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First build linear PSU help - load balancing outputs.
First build linear PSU help - load balancing outputs.
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Old 23rd September 2018, 01:44 PM   #1
atca is offline atca
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Question First build linear PSU help - load balancing outputs.

Hello

This is my first project. I am planning to build a linear power supply to power a Raspberry Pi 3 and also an Allo Katana DAC.

The RPi needs 5v 2.5A, the Allo Katana needs 5v 1A.

I plan to buy two of these modules, which will give me 4 DC outs:

0.56uV Ultralow noise DAC power supply regulator 3.3/5/7V 2A*x2 - DIYINHK

I appreciate this is minimal DIY but I don't have the knowledge / skill / time to completely design the whole PSU from scratch. I am also aware that additional LT3045 can be added to these circuits - but I don't have the skill / tools to solder SMT chips.

To get the 3A I need, I plan to parallel the output of 3 DC outs which individually produce 5v 1A, to create a 5V 3A rail for the RPi3. The 4th DC out will be used standalone giving 5V 1A for the Katana.

I understand to parallel the outputs that I should use resistors to load balance the three separate supplies. I understand that I need around [CORRECTION 250mOhm] on each output, typically consisting of lower value resistors.

I currently plan to parallel 4 x [CORRECTION 1Ohm] resistors on each output. Thereby 12 resistors in total, which will be placed in parallel.

Questions

1) Why is it recommended to use multiple [CORRECTION 1Oh] resistors as opposed to a single [CORRECTION 250mOhm] resistor?

2) Assuming 1Ohm resistor - what rating do they need to be?

5v3A = 15w . Which spread over 12 resistors ~1.25w per resistor. Not sure the math works like that but I assume I need resistors of >1w.

3) All the 250mOhm >1w resistors I've seen are wirewound, are these suitable for this application?

4) Would I be better with film - what is a suitable film resistor?

Thanks for any guidance.

Also does anyone know the AC input voltage to those power supply regulators? The website specs don't say.

Last edited by atca; 23rd September 2018 at 04:46 PM.
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Old 23rd September 2018, 02:37 PM   #2
atca is offline atca
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Basic example here, I've skipped all of the components around the LT3045 but hopefully this illustrates in principle what I said above.

Click the image to open in full size.
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Old 23rd September 2018, 03:50 PM   #3
FauxFrench is offline FauxFrench  France
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Quote:
Originally Posted by atca View Post
Basic example here, I've skipped all of the components around the LT3045 but hopefully this illustrates in principle what I said above.
I cannot load your photo.

The LT3045 chips can handle up to 20V on the input. I guess it is the same for the board. A 9Vac should be fine. Does the board include input voltage rectifiers?

1) Four 0.250 resistors in parallel do not bring 1 Ohm but close to 0.063 Ohm.

2) The loss in each resistor is the square of the maximum current in each resistor times the resistance value.
The 5V3A is the output power, no?

3) You have cement and metal-film types as well.

4) Metal film - 0.5W/1W/2W
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Old 23rd September 2018, 04:44 PM   #4
atca is offline atca
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Click the image to open in full size.

1) Thanks soory I have that back to front. The load balancing should be 0.250Ohm using 4* 1Ohm resistors in parallel.

Question still stands now corrected why are 4*1Ohm resistors better than 1*0.250Ohm resistor?

2) 5v out load balanced over 4 * 1Ohm resistors, v/r=i so 5v/1Ohm = 5A per resistor.

Power = I * V.
Power = 5A x 5V = 25W per resistor.

Power = I^2 * R
Power loss = 5A ^ 2 x 1Ohm = 25W per resistor.

This seems very high?! surely I don't need resistors rated for 25W for each load balancing resistor. Especially since given total power out across all three modules = 5v3A = 15W.

3) Yes I know there are ceramic and metal film. I understand ceramic are noisy. Am I better off using wire wound or metal film for this application?

4) eek need to reconcile this with my calcs in 2)
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Old 23rd September 2018, 07:43 PM   #5
FauxFrench is offline FauxFrench  France
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I still have problems with your photo but never-mind for now.

The description of the power supply boards is not the most precise I have seen but I arrive at the same conclusion as you: You get two boards, each with two outputs of 1A. On each board there is two LT3045 chips with two 0.5A outputs already "bundled" to 1A. So I agree with you, you will have 4 outputs of 1A.

1) Balance resistors. 0.25 Ohm for each 1A output seems likely though higher than what AD uses in their examples (20mOhm and 50mOhm).
With 0.25Ohm and 1A the power loss is 1A*1A*0.25Ohm= 0.25W. It is practical to use four 1Ohm metal-film resistors in parallel because 1Ohm is the lowest regular 0.5W value and metal-film resistors have little noise.

2) No, you are not short-circuiting the output. The balancing resistors go in series with the load. Luckily no 25W. Else, you would current-overload the 1A regulator modules heavily. The 0.25 Ohm balancing resistors add up to 0.25V on top of the 5V.

3) As you can use ordinary 0.5W metal-film resistors in parallel, it is the best. Even 1/3W or 1/4W resistors can be used in parallel.

4) You NEVER get 5V across the balancing resistors unless you short-circuit the output. Then, the regulator current limiters will be invoked and you won't get 5V anyway.

Last edited by FauxFrench; 23rd September 2018 at 07:50 PM.
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Old 23rd September 2018, 07:48 PM   #6
atcapollo is offline atcapollo  United Kingdom
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Try psrr load balance load example - Imgur

My plan was 4 balancing resistors per 1A output so 12 resistors across the 3 outputs.

Fourth output does not need balancing as used directly.

Quote:
Originally Posted by FauxFrench View Post
I still have problems with your photo but never-mind for now.

The description of the power supply boards is not the most precise I have seen but I arrive at the same conclusion as you: You get two boards, each with two outputs of 1A. On each board there is two LT3045 chips with two 0.5A outputs already "bundled" to 1A. So I agree with you, you will have 4 outputs of 1A.

1) Balance resistors. 0.25 Ohm for each 1A output seems likely though higher than what AD uses in their examples (20mOhm and 50mOhm).
With 0.25Ohm and 1A the power loss is 1A*1A*0.25Ohm= 0.25W. It is practical to use four 1Ohm metal-film resistors in parallel because 1Ohm is the lowest regular 0.5W value and metal-film resistors have little noise.

2) No, you are not short-circuiting the output. The balancing resistors go in series with the load. Luckily no 25W. Else you would current overload the 1A regulator modules heavily. The 0.25 Ohm balancing resistors add up to 0.25V on top of the 5V.

3) As you can use ordinary 0.5W metal-film resistors in parallel it is the best. Even 1/3W or 1/4W resistors can be used in parallel.

4) You NEVER get 5V across the balancing resistors unless you short-circuit the output. Then the regulator current limiters will be invoked and you don't get 5V anyway.
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Old 23rd September 2018, 07:49 PM   #7
atcapollo is offline atcapollo  United Kingdom
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Where can I see the AD examples?
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Old 23rd September 2018, 08:40 PM   #8
FauxFrench is offline FauxFrench  France
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Quote:
Originally Posted by atcapollo View Post
Where can I see the AD examples?
No need to balance the last 5V/1A output - correct.

Examples:
LT3045 Parallel Devices for 5Vin to 3.3Vout @ 1A with Ultralow Noise Circuit Collection | Analog Devices
Paralleling Multiple LT3045s for 2A Output Current Circuit Collection | Analog Devices
Paralleling Multiple LT3042s for Higher Output Current Circuit Collection | Analog Devices

Last edited by FauxFrench; 23rd September 2018 at 08:43 PM.
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Old 23rd September 2018, 09:23 PM   #9
atcapollo is offline atcapollo  United Kingdom
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That's great thanks really helpful so they use a single 20mOhm resistor.

Wonder why this article, third diagram, High Current Voltage Regulation | REUK.co.uk suggests using 4 resistors per output?

What voltage loss will I see from a 20mOhm resistor on the 5v out?

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Old 23rd September 2018, 09:30 PM   #10
atca is offline atca
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Quote:
Originally Posted by FauxFrench View Post
... The 0.25 Ohm balancing resistors add up to 0.25V on top of the 5V.
Can you explain this a bit more?

How does 0.25Ohm resistors add up to 0.25V ?

How can it be on top of the 5V the LT3045 modules only output 5V ?
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