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This question is related to this question: udn2981-maximum-output-current-too-less.

In the UDN2981 datasheet, the picture below shows the maximum current (page 14, bottom left picture).

Since if I would would use 8 mosfets/transistors to switch, I can get 200 mA per pin (using e.g. a 2N7000 per pin), and an UDN2981 can 'only' handle 400 mA in total, and even less if multiple pins are used (like with 8 pins, only about 120/8 = 15 mA (at 100% duty cycle).

Why is there a relation between the pins? Is it related to heat?

enter image description here

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Yes, it's because of heating. All the transistors are on one die so they all contribute to heating.

Note the "at temperature 50°C" on the Y-axis. If your maximum ambient could exceed the rather low value of 50°C you should derate those numbers.

Also this graph will apply specifically to the stated package type, as thermal behavior is different for different packages.

You should stay well away from "maximum recommended" values if you care about long-term reliablity.


A physically small MOSFET with low Rds(on) can handle much more current because the voltage drop across it when 'on' is much less than with the Darlington output stage on the UDN2981. For example, an AO3400 has less than 50m\$\Omega\$ Rds(on) with 4.5V drive, so at 1A the power dissipation is \$I^2R\$ or 50mW. The UDN2981 at 350mA can drop as much as 2V so 700mW.

At equal current (say 225mA) typical dissipation on the UDNxxx is 383mW (one output) vs. about 1mW for the AO3400! (note that I'm comparing N-channel MOSFETs with a high-side driver, but a P-channel would be very similar)

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  • \$\begingroup\$ Thanks for this clarification ... I guess I will solder a bit more and just use mosfets. \$\endgroup\$ – Michel Keijzers Mar 7 '19 at 12:22
  • \$\begingroup\$ I created a new question (electronics.stackexchange.com/questions/426084/…), however, is my calculation right? only 15 mA per pin? or is it around 120 mA per pin I can 'sink' ? \$\endgroup\$ – Michel Keijzers Mar 7 '19 at 13:55
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There are two factors involved here. One is the total power dissipation allowed for the package, because this power is transformed to heat. Because it is the average power dissipation that matters, over the entire package, the package power is proportional to duty cycle, times the number of simultaneous outputs switching.

The second factor is that all of the current that is sourced by the outputs must enter the package through a single pin, the Vs pin. So, the conductors that form that pin and the internal distribution of current have a limited current capability. Even if the current is below a hard-failure level, the IR drop on these paths can cause the device to fall out of spec, so the manufacturer has to put a conservative limit on the total amount of current sourced by simultaneously driving pins.

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  • \$\begingroup\$ Thanks for this answer; another reason (the single pin) to use separate mosfets instead of a dedicated source driver. \$\endgroup\$ – Michel Keijzers Mar 7 '19 at 12:23

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