I do not have visio installed yet (or any graphics software) so I'm hoping you'll understand the ascii diagram 
===== the circuit ===============
12V 3.12A transformer => bridged diodes => .22ohm 5W resistor => 10mF 16V => .22ohm 5W => 10mF 16V => LM1084 => 10mF 16V
=====================
LM1084 is similar to LM317 except that it has a lower voltage dropout and has a max current rating of 5A.
I have added the .22ohm resistors to drop some voltage to reduce the voltage that will bed dropped in the LM1084. It should also allow me to use a smaller heatsink. I will increase/decrease this .22ohm resistor as required.
The R2 (the one connected to the adj pin and to ground) in the LM1084 will be bypassed by a 47uF cap.
this circuit will drive an 8V 2.55A load.
thank you.
ps. please assume that the necessary resistors are connected to the LM1084 to produce the desired 8V.
===== the circuit ===============
12V 3.12A transformer => bridged diodes => .22ohm 5W resistor => 10mF 16V => .22ohm 5W => 10mF 16V => LM1084 => 10mF 16V
=====================
LM1084 is similar to LM317 except that it has a lower voltage dropout and has a max current rating of 5A.
I have added the .22ohm resistors to drop some voltage to reduce the voltage that will bed dropped in the LM1084. It should also allow me to use a smaller heatsink. I will increase/decrease this .22ohm resistor as required.
The R2 (the one connected to the adj pin and to ground) in the LM1084 will be bypassed by a 47uF cap.
this circuit will drive an 8V 2.55A load.
thank you.
ps. please assume that the necessary resistors are connected to the LM1084 to produce the desired 8V.
Jarthel-
What is your load? At 12V (X1.41) you will have ~17V at the first filter cap, and 16.5V at the second one, and across your LM1084, you will have 7.84V @ 2.55A for a power dissipation of 20.01W.
Basically, you have an R-C-R-C filter before the regulator, which is good for low-pass filtering. BTW, are your caps 10mF or 10mF? Or 1mF (1000mF)? This will, of course, affect the cut-off frequency of your LP filter.
Have you considered a SimpleSwitcher IC, such as the LM2576 (3A) or MC33167 (5A)? Both are extremely easy to implement and would offer considerably better than the current 54% efficiency. This is just a thought, but with proper filtering and shielding, you can realize a high quality power supply without all the heat dissipation and thermal issues associated with a linear power supply. In their simplest form, they would require one diode, one coil, and two caps (input & output). Variable versions use the same voltage divider technique as the LM1084.
Of course, if noise is of paramount consideration, then a linear will most likely be the way to go. Anywho, if you're interested, let us know, and we can post some links to datasheets on many of the switching regulators out there.
Cheers,
Steve
What is your load? At 12V (X1.41) you will have ~17V at the first filter cap, and 16.5V at the second one, and across your LM1084, you will have 7.84V @ 2.55A for a power dissipation of 20.01W.
Basically, you have an R-C-R-C filter before the regulator, which is good for low-pass filtering. BTW, are your caps 10mF or 10mF? Or 1mF (1000mF)? This will, of course, affect the cut-off frequency of your LP filter.
Have you considered a SimpleSwitcher IC, such as the LM2576 (3A) or MC33167 (5A)? Both are extremely easy to implement and would offer considerably better than the current 54% efficiency. This is just a thought, but with proper filtering and shielding, you can realize a high quality power supply without all the heat dissipation and thermal issues associated with a linear power supply. In their simplest form, they would require one diode, one coil, and two caps (input & output). Variable versions use the same voltage divider technique as the LM1084.
Of course, if noise is of paramount consideration, then a linear will most likely be the way to go. Anywho, if you're interested, let us know, and we can post some links to datasheets on many of the switching regulators out there.
Cheers,
Steve
N-Channel said:Jarthel-
What is your load? At 12V (X1.41) you will have ~17V at the first filter cap, and 16.5V at the second one, and across your LM1084, you will have 7.84V @ 2.55A for a power dissipation of 20.01W.
Basically, you have an R-C-R-C filter before the regulator, which is good for low-pass filtering. BTW, are your caps 10mF or 10mF? Or 1mF (1000mF)? This will, of course, affect the cut-off frequency of your LP filter.
Have you considered a SimpleSwitcher IC, such as the LM2576 (3A) or MC33167 (5A)? Both are extremely easy to implement and would offer considerably better than the current 54% efficiency. This is just a thought, but with proper filtering and shielding, you can realize a high quality power supply without all the heat dissipation and thermal issues associated with a linear power supply. In their simplest form, they would require one diode, one coil, and two caps (input & output). Variable versions use the same voltage divider technique as the LM1084.
Of course, if noise is of paramount consideration, then a linear will most likely be the way to go. Anywho, if you're interested, let us know, and we can post some links to datasheets on many of the switching regulators out there.
Cheers,
Steve
The load will be 8V 2.55A or 3.13ohms. The load will be tube heaters.
are you saying a 12V transformer is too much? maybe a 9V will do? I have asked about this in another forum and they have said 9V is "cutting it close". maybe they are wrong?
I have made a bridged rectifier powers supply before. It was a 24V 1A. And after the 1st cap, the voltage was around 30+V. but the current requirement is not as large as this new one.
I was looking at the hammond website and it says
============
average voltage after the 1st cap: .9 x Vsecondary
peak voltage after the 1st cap: 1.41 x Vsecondary.
================
At least according to them, 1.41 is only for peak and I would not like to build a PSU using peak values.
noise is not of importance but I already have the LM1084. and the switcher IC you mentioned aren't available locally.
Jarthel,
I gotta concur with cerrem. Heat dissipation alone will be a problem. We don't want to sound like we don't want to help, on the contrary, it's feedback like this from forum members that can prove to be a BIG help, and perhaps enable DIYers to see something from a different perpsective.
If you don't want to order an LM2678, that's fine, especially when National Semiconductor will sample them for free. That's the best price anywhere!
I gotta concur with cerrem. Heat dissipation alone will be a problem. We don't want to sound like we don't want to help, on the contrary, it's feedback like this from forum members that can prove to be a BIG help, and perhaps enable DIYers to see something from a different perpsective.
If you don't want to order an LM2678, that's fine, especially when National Semiconductor will sample them for free. That's the best price anywhere!
N-Channel said:Jarthel,
I gotta concur with cerrem. Heat dissipation alone will be a problem. We don't want to sound like we don't want to help, on the contrary, it's feedback like this from forum members that can prove to be a BIG help, and perhaps enable DIYers to see something from a different perpsective.
If you don't want to order an LM2678, that's fine, especially when National Semiconductor will sample them for free. That's the best price anywhere!
I tried getting samples for free months ago and I never got it.
I'll try again.
Hi,
It is very normal for capacitor input supplies to use the peak voltage value and design for it's consequences.
Your 9Vac transformer will develop 12.7Vpk before the rectifier.
A bridge will take out 1.4V to 1.6V leaving 11.3 to 11.1Vpk.
From this you need to subtract the ripple to find the minimum voltage into the regulator. If your caps are small then the ripple must be higher. Lets put in 1Vpp of ripple leaving about 10.1V (in the waveform trough) into the regulator. You have just 2.1V drop across the regulator. The average voltage drop is half the ripple above this i.e. 2.1V +0.5V=2.6V. Pd=2.6V*2.6A~=7W.
Unfortunately you have to now look at the peak voltage when the transformer is delivering the current you are actually using and re-do the calculations for that and check your regulator dissipation.
Then re-calculate for the maximum mains supply voltage and also calculate for the minimum mains supply voltage. This is the senario that may rule out the 9Vac transformer. 10Vac would probably do the job.
Where do you find a 10Vac transformer? Add a few turns to the secondary of a toroid if you must. But the jump to 12Vac might kill your regulator.
BTW the DC current out of your supply needs to be factored by two (2) to give you the nescessary AC current in your tranformer to take account the various factors that determine the VA rating you really need. So 2.6Adc needs 5.2Aac resulting in a 10Vac 52VA transformer or transformer winding.
I think this applies to a choke regulated PSU. It certainly does not apply to a capacitor input PSU.
applies to all capacitor input PSUs.
It is very normal for capacitor input supplies to use the peak voltage value and design for it's consequences.
Your 9Vac transformer will develop 12.7Vpk before the rectifier.
A bridge will take out 1.4V to 1.6V leaving 11.3 to 11.1Vpk.
From this you need to subtract the ripple to find the minimum voltage into the regulator. If your caps are small then the ripple must be higher. Lets put in 1Vpp of ripple leaving about 10.1V (in the waveform trough) into the regulator. You have just 2.1V drop across the regulator. The average voltage drop is half the ripple above this i.e. 2.1V +0.5V=2.6V. Pd=2.6V*2.6A~=7W.
Unfortunately you have to now look at the peak voltage when the transformer is delivering the current you are actually using and re-do the calculations for that and check your regulator dissipation.
Then re-calculate for the maximum mains supply voltage and also calculate for the minimum mains supply voltage. This is the senario that may rule out the 9Vac transformer. 10Vac would probably do the job.
Where do you find a 10Vac transformer? Add a few turns to the secondary of a toroid if you must. But the jump to 12Vac might kill your regulator.
BTW the DC current out of your supply needs to be factored by two (2) to give you the nescessary AC current in your tranformer to take account the various factors that determine the VA rating you really need. So 2.6Adc needs 5.2Aac resulting in a 10Vac 52VA transformer or transformer winding.
Inductor Values
Jarthel,
If those induators are in the value ranges specified by the National nomographs in the datasheet, then, yes, they will do.
The LM267X are available in the standard 5-pin TO-220 thru-hole mount, a 5-pin surface-mount version. This is basically the TO-220 with a much smaller metal tab for direct soldering to the main pc board and pins bent for surface mounting. I have used the surface mount version because they eliminate the need for the thru-hole drilling step of the fabrication process, and are easier to work with.
In answer to your heatsinking question, yes, you can use them (most of the time) without the need for a heatsink. My most efficient LM2677 ckt ran so cool and efficient, that the only time the chip got warm was when it picked up heat from my finger when I touched it.
If you are still having trouble with sampling from National, let me know, and I will sample a few from them, and send them to you in Australia. There's more than one way to skin a cat.
Steve
Jarthel,
If those induators are in the value ranges specified by the National nomographs in the datasheet, then, yes, they will do.
The LM267X are available in the standard 5-pin TO-220 thru-hole mount, a 5-pin surface-mount version. This is basically the TO-220 with a much smaller metal tab for direct soldering to the main pc board and pins bent for surface mounting. I have used the surface mount version because they eliminate the need for the thru-hole drilling step of the fabrication process, and are easier to work with.
In answer to your heatsinking question, yes, you can use them (most of the time) without the need for a heatsink. My most efficient LM2677 ckt ran so cool and efficient, that the only time the chip got warm was when it picked up heat from my finger when I touched it.
If you are still having trouble with sampling from National, let me know, and I will sample a few from them, and send them to you in Australia. There's more than one way to skin a cat.
Steve
Re: Inductor Values
1. I got some samples. thank you for telling me and thank you for the offer. It seems some parts are free shipping while others are not.
2. In regards to chokes, I assume I can't use chokes like common mode/RF and others in the circuit. right? I can only use "power chokes"? Looking at the 2 biggest electronic place in the country, the best I can find are SMT parts.
3. Do I need to make my own PCB?
4. for parts substitution: for example, the suggested Panasonic FC (560uF 50V) is not available locally. I can use a different manufacturer? I assume the ESR and ripple must be same or at least close to each other?
N-Channel said:Jarthel,
If those induators are in the value ranges specified by the National nomographs in the datasheet, then, yes, they will do.
The LM267X are available in the standard 5-pin TO-220 thru-hole mount, a 5-pin surface-mount version. This is basically the TO-220 with a much smaller metal tab for direct soldering to the main pc board and pins bent for surface mounting. I have used the surface mount version because they eliminate the need for the thru-hole drilling step of the fabrication process, and are easier to work with.
In answer to your heatsinking question, yes, you can use them (most of the time) without the need for a heatsink. My most efficient LM2677 ckt ran so cool and efficient, that the only time the chip got warm was when it picked up heat from my finger when I touched it.
If you are still having trouble with sampling from National, let me know, and I will sample a few from them, and send them to you in Australia. There's more than one way to skin a cat.
Steve
1. I got some samples. thank you for telling me and thank you for the offer. It seems some parts are free shipping while others are not.
2. In regards to chokes, I assume I can't use chokes like common mode/RF and others in the circuit. right? I can only use "power chokes"? Looking at the 2 biggest electronic place in the country, the best I can find are SMT parts.
3. Do I need to make my own PCB?
4. for parts substitution: for example, the suggested Panasonic FC (560uF 50V) is not available locally. I can use a different manufacturer? I assume the ESR and ripple must be same or at least close to each other?
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