Power Supply - Too much voltage?

[carcanhol]

Hi, i am building a dc power supply and dont know what to do.
I have a dc-dc converter to use with it but max input is 40V 2-3A max.
My capacitor gets 50+ volts.
Is there any way to reduce the voltage by 10 or more volts?
Or its just not worth it and split the transformer outputs into separate outputs with their rectifier and capacitor and make it a dual PSU?
Any help would be nice.

Thanks



Edit: on the falstad circuit sim, if i change the capacitor resistor to a higher value i get higher voltage, is that right? Does it need a load to be the right voltage?

[rickselectricalprojects]

usually you can lower the output voltage with diodes if the voltage if 1 or 2 volts over but not with 10v (you would need a lot of diodes) i would do what you said where you just split the transformer outputs into separate outputs
i hope this is helpful

[carcanhol]

usually you can lower the output voltage with diodes if the voltage if 1 or 2 volts over but not with 10v (you would need a lot of diodes) i would do what you said where you just split the transformer outputs into separate outputs
i hope this is helpful

I guess that would be what has to be done, i think the transformer has 2x18V ac, not sure, I'm not at the lab right now, im puzzling why it gets 50v or more on the cap, 2x18v=36vac when rectified it gets to 50?.   
Thanks for the help

[Delta]

36vAC *RMS* is the clue there!

Work out the peak AC voltage... 😉

[carcanhol]

36vAC *RMS* is the clue there!

Work out the peak AC voltage... 😉

AHA, it has 36V RMS but its 50V or so peak, am i right? im such a noob

[carcanhol]

I think i got this, when i power up it reaches 50V peak so the cap stays charged at 50V because it has almost no load it stays there, but when i wire in a bulb or resistor it lowers to 36V. tomorrow i will test the theory.

Sim after reset and after load on attachments.

[mariush]

Peak DC voltage is  Voltage AC RMSc x 1.414 - 2 x Forward Voltage Rectifier  = 36 x 1.414 - ~ 2v = 49v

Note that at low loads, transformers can output up to 5-10% more than the advertised voltage, which can explain your 50v value.

The capacitor charges to keep the minimum DC voltage as high as possible.  There's a formula that approximates the capacitance required, which is Capacitance =  Current DC / [ 2 x Frequency AC x  ( Peak DC Voltage -  Minimum DC Voltage)].

27000 uF (if the circuit you have is what's in the picture) is really too much considering the size of your transformer, you're just stressing the bridge rectifier and potentially your fuses when you're plugging your circuit in the mains socket. In general, most engineers design linear power supplies with a balance between transformer voltage and capacitor sizes, basically they use around 3300-4700uF for each 1A of current.  It's ok for the voltage to fluctuate a few volts because it's going to be regulated later anyway, no need to use such large capacitor to get such steady voltage even at high current load.

[carcanhol]

I give up, i'm doing what rickselectricalprojects sugested.   

Thanks for everyones help. 

[Kleinstein]

What type of DC/DC converter is this ?  In might be that such amodule is not the best way to go.  It may also word well with just half the voltage.

Normal loading will not reduce the voltage by much. So maybe 45 V or so - still too much. Only with an additional inductor (or resistor)  in series the voltage will drop under load

[klunkerbus]

I give up...   

Welcome to power supply design...

As Mariush suggests, transformer secondary voltage ratings are stated with the output under the rated load current. One way to measure the "quality" of a transformer is to compare the output voltage between full load and no load.   

In case you are receptive, I have a few more observations on your design.  The capacitor appears to have a high power resistor between the terminals.  I assume this was added to either put some load on the transformer to lower the voltage a bit, or to discharge the power supply output fairly quickly when the AC power is shut off.  Regardless, the load is going to reduce the net filtering effect of the capacitor.  If the resistor is there to discharge the cap, one thing I've done on some crude home projects is use an AC relay to switch in a similar load resistor when AC power is shut off.  Turning AC power on energizes the relay, opens the contacts the resistor is wired to, and the resistor load is removed from the capacitor. 

You also have the capacitor wired off to the side.  The benefit may be marginal, but I'd typically run the load off the capacitor, not back at the rectifier. This puts the capacitor more in-line with your load. Otherwise the inductance of the wiring to the capacitor will also be having a (small) negative effect on the filtering you really get.   

[Kleinstein]

The rectifier just on top of the transformer is not such a good idea. Both the transformer and rectifier produce some waste heat - putting them close togehter hinders heat removal. It's better to have the rectifier monted on the case - so this can word as a heat sink.

There should be a fuse (or 2) between the transfomer and the rectifier. With proper limiting of the inrush current the fuse before the transformer might be enough, if its not choosen to large.

At this size transformer its a good idea to have the resistor directly at the capacitor - just to descharge it in reasonable time. The effect on filtering is minimal. To limit current peaks an inductor or small resitor / longer thin cable before the capacitor is a good idea - this makes current peaks less sharp and increases the powerfactor.  So the resitance of the fuse even has a good side to it. And yes connects the load to the capacitor, not back from the rectifier.

[carcanhol]

What type of DC/DC converter is this ?  In might be that such amodule is not the best way to go.  It may also word well with just half the voltage.

Normal loading will not reduce the voltage by much. So maybe 45 V or so - still too much. Only with an additional inductor (or resistor)  in series the voltage will drop under load

no converter yet just a bridge rectifier

[carcanhol]

I give up...   

Welcome to power supply design...

As Mariush suggests, transformer secondary voltage ratings are stated with the output under the rated load current. One way to measure the "quality" of a transformer is to compare the output voltage between full load and no load.   

In case you are receptive, I have a few more observations on your design.  The capacitor appears to have a high power resistor between the terminals.  I assume this was added to either put some load on the transformer to lower the voltage a bit, or to discharge the power supply output fairly quickly when the AC power is shut off.  Regardless, the load is going to reduce the net filtering effect of the capacitor.  If the resistor is there to discharge the cap, one thing I've done on some crude home projects is use an AC relay to switch in a similar load resistor when AC power is shut off.  Turning AC power on energizes the relay, opens the contacts the resistor is wired to, and the resistor load is removed from the capacitor. 

You also have the capacitor wired off to the side.  The benefit may be marginal, but I'd typically run the load off the capacitor, not back at the rectifier. This puts the capacitor more in-line with your load. Otherwise the inductance of the wiring to the capacitor will also be having a (small) negative effect on the filtering you really get.

The resistor is there to discharge the cap but slowly, and the AC relay idea is great, i will give it a try for sure, thanks.

Loads connect to the cap directly.

[carcanhol]

The rectifier just on top of the transformer is not such a good idea. Both the transformer and rectifier produce some waste heat - putting them close togehter hinders heat removal. It's better to have the rectifier mounted on the case - so this can word as a heat sink.

There should be a fuse (or 2) between the transformer and the rectifier. With proper limiting of the inrush current the fuse before the transformer might be enough, if its not chosen to large.

At this size transformer its a good idea to have the resistor directly at the capacitor - just to discharge it in reasonable time. The effect on filtering is minimal. To limit current peaks an inductor or small resistor / longer thin cable before the capacitor is a good idea - this makes current peaks less sharp and increases the powerfactor.  So the resistance of the fuse even has a good side to it. And yes connects the load to the capacitor, not back from the rectifier.

So a 3, 4A fuse should work between the transformer and rectifier?
 

[Zero999]

Hi, i am building a dc power supply and dont know what to do.
I have a dc-dc converter to use with it but max input is 40V 2-3A max.
My capacitor gets 50+ volts.
Is there any way to reduce the voltage by 10 or more volts?
Or its just not worth it and split the transformer outputs into separate outputs with their rectifier and capacitor and make it a dual PSU?
Any help would be nice.

Thanks



Edit: on the falstad circuit sim, if i change the capacitor resistor to a higher value i get higher voltage, is that right? Does it need a load to be the right voltage?

What's the minimum operating voltage for the DC-DC converter to work properly?

You might be able to connect the the transformer secondariness in parallel to give half the voltage.

[Kleinstein]

With bridge rectifier an capacitor the AC current is something like 1.8-2.5 times higher than the DC current, details depend on the resistance of the transformer and size of the capacitor.  So for up to 3 A of DC current the correct fuse would be something like 8 A (slow blow)  Of cause the fuse should not be much higher than transformer ratig unless there is an additional temperatur supervision.

[carcanhol]

With bridge rectifier an capacitor the AC current is something like 1.8-2.5 times higher than the DC current, details depend on the resistance of the transformer and size of the capacitor.  So for up to 3 A of DC current the correct fuse would be something like 8 A (slow blow)  Of cause the fuse should not be much higher than transformer ratig unless there is an additional temperatur supervision.

Thanks for the info. i will do that this afternoon.

[carcanhol]

Hi, i am building a dc power supply and dont know what to do.
I have a dc-dc converter to use with it but max input is 40V 2-3A max.
My capacitor gets 50+ volts.
Is there any way to reduce the voltage by 10 or more volts?
Or its just not worth it and split the transformer outputs into separate outputs with their rectifier and capacitor and make it a dual PSU?
Any help would be nice.

Thanks



Edit: on the falstad circuit sim, if i change the capacitor resistor to a higher value i get higher voltage, is that right? Does it need a load to be the right voltage?

What's the minimum operating voltage for the DC-DC converter to work properly?

You might be able to connect the the transformer secondariness in parallel to give half the voltage.

i have split the outputs to 2x18v and will use 2 regulators.
This is the converter i'm using right now: http://www.ebay.com/itm/5Pcs-Output-1-23V-30V-DC-DC-Buck-Converter-Step-Down-Module-LM2596-Power-Supply-/321476590827?hash=item4ad97f7ceb

but i guess LM2596 has 40V max input.
http://www.ti.com/lit/ds/symlink/lm2596.pdf

[carcanhol]

First test: https://youtu.be/GXCToY0WnCA

The vibration was so bad the multimeter reset itself, possibly the batteries moved in the case.