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LP2980IM5-3.3 current limitations

Tim Anderson1
Prodigy 120 points
Other Parts Discussed in Thread: LP2980-N

First of all, thank you in advance for any help you can provide.

I am interested in using the LP2980IM5-3.3 (3.3-volt) in a product we are developing because of its very low dropout voltage and very small (SOT-23) size. Unfortunately, its current limitation is 50ma. However... according to the data sheet (Figure 7) this can be increased using pass transistor. We only need a maximum of 60ma, but in our application the input voltage could vary from 0 to 16-volts DC. Assuming worst case (16V) if a pass transistor were used does that mean it would have to dissipate 960mw (16V x 60ma)? or would the LP2980 handle the 50ma and the pass transistor the difference. If so, could we use an MMBT4403 (also a SOT device)? I apologize if this is a stupid question, but I am not an expert in this area. As an alternative, could we used two LP2980 devices in parallel (common ground and common output and let them share the load?

Again, thank you for any help/guidance you can provide.

Best regards, Tim

  • 0
    TI__Mastermind 22610 points

    You actually have two problems here: 1) Thermal, and 2) Current.

    The thermal problem is, by far, the biggest issue. The current problem, although inter-connected to the thermal problem, is minor by comparison.

    By far, the easiest solution is a different device with a package have better thermal performance (θja < 100°C/W), and higher output current, like : LP2950CDT-3.3.

    But, for this discussion I will presume that you are locked into the LP2980 and the SOT-23 package.

    If all the current goes through the LP2980 the max dissipation during normal operation would be:

     ( (Vin-Vout) x Iout  ) = ( (16V-3.3V) x 0.06A ) = 762mW.

    The value of θja for the SOT-23 package is 220°C/W (per Note 3 in the LP2980-N data-sheet,SNOS733L) , so with external ambient temperature of 25°C, the internal die temperature (Tj) would reach:

     Tj = (Ta + (θja x Pdiss) = (25°C + ( 220°C/W x 0.762W) = 192°C

    Yes, you can add a parallel pass transistor, but the biasing of the Vbe depends on a series resistor on the input and some targeted minimum amount current through the LP2980 (input pin to output pin). The end result is that the low dropout voltage specification of the LP2980 is no longer valid since a series IxR element (up to ~700mV for pnp Vbe has been added on the input. This takes away a key feature that you highlighted. Dissipation will be shared between the LP2980 and the external transistor. But, as the transistor heats from the initial dissipation, and Vbe drops, more of the current shifts to the transitor, etc, etc.

    Adding a parallel LP2980 might be a better choice, but I don't reccomend that the two outputs simply be tied together as this will result in the one LP2980 with the lower output voltage (even by a few milli-volts) supplying the vast majority of the the output current. Typically this current sharing problem is mitigated with a series ballast resistor on each output pin. The downside to using the ballast resistors is that they introduce a load regulation issue which might, or might not, be a concern for your application.

    http://www.ti.com/lit/an/slva250/slva250.pdf

    Even with ~robust~ ballast resistor values it is unlikely that better than 60%/40% current sharing (40mA through the LP2980 with the lower Vout, and 20mA through the LP2980 with the higher Vout) can be achieved, unless you can match the LP2980 output voltages to within a few milli-volts.

     

  • 0 in reply to Donald Jones
    Prodigy 120 points

    Hi Donald,

    Thank you very much for your speedy reply. Your explanation is very clear. I understand the math and it is very helpful. It appears that if we want to use the LP2980 and our input voltage reaches the 16-volt maximum, the device can handle about 36ma before it goes into thermal shutdown. We may be able to live with that for one application.

    We would use this regulator to provide controlled output for a small PIC that drives several LEDs. The PIC itself only draws 1/2ma so we could safely drive it and two 20ma LEDs if their current is limited to ~16-17ma each. In cases where we would have a third or fourth LED, I'm thinking we could run 2 of the LP2980s in parallel with their inputs and ground common, but the outputs driving LED anodes separately. The LED cathodes would be controlled by the PIC. Bench testing is definately in order to see if this is a viable approach, as I am unsure if their would be any feedback/leakage between the 2 LP2980s through the LEDs and the PIC (common ground).

    Again, thank you for your help, I really appreciate it.