Has anyone had experience with this? I want to power an audio opamp with a 12 v wall wart, either AC or DC, I've got some of each. I can only use a single end power supply if i run the 12 v into a voltage divider and apply only +6 vdc to the V+ and ground V-. What if I run the 12 v into a voltage doubler circuit first, increasing it to 24 v, use that for PS reference then go to the voltage divider and get 12 vdc to apply to the opamp? Is trying to get something for nothing?
John K
Right, thanks, Phil. That gives me the incentive to actually solder the thing up. I'm incorporating a socket for the opamp so that the prototype chip 1458 which I have can be swapped for a better one NE5532, which I don't have. Am I right to say they are interchangable?
No need for all that.
Just AC couple your input and output signal. between the coupling caps, bias the signal up the 6V and you should be fine.
Ok great. I wouldn't have gone as high as 10,000 uF for the caps, so thanx to all. And I may just hook up the 6 vdc to start, which may be plenty of voltage anyway.
You're referring to the capacitively coupled charge pump arangement to get an extra rail ?
Sure . It works. I use it sometimes to generate the unreg volts for an audio 'phantom power' 48V supply when an extra winding on the TX is inconvenient.
Don't use that method for too much current but it works just fine for sure.
Graham
"Porky"
** Use the 12 volt AC supply and make a voltage doubler with two diodes and two filter caps in series.The mid point of the two caps becomes the 0 volt line and you will have about +/- 18 volts DC on the ends.
......... Phil
"Porky"
** Use the 12 volt AC supply and make a voltage doubler with two diodes and two filter caps in series.The mid point of the two caps becomes the 0 volt line and you will have about +/- 18 volts DC on the ends.
** They have the same pin connections and generally will be compatible.The 5532 has far better high frequency performance and a lot less noise.
BTW here is an example of the "voltage doubler" arrangement I mentioned:
......... Phil
18V is too >high.
The data sheet says 22 volts. Don't worry I won't get anything over 12, and only if 6 is unsatisfactory.
"Porky"
** Learn how to post properly !!!!!Always leave previous material visible and the name of the poster too.
See how others do it - do the same.
** You do not - as you do not need 1 amp............ Phil
You'd better check the voltage rating on that NE5532. +/- 18V is too high.
6V may be a little on the low side for the MC1458 considering line variation of +/-15% and unregulated operation. Something like +/-9V derived from the 12VAC plug-in and regulated by a shunt zener would be much better. Taking one of the smallest transformers from Radio Shack 12.6VAC @300mA as an example source, you would first determine the circuit current draw. The LM1458 lists 5.6mA as a maximum for the entire IC, and allowing a total load to GND off of either rail of 1K gives a circuit draw of 15mA from a single rail. Then choosing the 1N4739A, a 1W 9.1V zener diode, and allowing for 20mA nominal bias current, this gives 35mA total loading of each rail. One of the simplest approximations you can make is to design for a diode conduction angle of +/-45o about the input voltage peak on the waveform. This makes the input Irms=2.2 x Idc or 2.2 x 35mA= 77mA. It also makes the peak input current pi/cos(45)*Idc or 4.44 x 35mA= 160mA. Assuming you will be using 1N4001 for the diodes, you will have ~0.8V of average forward voltage drop over the conduction interval at this current. Getting back to the transformer, the 12.6VAC@300mA is usually at 10% regulation, so this means the open circuit transformer voltage is 1.1 x 12.6=13.9Vrms and an equivalent series resistance of 0.1*12.6/300mA=4.2 ohms. The peak open circuit voltage is 1.414*13.9=19.7V. At 45o, the voltage is 0.707 x 19.7=13.9V and the mean DC developed across the filter capacitor will be 13.9-0.8=13V. Since we know the peak input current is 160mA, this means the transformer requires a series resistance of (19.7-0.8-13)/0.160=36.2 ohms, and subtracting the 4.2 already there makes for 33 ohms external resistor. Getting back to the 35mA circuit load at 9.1V, the series circuit resistor must be (13-9.1)/0.035=100 ohms. Going for 10% ripple at this loading means 15ms of discharge at 35mA results in less than 0.2*13.=2.6V so that C= 0.035mA*0.015ms/2.6V=200uF, so make this 220uF at 25WVDC. This pretty much is all you need and the circuit looks like this: View in a fixed-width font such as Courier.
. . . . +--------+ 5W 1N4001 1/4W . | |--[33]---+-|>|------+--[100]----+---+-> (+)9V . | | | | | | . | | | | 1N4739A| | . |12.6VAC | | + | --- | . | 300mA | | === // \\ === . | | | |220u --- |0.1u . | | | |25WVDC | | . | | | | | | . | |--------------------+-----------+---+----+ . +--------+ | | | | -+- . | | 1N4739A| | /// . | + | --- | . | === // \\ === . | |220u --- |0.1u . | |25WVDC | | . | 1N4001 | | | . +-| (-)9V . 1/4W . . .
That should be Irms=3.1 x Idc at 45o for 110mA,rms at Idc=35mA. The power dissipation in the 33 ohm series resistor is the 2 x Irms^2*33 =0.8W.
Phil Allison wrote:
No- they were derived by brute force integration assuming the capacitor charge up was of negligible consideration- the 45o ratio is 2.0 and not
3.0- still can't get it right: View in a fixed-width font such as Courier.. . . -I->
. . V >---[Rs]-|>|--+----+ . in | | . C === [RL] . | | . - - . . . || . Imax- _ _ . / \\ / \\ . | | | | . | | | | . | | | | . 0----/ \\-----------/ \\-------- . . | | . . 90-A 90+A . . . Imax pi . Idc=---- * sin(A) and Imax= ------ x Idc . pi sin(A) . . . --------------------- . |A x pi/180 +sin(2A)/2 . Irms=Imax x |--------------------- . \\ 2pi . . . . --------------------------- . Irms | A x pi/180 +sin(2A)/2 . ---- = | pi x --------------------- . Idc \\ 1- cos(2A) . . . . . ------------------ . | | . | A | Irms | . | | ---- | . |(degs) | Idc | . +-------+-----------+ . | 0 | oo | . |-------+-----------| . | | | . | 15 | 3.5 | . |-------+-----------| . | | | . | 30 | 2.5 | . |-------+-----------| . | | | . | 45 | 2.0 | . |-------+-----------| . | | | . | 60 | 1.8 | . |-------+-----------| . | | | . | 75 | 1.6 | . |-------+-----------| . | | | . | 90 | 1.6 | . ------------------- . . .
The circuit shown with Rs=33 buys you a few things besides maintaining regulation over an input of 105-135VAC and 50-70Hz. If you went for Rs=0 and straight charged the capacitor, besides introducing significantly higher current harmonics, the RMS/DC ratio goes to something like 3, up from slightly less than 2 with Rs=33, making the RMS 50% larger. The Rs=33 ohm circuit dissipates 320mW in Rs and 80mW in the two zener resistors for a total of 480mW. The Rs=0 circuit dissipates about 580mW in its two zener resistors for the same bias.
"Fred Bloggs"
** Were those 2.2 and 3.1 figures ( rms to DC ) were arrived like this ? 90/360 = 0.25sqrt 0.25 = 0.5
recip 0.5 = 2
add 10 % = 2.2 (for the curvature of the wave )
AND
45/360 = 0.125sqrt 0.125 = 0.3535
recip 0.3535 = 2.83
add 10% = 3.11
......... Phil
Your assumptions about this transformer are overly optimistic. The published parameters are: 120 VAC in gives 16.38 VAC out, unloaded, and 12.6 VAC out with a 300 mA load, which is 30% regulation (exactly 30.00000%; I think that must have been their design goal).
The actual transformer I have measures 16.48 VAC unloaded with 120 VAC in, and
11.75 VAC out with a 300 mA load, which is 40% regulation. The measured primary DCR is 316.4 ohms, and the secondary DCR is 7.75 ohms (at 20 degrees).This changes your design a bit.
If you assume +/- 45o (about the peak) diode conduction angle (total of 90o conduction angle), then the diode is *not* conducting for 270o, which is 270/360/60 or 12.5 mS.
That's ridiculous- nobody makes a transformer that bad.
Actually, it does not. Leave all component values the same but up the
100 ohm zener series resistors to 1/2W rating, and up the filter caps to 35WVDC. The supply performs *much* better now that it has all that much more open circuit voltage to work with:-)The big e-caps are +80/-20%, unless they are RS rejects, so I'm not going to quibble about ripple discharge for even +/- one period, this is half-wave. SPICE returns a conduction angle of 110o at nominal line and load conditions.
Fred Bloggs wrote:
Ok, Fred, you are probably right about the 6V being inadequate, butt.. i would just take that at face value, not that the math is entirely meaningless. I went ahead and built an op amp circuit aruond the 1458. The power source is a Canon 1A wall wart which puts out 16.8 vdc unloaded. I used a simple voltage divider to deliver about 8.4 vdc to the V+ of the chip, grounding the V-. This works fine, buttt..... there is too way much noise generated by this circuit. Attempting to boost the weak signal from a salvaged reverb circuit (4 transistors and a tiny spring tank), using the same power supply wall wart, two discrete transistors seem to do the job better than the op amp, quieter. Of course my design rationale is faulty. I just try things and compare the results. Next time I may use a proven op amp design, copied from Roy Mallory's research. For this project, I'll just go with transistors.
Until superconducting wire becomes commonly available for transfomer construction, there is no choice but to make small transformers "that bad". The center leg of this transformer is .95 x .95 cm. The winding window is .7 cm x 2 cm. If we calculate how many turns would be required to give a peak flux density of 15000 gauss with 120 VAC applied, the required number is 4287 turns. The area of the window is 1.4 cm^2; making some allowance for packing factor, the wire size that will allow 4287 primary turns to fit in half the window (.7 cm^2) is 36 gauge. The mean length of turn is about
5.4 cm., so we will end up with about 759.5 feet of wire if the primary is wound on first. Since 36 gauge copper wire has a resistance of about .4148 ohms/ft (at 20 degrees), 759.5 feet will give about 315 ohms for the primary, a value very close to the measured value.A similar calculation for the secondary will give a winding resistance which is close to the measured value.
It does if you consider component ratings part of the design.
I will plug in a 5532 when I get my hands on one. Not giving up on it quite yet.
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