# can this 9VAC/60Hz Voltage Doubler supply total of 180mA? (pic inside)

#### azone

Joined Feb 1, 2008
15
I'm modifying an existing piece of hardware and can't redo the power supply. I need to add a couple additional 12v sources and the only way is to use a voltage doubler since the power supply is 9VAC. I'm wondering what are the limitations of voltage doubling? Can I just buy an AC adapter with double the current rating that I need (9V/400mA in this case)? Is there anything I need to take into account regarding the filter capacitors?
I'd really just like to know if this power supply can handle 200mA of current draw? At some point would the current draw exceed the filter capacitors ability to regulate?

Joined Jan 10, 2006
614
I see a bridge rectifier, where is the doubler..?
When doubling an AC voltage you will only get half wave rectification or less, so pick filter caps to suit. Doublers are typically used in low current apps, and may introduce some hum into audio circuits.

#### studiot

Joined Nov 9, 2007
4,998
In theory a voltage doubler will supply half the current, but it usually works out at about 80% of that.

#### azone

Joined Feb 1, 2008
15
I made a mistake...fixed now. The 7812 and LM317 branch should be doubled.
I guess what I don't understand exactly is why it is only suitable for low current application? I could always use a bigger supply. Is it the caps?
I'm also not sure what constitutes low current. I deal mostly with audio and control circuits where 200mA is a pretty good chunk of power.

The application is an audio circuit. Most of the stuff powered by this is analog control signals and not part of the audio path but the LM317 branch does power some op amps that are part of the audio circuit...so noise would be a concern.

#### studiot

Joined Nov 9, 2007
4,998
Dont forget your addition is unbalancing the bridge which won't now charge evenly on alternate half cycles.

#### azone

Joined Feb 1, 2008
15
Dont forget your addition is unbalancing the bridge which won't now charge evenly on alternate half cycles
yeah I can see that in my simulations - total ripple is about 1.3v taking into account both halfs. Still a few mV after the regulators.

#### studiot

Joined Nov 9, 2007
4,998

The added doubler is an ingeneous use of two of the bridge diodes, the two 2200UF caps and earth in the attached doubler configuration.

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#### nomurphy

Joined Aug 8, 2005
567
I think you're missing some components, see attached:

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#### Papabravo

Joined Feb 24, 2006
19,344
You do know that the peak voltage input to the bridge is
Rich (BB code):
9 VAC * SQRT(2) = 12.726 Volts
and that you will loose approximately 0.7 V in the bridge rectifier
which means the input to the 7812 will be just over 12V.

Has it escaped everyones notice that the 7812 input voltage is below the dropout level?

Has it also escaped everyones notice that the 7805 input is connected to the wrong spot on the bridge rectifier to produce a +5VDC output?

#### nomurphy

Joined Aug 8, 2005
567
The original circuit will provide ~ +12V to the 7805 reg, which is connected correctly.

With the modified version I supplied above, there will be ~22V to the 7812 and LM317 regs.

Attached is a simulation, without the regulators.

.

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#### SgtWookie

Joined Jul 17, 2007
22,227
You can shave about 9.6mA off of your power consumption, if on the LM317 you use a 1.2k fixed and a 10k pot for the voltage divider network instead of the 120 Ohm and 1k pot. Keep the .1uF ceramic output caps and put a 1uF to 10uF in parallel with them.

I tried the circuit using an older simulator than yours; it didn't work so well. Instead of using diodes and caps for voltage doubling, you might have to go to a boost circuit - but that will induce audible noise unless it's running above audible range.

The Vregs have a minimum Vdrop across them of 1.7v. You might look at getting some more modern low-drop regulators, which would help quite a bit. Otherwise, you're just consuming power in the regulators.

#### azone

Joined Feb 1, 2008
15
Nomurphy your were right - it looks like I was missing some parts.

SgtWookie thanks for the additional comments. I was thinking the same thing with the LM317 - although I only scaled it to 330/3.3k with my latest simulation. Even better would be to decide on a fixed voltage instead of the 317.

I did some measurements on the actual thing and found some of the current draw is less than what I though.

7805: 40mA
7812: 60mA
LM317: 20mA

I'd like to avoid implementing a boost circuit because of the concerns with noise. Efficiency aside will this work as is - or is there a problem with the current requirements? I always read voltage doublers/triplers etc.. are usually used for low current circuits - or at least stuff without constant current requirements. In my case there will be a pretty constant draw of 60mA or so from the doubler. I'm only doubling - not tripling etc.. so maybe it's not as huge an issue?

#### Ron H

Joined Apr 14, 2005
7,014
I think you're missing some components, see attached:
Nomurphy, I don't see what the parts you've added do, except drop the voltage doubler output by a diode drop. Probe the voltage on the anode of the diode you added.

#### nomurphy

Joined Aug 8, 2005
567
If the voltage doubler were 100% efficient then you would need:
(20mA + 60mA) * 2 = 160mA
160mA + 40 mA = 200mA

however:
200mA * 2 (headroom) = 400mA

So if you want some good wiggle room, get a 9V 500mA wall wort.

Cs, the series cap, is the doubler that feeds the required peak detector and filter of Dx/Cx.

#### azone

Joined Feb 1, 2008
15

Other than being highly inefficient is this concept scalable if I need more power, assuming I use the correct size transformer and select the correct value capacitors? For example say I need twice the power versus the scenario above - would I simply use a 9V/1.2A transformer and I'd be OK?

#### Ron H

Joined Apr 14, 2005
7,014
Behold a sim of Azone's original schematic (I think).

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#### SgtWookie

Joined Jul 17, 2007
22,227
SgtWookie thanks for the additional comments. I was thinking the same thing with the LM317 - although I only scaled it to 330/3.3k with my latest simulation. Even better would be to decide on a fixed voltage instead of the 317.
I did a quick check in my simulation; the 78xx series regulators have a fixed voltage divider network in them, so that causes them to use an extra 3 to 5 mA each. The LM317, having an external divider, can be "tweaked" to reduce the current consumed. However, the reduction in current will make the regulator more susceptible to noise.

You might consider trading places - use an LM317 for both of the regulators that are currently 78xx series, with higher value resistors in the network - and use a 7812 for the new audio circuit, or an LM317 with standard value resistors, or double value resistors. Or, investigate some of the newer low-dropout (LDO) regulators, which seem to be perfect for this application.

Check out Linear Technology's offerings for the 4-to-80v, 250mA range:
http://www.linear.com/pc/viewCategory.jsp?navId=H0,C1,C1003,C1040
In particular, this one:
http://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1040,P13772
I did some measurements on the actual thing and found some of the current draw is less than what I though.

7805: 40mA
7812: 60mA
LM317: 20mA
That's good news. It's usually the other way around - out of power budget, time and money all at once
I'd like to avoid implementing a boost circuit because of the concerns with noise. Efficiency aside will this work as is - or is there a problem with the current requirements? I always read voltage doublers/triplers etc.. are usually used for low current circuits - or at least stuff without constant current requirements. In my case there will be a pretty constant draw of 60mA or so from the doubler. I'm only doubling - not tripling etc.. so maybe it's not as huge an issue?
Well, the closer you come to optimizing your design, the more headroom you'll have left over for the rest of the circuit. Getting rid of the minimum 1.7V dropout that the 78xx and LMx17 series regulators will give you a big boost. The LT3012 I gave you a link to above has a 0.4v dropout; it's like adding 1.3V to your rectified output while reducing heat dissipation in your power supply regulator ICs. That would effectively give you 10% more available reserve in your 12v output, and 26% more on your 5V circuit.

Or, just keep LDO's in mind next time a linear power supply design opportunity comes your way

#### Ron H

Joined Apr 14, 2005
7,014
LM317 has a minimum load current of 10mA, in order to guarantee regulation.

#### azone

Joined Feb 1, 2008
15
thanks for the responses, there's some good info here.

I'm checking out some LDO's including the ones mentioned above.
For the 12v regulator I found a decent priced part that seems readily available... Fairchild KA278R12C - looks like it would be a good part substitution. Any thoughts?

For the 5v regulator there seems a lot more available including the LT stuff and many others.

#### SgtWookie

Joined Jul 17, 2007
22,227
Fairchild KA278R12C - 2A Fixed 12V Low Dropout Voltage Regulator (LDO)

Sure, that would work. Keep in mind, it has a fixed internal divider network of R1=1.8k, R2=600 Ohms, resulting in a constant 5mA load. (see page 16, fig 2) You can "tweak" the output by adding R3 and R4, but that results in even more load.

If you wish to have the maximum reserve power available, you should select an adjustable regulator, as then you can select the resistor network to reduce that current while keeping the output stable.

TO-220 cased devices are somewhat overkill for your power needs, but then I've been known to use sledgehammers to swat flies. Might be easiest to select just ONE adjustable device, and use it for all three regulators. You could even use a 10k 21-turn trimpot to set each output voltage, so you only have to add a single component vs the fixed regulators. For the 12V outputs, a 10k pot would use 1.2mA current in the divider network.

You could also use 5v fixed devices for everything. For 12v, just add a pot between the ground terminal and ground. The addition of a pot will change the relationship of the internal voltage divider. The greater the resistance you add, the higher the output voltage (within reason, of course).