Use Voltage Divider Instead of Transformer?

[hamfiles]

I'm building a regulated power supply, and I'm trying to save money by not buying a transformer. I got to thinking, why not simply divide the voltage. My mains outlet is 120 Vac. I want 60 volts dc output to a high voltage regulator. With 2 2.4 k resistors, the current draw should only be 25 milliamps. And when I tap into this, won't the circuit only see one resistor as Ri, and be able to draw as much current as needed (minus the 25 millilamp drop)?

Are there any drawbacks to using a voltage divider this way? I would not actually use two 400 watt resistors, I would use several smaller ones in parallel to get equivalent values in circuit, whatever combo is cheapest. Maybe some 50 watt ones. Thanks in advance for any info.
**broken link removed**

 

[Phasor]

Drawbacks:

1) As load increases, voltage drops. OK if you are running a constant load, but you have to know the resistance of the load, and take it into account in your calculations.

2) HIGHLY inefficient.

3) Resistors won't rectify your AC mains to DC.

And how did you come up with 400W??? A 400W resistor will set you back more money than a transformer, and will probably weigh more.

 

[hamfiles]

I could not get image to load, but here is circuit.

http://www.geocities.com/hamfiles/V-Divider.htm

Actually, I believe only one of the resistors would have to be 400 watt, the one used to drive the output. If my load draws 3 amps, 360 watt resistor is needed, so I would select 400 just to be over a little.

Now that I think of it, though, this does not work, because like you said, the resistance will change, and will no longer divide evenly.

 

[Phasor]

And besides - going back to ohm's law, if you want to draw 3A through 2400 ohms, you need 7.2kV across the resistor. And that works out to 21600W, not 360.

V = IR
V = 3 x 2400
V = 7200 V

P = I^2 x R
P = 3 x 3 x 2400
P = 21600 W

Now, was this supposed to be for an instantaneous hot water heater?

 

[hamfiles]

Yep, you are right. Don't know what I was thinkin', guess I was trying to get something for nothing...

 

[bogdanfirst]

just a suggestion, if you want a transformer less power suply, wich has a constant and quite little load, you could use a cap in series. because the current and voltage trough the cap are at 90 degrees apart(pi rad) then no active power is lost on it. if you plan to power a led or something, it is ok. but otherwise........

 

[Sebi]

You can considering some TV power supplyes. Most of cases it contains a buck switching regulators (only one IC +inductor + low ESR cap and scottky diode) If no need potential-free PSU, apply this.

 

[abhishek singh]

isolation?

hi,

your register network also does not has the isolation from the HT mains. any inductive kickback from any inductive appliance attached to the same mains line will create thousands ove volts destroying everithing.


abhishek

 

[hamfiles]

Thank you for all your replies. I went ahead and bought a transformer.

One more quick question, If I use a inductor/capacitor filter combo, will I need to bypass the rails at any point w/ a diode to prevent back emf?

I do not think that there will be any back emf, as all of the voltage should go back through the capacitor and to ground.

I'm using a bridge rectifier, followed by an inductor in series, bypassed by a capacitor to ground.

L=250 millihenrys C=10,000 microfarads. Sorry about the poor diagram. This is the voltage output rails, after the rectifier.



+---inductor--------------------Positive rail, output
.....................c
.....................a (capacitor connected across rails, after inductor)
.....................p
.....................a
.....................c
......................i
......................t
......................o
......................r
----------------------------------ground rail

At 10,000 microfarads, the capacitive reatance at 120 hertz ( I'm using a bridge diode rectifier) is about 132.6 milliohms.
At 250 millihenries, the inductive reatance at 120 hertz is about 188 ohms. Because the inductive reactance is much higher than X of C, nearly all of the ripple voltage should return through the capacitor to ground, filtering out the ripple voltage. I do not believe there will be any problem with back emf because of this, so I am not adding a protection diode. I believe you would only get back emf when you have a 90 degree phase difference, like a resistor and inductor combo, or with an inductor and no other component. What do you all think?

Also, the output from this stage is going into an LM 350, 3 pin regulator.

 

[Sebi]

You need only the regular 10...100nF condensators nearby the LM350 pins, to prevent the HF oscillation.

 

[hamfiles]

Thanks, Sebi. I was worried about filtering out the voltage ripple. I am using this supply to build and test audio amplifiers, and I really hate it when you get a hum from the power supply.

But I just breadboarded the entire circuit together, and I am very happy with the LM 350's ripple rejection. At 30 volts dc output, I am only getting .1 millivolts rms of ripple. This is without the inductor, so I am not going to use it. I used a 10 mfd on the 350's pins, and 22 mfd bypassing the output of the LM 350.

The inductor does help lower the ripple before it gets to the LM 350, but there's no significant difference on the final output of the LM 350. This regulator chip apparently has an excellent ripple rejection.

I did not use a 10,000 mfd bypass cap between the rectifier and LM 350 across the rails. Could not buy one on Sunday. I only used 470 mfd, and it works fine. When I remove the 470 mfd cap, the ripple does go up, but only to about 5.5 millivolts. Tomorrow I will buy a 1000 mfd cap and use it on the final build.

My power supply is almost finished. I have only to etch a board and soldier it together; I already have a metal box w/ all the holes drilled, and output test lead sockets mounted. I did a little drilling and bolted an old Pentium heatsink to the LM 350, which is probably overkill, but the cooler, the better. 8)

Thanks everyone for all of your input.

 

[Sebi]

In my experience, - when apply a regulator IC - for 1A output 1000...1500uF filter cap quite enough. For 3A output apply 3300...4700uF to avoid the high AC ripple on filter cap.(The IC can work also with 470uF...) Keep in mind: the electrolytic condensators don't like higher AC as 10...15% of DC.

 

[hamfiles]

Thanks, Sebi. I will keep this in mind.


[update] Man, I just connected this pwr supply to an FM radio w/ a 1 watt amplifier, and I still had hum, even with only a 100 microvolt ripple!


What I finally did was put a 30 millihenry inductor on the Final stage, after the LM 350, bypassed this last output stage w/ a capacitor, and then the hum went away.

I don't understand. I always thought that the hum from a pwr supply was the ripple voltage. Anyway, as far as I can tell, this supply is now nearly totally hum free, that is to say, there is no hum that I can hear.

 

[Sebi]

Yeah, maybe this is a very cheap FM receiver with bad AM suppression. Most of case help, when apply parallel 2...22nF ceramic cond. for each diodes of bridge.

 

[hamfiles]

That's a good idea, but I'm using the lazy man's diode bridge, a rectifier chip.

 

[Sebi]

This is no problem, because all pins available....

 

[hamfiles]

That's right. I'll try it. Thanks.