LM138: 5A constant current over a long cable.

Hello Long,

                 I am looking into this, I will get back to you by the end of the day tomorrow. Thanks!

Regards,

Srikanth

Hi Srikanth, thanks for the info.  In that case, is there another TI product or some other method that would allow me to achieve my design goal?  It seems to me like any design would see this high inductance load @ start up yes?  by the way, my load would just be an resistive load (like a 1Ohm resistive load for example).   Not sure that would change the recommendation since we would still have the long inductive cable in between.

Would it help if we have a zener diode on the line/cable to clamp the initial spike?  (just a thought, assuming we could find one that works).

regards,

-Long

Hello Long,

                  The principle issue with the design is the power dissipated across the Die. With V_IN = 28V, the voltage across the load being 5V (5A*1Ω), the drop across the cable being ~9V implies that the output voltage of the LDO is ~14V. The power dissipated across the device is ~70W ((28V - 14V)*5A), which is extremely large. Would it be feasible for you to connect the LDO closer to the 1Ω load end of the cable? If so, then the inductance issue would be bypassed and one possible solution would be to step down the Input voltage of 28V using a Buck converter and feed this voltage to the Input of the LDO. This  will reduce the input voltage (and power dissipation) of the LDO.

                   LM138 requires minimum dropout of 3V, which implies that the power dissipated across the device would be at least 15W (3V*5A). This is still way too large for this device. LM1084 requires 1.5V of dropout (significantly lower than LM138), and using the NDE package on a board with good layout and using heatsink(s) after stepping voltage down with a Buck converter, it may be possible to dissipate power while keeping the Junction temperature less than 125°C. Thanks!

Regards,

Srikanth   

hi Srikanth, I understand the info you provided above regarding the power dissipation and all but I'm a bit confused now of the capability of this part.  

1) The part datasheet indicates that it is a 5A regulator and if a minimum of 3V drop is required, then regardless of any application or loads, the power dissipation is always at the minimum 15W (3V * 5A).   Can the part handle 15W of power dissipation?

2) Figure 1 of the datasheet shows the current limit at different Vin-Vout.   Even at Vin - Vout = 10V (ie the difference between input and output is 10V), the output current is ~8A.   This is peak current @ roughly 100ms or so I guess it is NOT a constant current right?  Figure 2.0 seems to suggest it could be a 'sustained' current but can you confirm?  

3) The application the I am considering is the Constant current as shown in Figure 35.   Since the regulator will do whatever it can to keep the Vout about 1.25V above the Vadj (across the R1v 0.24 Ohm resistor), can you let me know the Vout in this application?  (For two cases: load is next to R1 and load is far away from R1).

***[LN] In this case I think you're saying it would depend on my load and the drop across the cable, hence your answer above.  I think I get this point so you can ignore this question #3.  Thanks.

***I will take a look at the LM1084 datasheet.  Thanks!

Q: Can I use LM1084 as a 5A constant current source in the same manner as the LM317? 

Anyhow, I tend to agree with you that I need to lower the input voltage to reduce the power consumption but I need your clarification above.  

thanks,

-Long

Hi Long,

Srikanth is out today. He will be back tomorrow to carry on the conversation. 

Regards,

Nick

Thanks Nick!  

Thanks Srikanth, will do that.  Could you confirm if I can use LM1084 in the same manner as shown in Figure 37 of the LM138 datasheet?  I would assume so but just want to confirm.

thanks!