# LT3012

#### Martino Chiro

Joined May 1, 2015
128
The specs of the LT3012 shown a 250mA max output current.
There is a solution to get more current from this device ?
It is possible to parallel more LT3012 ? Or, put some power transistor in the basic configuration
shown in the data shhet that boost the 250mA limit ?

Thank You

#### crutschow

Joined Mar 14, 2008
33,350
Why not just use a regulator that has a higher current capability?
How much current do you need?
What is the input and output voltage?

#### Martino Chiro

Joined May 1, 2015
128
Thank You Crutschow.
I need the high input voltage (80V) of the LT3012 (48V minimum in my case) and 1A of output current.

#### Alec_t

Joined Sep 17, 2013
14,007
An external transistor could be used, but would have to dissipate, for example, (80-5)*1 = 75W for an 80V input and 1A output at 5V .

#### Dodgydave

Joined Jun 22, 2012
11,149
What output voltage do you need, and what input range do you have?

#### Martino Chiro

Joined May 1, 2015
128
The input voltage is 48V (+/- 10%, and the output voltage required is 24V and 1A continuous.

#### Alec_t

Joined Sep 17, 2013
14,007
So power dissipation by the outboard transistor would be 24W +/- 10%. Still a lot to handle!
A switch-mode converter would be a lot more efficient.

#### mcgyvr

Joined Oct 15, 2009
5,394

#### Martino Chiro

Joined May 1, 2015
128
An external transistor could be used, but would have to dissipate, for example, (80-5)*1 = 75W for an 80V input and 1A output at 5V .
But how it need to be connected to the LT3012 ?

#### mcgyvr

Joined Oct 15, 2009
5,394
But how it need to be connected to the LT3012 ?
How do you intend to handle the dissipation?
Its a total waste trying to use that IC for your needs..

#### Dodgydave

Joined Jun 22, 2012
11,149
What about a Lm2596, max input 40V, you just need to drop from 48 to 40v, takes 3amp output.

#### Martino Chiro

Joined May 1, 2015
128
Interesting the use of he LM2596. But again i need to put a transistor to get a drop of at least 8 - 10V.
It is not possible to parallel more LT3012 ?

#### spinnaker

Joined Oct 29, 2009
7,830
What about a Lm2596, max input 40V, you just need to drop from 48 to 40v, takes 3amp output.
The lm2596 can only have 40 in as I recall but it would not surprise me if TI had a chip capable of 48v in.

#### Alec_t

Joined Sep 17, 2013
14,007
It is not possible to parallel more LT3012 ?
In theory you could, but in practice the ICs will not be exactly matched (even if from the same batch) so they won't share the current equally. One will be taking more than its fair share of the load. You will have to ensure that this share does not cause the IC to go into thermal shut-down. The total power dissipated will still be 24W, which is going to need very effective heat-sinking. As mcgyvr asked, how are you going to handle that? Unless you have a sackful of 3012's looking for a use it will be cheaper and more efficient by far to use a switched-mode supply.

#### benta

Joined Dec 7, 2015
101
The TI LM317HV would fit your needs, I think.

Regards,

Benta.

#### Martino Chiro

Joined May 1, 2015
128
Both the L4978 and the LM317HV are good solutions for me.
When Vin - Vout = 24V the L4978 seems can deliver 2A of output current max.
When Vin - Vout = 24V the LM317HV seems can deliver 1A of output current max
from the Figure 17 "Current Limit" for T and K package type, in the TI data sheet.

#### benta

Joined Dec 7, 2015
101
I don't know the L4978, but I've worked a lot with the LM317/317HV.
Your main concern is thermal management = cooling, and this is true for both devices.
You'll need to dissipate ~24 W to the surrounding air, and that means a relatively large heatsink.
Thermal resistance junction-case is 0.9 K/W for the "T" version, you need to add insulation (mica or silicone) which is around 1 K/W (worst case), let's say 2 K/W from junction to heatsink. This means a temperature rise of 48 Celsius.
Say your maximum surrounding temperature is 35 C (I'm guessing here).
To keep the junction temperature below 125 C, your heatsink needs to have a thermal resistance of:

125 - 35 - 48 = 42, meaning your heatsink must have a thermal resistance of no more than 42/24 = 1.75 K/W

That is quite a big chunk of aluminium, please check the heatsink suppliers' websites for example products.

A solution with a switching regulator or a DC/DC-converter module might be better.

Best Regards,

Benta.