Hi,
I have a power supply that outputs 13.8vdc and I use it to power various devices.

I wish to add another device but need to limit this device to its max of 5a at 12vdc.

I have been toying with the idea of limiting this device with a resistor instead of a buck convertor but need my calculations checked.

So, I have come up with a resistor value of 0.36ohms or 360milliohms ((13.8 - 12)/5).

I then used an online calculator to find the power rating for the resistor and got 4w

Do these figures look correct?

Many thanks.

 

1.8V @5 amp is 9 W,

Personally i would modify the psu to give out 12V.

 

I wish to add another device but need to limit this device to its max of 5a at 12vdc.

Using a resistor to drop voltage only works if the load is constant. Is that the case with your 12V device?

Are you certain that your 12V device won't operate at 13.8V?

 

Two diodes in series to knock the voltage down to 12.4V?
http://www.mouser.com/Semiconductor...Z1yuobgsZ1yuoc28Z1yuoan2Z1yuoc70&Ns=Pricing|0

Or 11.92V with a bridge rectifier?

 

Using a resistor to drop voltage only works if the load is constant. Is that the case with your 12V device?

Are you certain that your 12V device won't operate at 13.8V?

It is rated up to 15v but as the voltage increases so does its current draw & I really need to keep it to 5a

Two diodes in series to knock the voltage down to 12.4V?
http://www.mouser.com/Semiconductor...Z1yuobgsZ1yuoc28Z1yuoan2Z1yuoc70&Ns=Pricing|0

Or 11.92V with a bridge rectifier?

Thanks for the info

 

Look into "Constant Current devices" or "Current Mirror's".

https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/current-mirrors/

 

Thanks to everyone for your input.

I'm now looking at either LM1084-12 or a 78h12 regulator.

As it's a TEC1-12705 device I am trying to regulate for, I have a heat-sink readily available.

Problem I'm finding is these items look like they're obsolete and the 78h12 needs to be the either the ASM or ASP version for below freezing use.

 

Also looking at this: https://uk.rs-online.com/mobile/p/rectifier-schottky-diodes/6311019/

Would this be suitable for my requirements?

 

First, we don't know your requirements. You mentioned something about freezing temperatures, so that's something you'll have to confer with the spec sheet. You can google it and probably get some good enough information as to what it's characteristics are and what temperatures it's rated to perform under. Second, I've not heard of using a diode as a regulator - unless you're talking about a zener diode. I suppose you could use a zener designed for 12 volt regulation (meaning anything over 12 volts will be dumped to ground through a properly sized resistor, sized for resistance and wattage. But there you're wasting energy. Still, you'll have your 12 volts.

If we knew more about what you're trying to accomplish I'm confident there are many people here who could steer you in the right direction. Heck! I might even be able to help. But to help, I'd need to know what you have, what you want and how you expect to get there. So if you care to share a bit more information on what you're doing I'm sure someone here can help you. HEY! That's what we do. Help. No, we don't design for others, we help others figure out their designs, help them understand the flaws and guide them to a happy ending.

 

First, we don't know your requirements. You mentioned something about freezing temperatures, so that's something you'll have to confer with the spec sheet. You can google it and probably get some good enough information as to what it's characteristics are and what temperatures it's rated to perform under. Second, I've not heard of using a diode as a regulator - unless you're talking about a zener diode. I suppose you could use a zener designed for 12 volt regulation (meaning anything over 12 volts will be dumped to ground through a properly sized resistor, sized for resistance and wattage. But there you're wasting energy. Still, you'll have your 12 volts.

If we knew more about what you're trying to accomplish I'm confident there are many people here who could steer you in the right direction. Heck! I might even be able to help. But to help, I'd need to know what you have, what you want and how you expect to get there. So if you care to share a bit more information on what you're doing I'm sure someone here can help you. HEY! That's what we do. Help. No, we don't design for others, we help others figure out their designs, help them understand the flaws and guide them to a happy ending.

OK, I'll try to explain in more detail.

I have a PSU that output 13.8vdc and this is powering some astrophotography equipment.

No I am constructing a Peltier cooling device for one of my CCD cameras using a TEC1-12705.

This TEC1 runs at 12v & draws 5a, although it is rated up to max 15v but the consequence of higher voltage is a higher current draw & possible shortening of the devices lifespan.

I wish to fix the TEC1 supply to 12v but do not want to alter the output from my main PSU as that is required for the rest of my equipment.
Also, because of space and weight restrictions, I do not want to build a seperate PSU just to drive the TEC1.

The TEC1 is mounted on an aluminium plate, which in turn is bolted to the housing of my CCD, this aluminium plate acts as a heat sink.
Also, a heatsink with fan is fitted to the hot side of the TEC1 to further aid cooling.

So all I am looking for is a straight forward fix to either keep the current to the TEC1 at 5a/12v.

I have looked at Buck converters and yes, after a fashion they could do the job but I would like something a little more elegant and space saving.

Thanks

 

I have a power supply from Radio Shack (DON'T HATE ME, they were going out of business and sold a hell of a lot of stuff for stupidly low prices). My supply is rated for 13.8 volts at (I think, not at my work desk) 19 amps. When I run a load my 13.8 volts drops a bit. Depending on how big a load I put on it the voltage drops more with the heavier loads.

In a post where someone asked about making circuit boards out of paper I posted a link to a video I did using an old carbon filter from a RO (Reverse Osmosis) system (water filtration). I turned the filter into a HUGE resistor. HUGE in wattage capabilities - though I don't know how many watts. Powered a 12 volt auto lamp directly from the supply and indirectly sliding it up and down the length of the carbon filter. The further away I got from the positive the dimmer the bulb got. But at full brightness on the bulb (if memory serves me) the voltage dropped down to 12.8 volts under load. The bulb was (I think) 35 watt 12 volt. Meaning it drew 2.9 amps.

Where I'm going with this is that under the load of whatever you're main purpose for the power supply is, it's very possible that under load you're not seeing a full 13.8 volts. With the Peltier on there it may fall even lower than mine did. Since your Peltier is rated for voltages up to 15 volts, why not just power the whole thing up and see what your final voltage across the Pelt is. You may find you're close enough to your goal to not have to worry about it.

 

Right, I think I am going for this: https://uk.rs-online.com/mobile/p/low-dropout-voltage-regulators/6868837/

Looks to do everything that I require in a small usable package.
I can mount it on my heatsink with pins entering through the side of a small mounted enclosure where all connections are made to the TEC1 & fan assembly from my remote PSU.

 

I have a power supply from Radio Shack (DON'T HATE ME, they were going out of business and sold a hell of a lot of stuff for stupidly low prices). My supply is rated for 13.8 volts at (I think, not at my work desk) 19 amps. When I run a load my 13.8 volts drops a bit. Depending on how big a load I put on it the voltage drops more with the heavier loads.

In a post where someone asked about making circuit boards out of paper I posted a link to a video I did using an old carbon filter from a RO (Reverse Osmosis) system (water filtration). I turned the filter into a HUGE resistor. HUGE in wattage capabilities - though I don't know how many watts. Powered a 12 volt auto lamp directly from the supply and indirectly sliding it up and down the length of the carbon filter. The further away I got from the positive the dimmer the bulb got. But at full brightness on the bulb (if memory serves me) the voltage dropped down to 12.8 volts under load. The bulb was (I think) 35 watt 12 volt. Meaning it drew 2.9 amps.

Where I'm going with this is that under the load of whatever you're main purpose for the power supply is, it's very possible that under load you're not seeing a full 13.8 volts. With the Peltier on there it may fall even lower than mine did. Since your Peltier is rated for voltages up to 15 volts, why not just power the whole thing up and see what your final voltage across the Pelt is. You may find you're close enough to your goal to not have to worry about it.

I will do that, thanks.
Ian trying to cover all eventualities and have everything in place at once in order to get the job done.

I'll post back my results

 

I have a power supply from Radio Shack (DON'T HATE ME, they were going out of business and sold a hell of a lot of stuff for stupidly low prices). My supply is rated for 13.8 volts at (I think, not at my work desk) 19 amps. When I run a load my 13.8 volts drops a bit. Depending on how big a load I put on it the voltage drops more with the heavier loads.

In a post where someone asked about making circuit boards out of paper I posted a link to a video I did using an old carbon filter from a RO (Reverse Osmosis) system (water filtration). I turned the filter into a HUGE resistor. HUGE in wattage capabilities - though I don't know how many watts. Powered a 12 volt auto lamp directly from the supply and indirectly sliding it up and down the length of the carbon filter. The further away I got from the positive the dimmer the bulb got. But at full brightness on the bulb (if memory serves me) the voltage dropped down to 12.8 volts under load. The bulb was (I think) 35 watt 12 volt. Meaning it drew 2.9 amps.

OK, I'll try to explain in more detail.

I have a PSU that output 13.8vdc and this is powering some astrophotography equipment.

No I am constructing a Peltier cooling device for one of my CCD cameras using a TEC1-12705.

This TEC1 runs at 12v & draws 5a, although it is rated up to max 15v but the consequence of higher voltage is a higher current draw & possible shortening of the devices lifespan.

I wish to fix the TEC1 supply to 12v but do not want to alter the output from my main PSU as that is required for the rest of my equipment.
Also, because of space and weight restrictions, I do not want to build a seperate PSU just to drive the TEC1.

The TEC1 is mounted on an aluminium plate, which in turn is bolted to the housing of my CCD, this aluminium plate acts as a heat sink.
Also, a heatsink with fan is fitted to the hot side of the TEC1 to further aid cooling.

So all I am looking for is a straight forward fix to either keep the current to the TEC1 at 5a/12v.

I have looked at Buck converters and yes, after a fashion they could do the job but I would like something a little more elegant and space saving.

Thanks

Where I'm going with this is that under the load of whatever you're main purpose for the power supply is, it's very possible that under load you're not seeing a full 13.8 volts. With the Peltier on there it may fall even lower than mine did. Since your Peltier is rated for voltages up to 15 volts, why not just power the whole thing up and see what your final voltage across the Pelt is. You may find you're close enough to your goal to not have to worry about it.

Just because it CAN take 12v and/or it can take 5A doesn't mean it SHOULD. if you do what you're suggesting that TEC1 will dissipate 60watts of heat right where you want it to cool. It won't work. You will have to limit the voltage and current to much lower levels to get an effective heat pump working. I suggest using an lm 317 (the 1.5 amp rating should be enough) and use a pot to set the output voltage to the minimum (1.25v) to about 5 or 6 volts. The 317 plus the pelter cell will both need heat-sinks. If you don't get the degree of cooling you want lower the voltage, use thermal insulation, bigger heat-sinks and wait longer. Adding more power (an therefore heat) will not solve the problem.

 

I'm not familiar with Peltiers but I have been inside those RS power supplies. They are dead simple.
The diode method I named will work. I think you have an imaginary problem, but a couple of diodes wasting power is simple and it won't hurt anything. The only warning is, 1.8V x 5 amps is 9 watts. Those diodes will need a heat sink. The quad rectifiers come in nice big housings that easily attach to an aluminum plate. Do not use the graphs to find the diode voltage loss at 0 centigrade. The diodes will not be at zero. They will self heat in less than a minute.

 

I'm not familiar with Peltiers but I have been inside those RS power supplies. They are dead simple.
The diode method I named will work. I think you have an imaginary problem, but a couple of diodes wasting power is simple and it won't hurt anything. The only warning is, 1.8V x 5 amps is 9 watts. Those diodes will need a heat sink. The quad rectifiers come in nice big housings that easily attach to an aluminum plate. Do not use the graphs to find the diode voltage loss at 0 centigrade. The diodes will not be at zero. They will self heat in less than a minute.

And if the thread starter ignores your diode suggestion and goes with the LDO, heat sinking will still be every bit as important! At 9W, the proposed LDO junction will be 27*C above the heat sink... or 450*C above ambient if run without a heat sink!!!

 

I'm not familiar with Peltiers but I have been inside those RS power supplies. They are dead simple.
The diode method I named will work. I think you have an imaginary problem....

Imaginary problem?? Very useful answer, thanks!

And if the thread starter ignores your diode suggestion and goes with the LDO, heat sinking will still be every bit as important! At 9W, the proposed LDO junction will be 27*C above the heat sink... or 450*C above ambient if run without a heat sink!!!

I take all suggestion on. board and weigh up options. Also, I have stated that the LDO *WILL* be attached to a heatsink. Thanks for replying

 

Just because it CAN take 12v and/or it can take 5A doesn't mean it SHOULD. if you do what you're suggesting that TEC1 will dissipate 60watts of heat right where you want it to cool. It won't work. You will have to limit the voltage and current to much lower levels to get an effective heat pump working. I suggest using an lm 317 (the 1.5 amp rating should be enough) and use a pot to set the output voltage to the minimum (1.25v) to about 5 or 6 volts. The 317 plus the pelter cell will both need heat-sinks. If you don't get the degree of cooling you want lower the voltage, use thermal insulation, bigger heat-sinks and wait longer. Adding more power (an therefore heat) will not solve the problem.

Lower the voltage to get more cooling?

So why not just short the device (i.e., zero volts) to get maximum cooling?

Are you familiar with how a Peltier works?

The device is specifically designed so that, over the intended range of operation, the Seebeck effect dominates the Ohmic resistance and thermal conductance effects so that the more power you put to the module, the more cooling power you obtain. They would be pretty useless otherwise.

 

Lower the voltage to get more cooling?

So why not just short the device (i.e., zero volts) to get maximum cooling?

Are you familiar with how a Peltier works?

The device is specifically designed so that, over the intended range of operation, the Seebeck effect dominates the Ohmic resistance and thermal conductance effects so that the more power you put to the module, the more cooling power you obtain. They would be pretty useless otherwise.

I did wonder about that post myself.

Possibly the poster did not read my post where I stated I have a heatsink & fan attached to the hot side to further draw the heat away.

 

Imaginary problem?? Very useful answer, thanks!

Are you familiar with how a Peltier works?

And you were proven wrong in 6 minutes. Good-bye.