LM317, 1A 12VDC Regulator

[drkidd22]

Hello,

I'm working on a design for a 12VDC Regulator that can be able of handling 1A.
I've already picked my resistors and they will be:
174Ohm for R1
1.5K for R2.

According to the data sheet I should get:
Vout = Vref (1+(R2/R1)) + IadjR2.

I'm not sure what the IadjR2 is, but I've seen many people just using Vout = Vref (1+(R2/R1)).

So for my design my calculated Vout should be: 12.03VDC

120VAC is input to a 10VA transformer's primary coils which are in parallel.
The secondary of the transformer are 14V each and are wired in parallel as well.
The output from the transformer is fed into DF10M rectifier and 470uF capacitor and 0.1uF are used at the input of the LM317.

My biggest concern right now is that I need help calculating the power dissipation of the LM317 at a load of 1A.

The data sheet states the below formula:
PD = ((VIN − VOUT) × IL) + (VIN × IG)

Vin = voltage from df10m rectifier
Vout = my calculated Vout: 12.03VDC
IL = My current load requirement?? 1A?
Not sure hot to calculate IG from.

TR(MAX) = TJ(MAX) − TA(MAX)
For my design:
TR(MAX) = 125C - 40C
TR(MAX) = 85C

θJA = (TR(MAX) / PD)
For my design:
Need help with this part, need to figure out what Pd is first.
If you guys can give me a hand calculating Power Dissipation for this circuit I will appreciate it.

 

[alec_t]

An approximation, ignoring winding resistance and ripple, goes like this.
Peak voltage out from the secondaries is 14 x 1.414 = ~ 20V. The rectifier bridge drops ~ 1.2V. So the smoothed IC input is ~ 20 - 1.2 = ~ 19V, and the voltage dropped by the IC ~ = 19 - 12 = 7V. At 1A current the regulator will therefore be dissipating ~ 7 x 1 = 7W.

 

[ericgibbs]

hi d22,
You say its a 10VA rated transformer, with a double 14Vrms winding.?

You will not be able to draw 1A at 14Vrms from the transformer.

I would estimate ~0.7Amax so at full load the smoothed DC input will be ~14V less 2 diode voltage drops.

Can you confirm that its rated at 10VA.??

 

[alec_t]

You say its a 10VA rated transformer, with a double 14Vrms winding.?
You will not be able to draw 1A at 14Vrms from the transformer.

Oops, I missed that. Well spotted Eric.

 

[drkidd22]

It is a 10VA transformer. 14A-10R-28 from signal.
You are correct, only ~715mA can be sourced from the transformer.
I just reviewed the circuit and what it is going to be used for, the 1A is just a max rating and it's 450mA over the load that will really be constantly applied.
So I only really need 550mA draw from the transformer and I will have 100mA available extra.
Same goes for the 12V since it's going to be where the load will be connected to.

 

[alec_t]

14A-10R-28 from signal

Would you care to explain that, Drkidd?

 

[drkidd22]

That;s the Signal transformer part number.
So I recalculated the numbers.
Vin ~ 19V - 12V = 7V
7V * 0.550A = 3.85W
Guess I will need a heat-sink for this.

Or I can change the transformer to a dual 12V output one, but the power dissipation will still be kinda high.
I'm trying to avoid having to use a heat sink, what's the maximum power dissipation that the LM317 can handle without a heat sink.

With the 12V dual secondary transformer I would get.
12 * 1.414 = 16.97V
16.97 - 1.2V = 15.768
15.768-12V = 3.768
3.768V * 0.550A = 2.0724

 

[ericgibbs]

hi Dr22,
The 10VA txr will not be able to maintain 19V-2*V diode drops at 550mA.

The 19V -1.4V = 17.6 is the Vpeak you could expect when the load current is close to zero, at 550mA I would expect the input to the LM317 to ~15.5V

The 470uF IMO should be increased to 680uF or 1000uF at 25Vwkg.

You will need a h/s for the LM317, a black finned type about 3cm square, with central mounting tab for the fixing screw.

If you have the regulator project in a metal enclosure , you could mount the LM317 onto the enclosure for the h/s.