Full wave rectifier question for a power supply

[zalmaflash]

This may seem simple or moronic but I honestly don't know.

I made a power supply [24v] to run some of my battery tools with bad batteries.

My transformer [120 X 24] is plenty big, I convert to 24vdc with a KBPC2510 Bridge rectifier, heat sink w/paste, and a 24vdc fan.

The rectifier still gets hot after a few min. with my sawzall. I stop and let it cool, and then continue.

I tried another identical rectifier and the same thing.

Here is my question. Can I put two Bridges [identical] in parallel to help eliminate the heating?

[budm]

Re: Full wave rectifier question for a power supply

How much current are you pulling?
By the way, using 24Vac transformer and filter cap, the DC output will be around 34VDC.
KBPC2510 has Vf of 1.1V @12.5A, so that will be total Vfdrops of 2.2V x Current draw, that will the power dissipation that it has to dissipate so you will need proper heatsink for it.

[eccerr0r]

Re: Full wave rectifier question for a power supply

Many of these power tools draw a fairly substantial amount of current, I suspect it can be around 10-20A at least.

Sticking bridge rectifiers or simple diodes in parallel is not a good idea - it won't have the intended effect as the one that conducts first will still kill itself. If you have perfectly matching bridges/diodes then this would be OK but chances are you will not find two bridges that perfectly match.

You could add resistors in series to limit this effect. Add a ~0.047 to 0.1 ohm resistor to the output of each bridge and then tie them together. Of course you'll have some loss here too.

[budm]

Re: Full wave rectifier question for a power supply

many of these power tools draw a fairly substantial amount of current, i suspect it can be around 10-20a at least.

Sticking bridge rectifiers or simple diodes in parallel is not a good idea - it won't have the intended effect as the one that conducts first will still kill itself. If you have perfectly matching bridges/diodes then this would be ok but chances are you will not find two bridges that perfectly match.

You could add resistors in series to limit this effect. Add a ~0.047 to 0.1 ohm resistor to the output of each bridge and then tie them together. Of course you'll have some loss here too.

+1

[mariushm]

Re: Full wave rectifier question for a power supply

Did you use a capacitor after the bridge rectifier?

Let's take things step by step.

Your transformer takes your 100-110v AC down to 24v AC. The transformer should also have a rating on it, listed in VA. That's the amount of power it can provide to whatever you connect to it.

When you use a bridge rectifier, the 24v AC at the transformer output is converted into DC voltage, and the peak DC voltage will be :

Vdc peak = Vac x 1.414 - 2 x Vdiode

where Vac is the AC voltage of the transformer and Vdiode is the forward voltage of a single diode in the bridge rectifier.
The value is inside the datasheet and varies with the amount of current going through he bridge rectifier but it's usually about 1v-1.2v

So, Vdc peak = 24 x 1.414 - 2 x 1.1v = 31.7 v ... for simplicity let's say 32v

The maximum current your transformer can provide can be determined if you have that VA rating.

Maximum current = ~ 0.62 x VA rating / Vac

So for example, let's say your transformer is rated for 75VA, then the maximum current would be 0.62 x 75 / 24 = about 1.9 A .. for simplicity let's round it up to 2A

Now going back to voltage ... remember, it was only a peak voltage, you don't have a constant DC voltage at the output..
Your AC voltage was something that 60 times a second was going to -24v and then back to +24v and all over again forever.
The bridge rectifier just flips the negative parts and makes them positive, so now for 2 x 60 times a second you have a voltage going slowly up from 0v to your peak dc voltage of about 32v and then again slowly back down to 0v.

If you want your output voltage to never go below a certain value, you need to add some capacitance.
How big of a capacitance depends on the maximum current you expect to have and what's the minimum voltage you want to see on the output.
For example, let's say we're going to have a maximum current of 2A and that we want to always have at least 26v at the output, so that our output voltage will be between 26v and 32v.
A formula that approximates the needed capacitance is then:

C (Farads) = Current / [ 2 x Frequency Mains x ( Vdc peak - Vdc minimum) ]

So we have C = 2 A / 120 x ( 32v - 26v) = 2 / 720 = 0.002777 Farads or 2777 uF

which means if we want at least 26v at the output all the time, we need at least 2777 uF capacitor, rated for a voltage of at least 50v (because a 35v rated capacitor would be too close to 32v, and therefore not safe for the capacitor). 2777 uF is not a standard value, so you could use instead a single 3300uF capacitor, or two 1500 uF capacitors in parallel (for a total of 3000uF)
Since you want to use this output with 24v devices, you don't really want to install too much capacitance .. the more capacitance, the more time the output voltage will be closer to the maximum 32v we determined above.

Now as you can see, this will make your DC output somewhat stable, but it will still hover between 26v and 32v. If you want it to always output a fixed value, now you install a linear regulator which will take a DC voltage at least about 1-2v higher than the output voltage and output the desired voltage.

For example, a LM317 will work just fine: https://cdn.badcaps-static.com/pdfs/...4077910403.pdf

It accepts any voltage up to about 40v and can output up to about 1.5A at the voltage you want, as long as the input voltage is at least about 2v above the output voltage (in our case since we did the math above for minimum 26v, everything should be just fine).

There's other linear regulators out there, for example classic 7824 that can output up to 1A.

-

Placing bridge rectifiers in parallel only lowers the current going through each rectifier, which may lower the voltage drop on each diode inside the bridge rectifier. For example, instead of a 1.1v drop, you may have only 1v so if your device uses 10A you initially had 2x1.1x10 = 22 watts lost in the bridge rectifier and now with two bridge rectifiers you have 2 x 2 x 1v x 5a = 20 watts.

[eccerr0r]

Re: Full wave rectifier question for a power supply

I have a 3.6V handheld vacuum cleaner that the batteries are toast. I tried to power it with a single 3.5Ah 26650 lithium ion battery to no avail - the original sub-C NiCds 3x1.3Ah are the only batteries capable of sending the ~20A starting current and maintaining the ~6A steady state current draw.

The 20A pulse at the beginning caused the protection circuitry to shutdown in the Li-ion cell.

[zalmaflash]

Re: Full wave rectifier question for a power supply

Thanks for all the answers. I have a lot to consider (and learn).

The only cap I have in my circuit is a smoothing cap on the primary.

My transformer is a 250 va.

I have my bridge rect. on a 4"sq. heat sink which is 5/8" thick, fins verticle, plenty of compound, and the fan is 1/2" from fins.

I just didn't want to burn up the bridge. (again)

[budm]

Re: Full wave rectifier question for a power supply

Thanks for all the answers. I have a lot to consider (and learn).

The only cap I have in my circuit is a smoothing cap on the primary.

My transformer is a 250 va.

I have my bridge rect. on a 4"sq. heat sink which is 5/8" thick, fins verticle, plenty of compound, and the fan is 1/2" from fins.

I just didn't want to burn up the bridge. (again)

It is obviously not good enough heatsink, you need to keep the junction temperature at or below 150c. To use proper heatsink you need to know how much current draw is.
BTW, what do you mean by having the filter cap in the primary side?

Thermal spec are all over the place or each manufacturers.
You should check the temperature at the case of the bridge to see what you get.
http://www.designworldonline.com/how...ble-heat-sink/

[budm]

Re: Full wave rectifier question for a power supply

Heatsink calculator: http://mustcalculate.com/electronics...1.4&rcs=1&rsa=

[zalmaflash]

Re: Full wave rectifier question for a power supply

BUDM - I mispoke about the capaitor across the main pwr. I should have said "varistor" across the mains.