output pin expect the maximum voltage produced at that pin to be Vcc - 1.5V. If you are consuming half an amp into an output pin expect that pin not to be at 0V but be dragged as high as 1.4V.

So if you wire two outputs as a H bridge drive and the load takes 0.5A, expect the total volt drop to be a maximum of 2.9V or typically 2.1V. What does this mean in terms of self-heating? 0.5A x 2.9V = power loss of about 1.5 watts (maximum) or 1 watt typically.

Your circuit uses two loads so double this for the total power loss.

Now do you see that you ought to consider a heatsink? Actually I'd avoid using the chip altogether - it's not very efficient at all and if you are running on the low supply voltage, there will be barely enough output voltage to drive a small/medium DC motor.

Look at the type of output circuit this device uses (lowest in picture below): -

Neither the upper or lower output transistor will ever really be turned-on properly by their drive circuits.

• I am using AVR chip to create logic and need a H-bridge to connect it directly. So, what do you suggest for H-bridge to drive 1-2 Amp stepper motors(L298?)? Commented Mar 16, 2014 at 21:57
• You need MOSFETs to do the power switching - try something like this ti.com/lit/ds/slvsar1c/slvsar1c.pdf and read page 6 - on resistance at 0.5A is about 0.2 ohm - that's 0.1V dropped per power transistor. Commented Mar 16, 2014 at 22:06

Just glancing over the datasheet I see the derating figure of $16.6\frac{mW}{^{\circ}{\rm C}}$. And I also see that it can drop up to $2V$ at $1A$. That's worst case, but with the bipolar stepper you are probably operating both coils at once. So you could get four drops to contend with. (The device shows BJTs wired with two $V_{be}$ drops apiece, high and low sides, so I'd expect perhaps on the order of perhaps $1.5V$ as a high end guess.) Note that with the derating figure applied you could have on the order of $\frac{4\cdot 1.5V\cdot 600mA}{16.6\frac{mW}{^{\circ}{\rm C}}} > 200^{\circ}{\rm C}$ change in temperature. You probably get less than that, but the order of magnitude is there. That's a lot.

You fell for the "absolute maximum ratings fallacy". NEVER use the figures in that section for design calculations. Those figures document the maximum stress a chip can SURVIVE for a SHORT (but undefined) TIME. Note that survive does NOT imply functioning.

Always base your design on the "(recommended) operating conditions" section. And as Spehro noted, don't use the 'sales talk' summary at the beginning of a datasheet. It's just that, sales talk. The figures might be OK, but they will omit the circumstances (like a 0-ohm drive, perfect cooling, typical, without long term drift, at 1% duty cycle, etc.).