Need Voltage Doubler Circuit

Voltage doublers give poor regulation, and regenerative sets want a good, steady input voltage. You'll get better performance off of you pair of wall warts, no matter how ugly things look.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

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Here is a standard voltage doubler circuit:

The 1N4004s will be fine. Use filter caps rated for 50V or more,

50-100uF, and consider adding a second stage of RC filtering by adding a resistor (dropping, say, 10% of your voltage at your current draw) followed by another filter cap (50-100uF again). For best results you should also regulate the voltage going into the detector stage - a simple resistor and zener will do. Many people also find that adding bypass caps (0.01 to 0.001) across the diodes helps eliminate power- supply radiated hum.

Mike

The diodes should work fine as long as the DC current load is less than about a half amp. A simple doubles connects two capacitors (negative side of one and positive side of the other) to one side of the 24 supply and two diodes (cathode or banded end of one and anode of the other) to the other side of the 24 volt supply. Then connect the free anode to the free negative capacitor end and the free cathode to the free positive capacitor end. The output is taken across those last two connections, so that the two capacitors are acting as a pair of stacked half wave supplies. Under no load, this will produce at least

24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep in mind that the 24 volts AC output may rise quite a bit under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line frequency ripple to the output. You may want to add an RC low pass filter stage with the series R selected to lower the voltage to closer to your desired 45 volts, and the capacitor from that output to the other side of the supply as an additional ripple filter. Or you may just lower the voltage under load by adding some resistance in series with the 24 volt AC output. This will not reduce the ripple as much, but will make the 24 volt transformer run a little cooler. I just looked up the transformer and it has several taps, so you can try different combinations to get very close to the desired output voltage.

For the two doubler capacitors, I would use something like

5,000 to 10,000 microfarads per amp of output load and rated for at least 16 volts DC. Your diodes and transformer will allow about 300 to 400 mA load current. If your load current is at or below 400 mA, that implies something around 2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get pretty close to 45 volts output with a bridge rectifier made with 4 of your diodes and a single capacitor, say, 2,200 to

4,700 uF, rated for 50 volts. This will give you a voltage around 40 to 45 volts, depending on load current that can be up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option. Sounds simpler, and I believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge rectifier, Smaller and cheaper than the one amp unit. I think the current draw of the radio is only about 8-10 mA. Transformer is M2860 @

ave

Another option is to use the 30V transformer in the voltage doubler circuit. Take your 40-45V off the lower filter cap (and add regulation for the detector), and take 80V off the upper cap for the power output stage that you talked about.

Mike

(snip)

Sounds like a plan. The light load voltage may be something like 33 volts, so the peak would be 33*sqrt(2)=47, but the bridge will waste about 2 volts. so that might do very nicely. At that low current, the 2,200 uF capacitor will produce a low ripple output. If the voltage comes out a little high, add a resistor in series with the either the 30 volt secondary, or in series with the primary.

--
Regards,

John Popelish

Is this how I would hook it up?

Sorry for the large image file, this image was emailed to me to help in building a power supply for a battery transistor radio.

What value resistor would I need to run a 2 volt LED off the 30 volts?

Would it be possible to run the tube filament (2 volts, 60 mA) off a

6.3 volt, 500 mA transformer, using something like resistors? I would like to eliminate using "D" cells. Thanks, Dave

"John Popelish" wrote in message news:I5KdnXOl8_a2Oo3VnZ2dnUVZ snipped-for-privacy@comcast.com...

You need to use capacitors rated at least as high as the maximum peak to peak unloaded output voltage of the transformer, so for 24 VAC that would be about 75 volts. Here is an LTspice ASCII showing a 24 VAC source boosted to 68 VDC with two diodes and two capacitors. The first capacitor can be rated at half the voltage of the second one. This circuit has poor regulation, but is inherently current limiting, so you can just add a zener (as shown) to the output for regulation. You need to adjust the value of C1 for the wattage of the zeners and load current. This is set for 45 VDC and about 50 mA max.

Paul

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xt -

Not for my suggestion. I'm suggesting, if you also want a higher voltage for a power output stage, that you use a voltage doubler like so:

The bottom of the two capacitors is your 0V or ground reference. The top of the two capacitors will give you 80-90V with a 30V transformer. The junction of the two capacitors will be at half that voltage or 40-45V. Thus you have low voltage (suggest additional filtering or regulation) for your detector and audio voltage amplifier, and higher voltage for your power stage. Each capacitor needs to be rated 50V or higher, since each one receives the full secondary peak voltage.

Also, you asked about AC on the filaments. That's likely to lead to hum. A good idea would be to turn the 6.3VAC into well-filtered DC and drop that to 1.4V with a resistor. The resistor will help protect your filaments from turn-on surge, which is why it's a better approach than building a 1.4V regulated source.

Mike

Umm, wrong link. Sheesh! :-O

If the 30 volts is DC , you just apply Ohm's law using the desired current and the amount of voltage the resistor must waste. For instance, if you had 30 volts DC and you wanted a resistor that would pass 10 mA (1/100 amp) to the LED while using up 28 of the 30 volts (leaving 2 volts for the LED), you plug those numbers into E=I*R and rearrange to solve for R. 28 V=(1/100 A)*R, or R=28*100=2800ohms. You could use thew standard values of 2700 or 3000 for a little more or less than 10 mA. But you would actually want to wait until the supply was built and measure the DC voltage before calculating the resistor.

It would be possible and a lot more efficient than using the

30 volt supply for the 2 volt filament. Same formula. AC voltages are measured in RMS volts, because that gives the equivalent DC voltage as far as heating effect on resistors goes. So you need to have a resistor that wastes 6.3-2 =4.3 volts while passing 0.06 amps. So 4.3=0.06*R, or R=4.3/0.06 = 71.67 ohms. Standard values close to that are 68 and 75. If the 2 volt filament can run off a D cell (1.5 volts) you can probably go with the higher resistance, especially if the transformer actually outputs a little more than 6.3 volts with this light load.

You also want a resistor that stands the heat that produces, so multiply amps through times volts across that resistor to solve for the watts it is producing.

4.3V*0.06A=0.258 watts.

A little too much for a 1/4 watt resistor to handle, so a

1/2 watt or larger is needed.

But check if you have a 6.3 volt center tapped transformer. You can get 6.3/2 or 3.15 volt from one end to the center terminal and lower the power you have to waste in the resistor.

--
Regards,

John Popelish

"Tim Wescott"

** The standard voltage doubler has exactly the *same* ( load ) percentage regulation as the same transformer would provide with a bridge rectifier and capacitor filter.

** Only way to get that is with a regulated supply. ** Not true.

Line voltage and load variations will affect them just like any unregulated supply.

The actual percentage load regulation depends on the size of the transformer and how much load is applied - ie light loads = good regulation.

..... Phil

Actually the wallwarts are regulated, I think they're M9926 @

text -

I see that I replied to your post by mistake. I meant to reply to John Popelish 's message about using a bridge rectifier and cap to boost the 30 Volts AC to 45 volts DC. Seems much more simpler to me.

Dave

"Dave.H"

** That is a switch mode power supply ( SMPS ).

... Phil

nd

Are we sure about that? Only one-half of the capacitor stack is charged each half cycle, but both capacitors are discharging each half cycle.

Which brings up a circuit I ran across years ago and was able to find again yesterday. This one claims to charge both capacitors each half cycle, making it a true full-wave circuit. I haven't tried it or analysed it - I just put it out for consideration.

Mike