newbie choosing toroid
« on: March 14, 2005, 03:43:04 PM »
Hey folks,

I'm salvaging some parts (mainly connectors, knobs, an input transformer, output transormer, etc...) to make my own api-style mic pre (all salvaged parts point to that sort of thing).

Basically, I want to build a ps for two channels, regulate down to +/-18v for the opamp, and maybe get phantom out of it somehow.

I guess I'm just wondering how to choose the right toroid for this project. Ultimately, I can get a transformer with dual primaries +/- 18v. It sounds too easy though. Would I no longer need voltage regs (which I've already got) for the 18v? Basically just rectify those primaries for +/- supply lines, then filter? Or would it be better to get 2*24v and voltage divide/filter?

In any case, what's a good size toroid? I'd probably be safe with .5 amps? This is the part where it would be very handy knowing how much current such a mic pre is going to draw. To do this, would I just take voltage/resistance at the opamp somehow and figure current requirements using ohm's law?

Sorry, I've been on the board for a couple hours searching posts.

Thanks folks!

kelly


PRR

newbie choosing toroid
« Reply #1 on: March 14, 2005, 04:22:38 PM »
> I can get a transformer with dual primaries +/- 18v. Would I no longer need voltage regs (which I've already got) for the 18v?

18 V RMS AC will rectify to 25.45 V DC.

newbie choosing toroid
« Reply #2 on: March 14, 2005, 04:49:02 PM »
Thanks very much! I will now research and find out why. Sorry for the dumb question...

kelly

mwkeene

newbie choosing toroid
« Reply #3 on: March 14, 2005, 04:59:13 PM »
If you wand to know why 18V RMS is rectified to 25.45V, it's simply because the rectified and filtered voltage will be pretty close to the peak voltage output of the transformer.

RMS voltage =  (.707) *  (Peak to Peak Voltage)

newbie choosing toroid
« Reply #4 on: March 14, 2005, 05:11:16 PM »
This isn't "full-wave" rectified, is it?? If you "full-wave" rectified it, wouldn't it be double that, or around 51.4??

And thanks, i just found RMS links off the meta...

kelly

mwkeene

newbie choosing toroid
« Reply #5 on: March 14, 2005, 05:18:48 PM »
no, because full wave rectifiers take the absolute value of the incoming sine wave, so where you had negative peaks before, you now have positive peaks.  Nothing about the amplitude of the voltage changes, just the frequency (there will now be twice as many positive peaks and no negative peaks, doubling our frequency).  Hope this helps...
-Mike

bcarso

newbie choosing toroid
« Reply #6 on: March 14, 2005, 05:22:33 PM »
Kelly, as far as sizing for current, 0.5 amp will typically sound like overkill.  However, toroids especially have a tendency to saturate on the peak charging currents associated with the typical diode/cap-input filter, and when they do they tend to radiate transient magnetic fields.  The charging transients also have more high-frequency components and the toroid cores tend not to work as well at higher frequencies, at least when they have been optimized for mains frequencies. These transient mag fields can get into the low-level circuitry and produce some nasty buzzing. And magnetic shelding is tricky and expensive.

I like to put some resistance in series with the transformer leads and use a few more volts over what would otherwise be plenty of headroom for the regulators.  This limits the peak currents and reduces the magnetic radiation effect.  Also, the higher current rated transformer helps as well.  With your 18VAC windings you should have a little room to do that already---as PRR pointed out, with ideal rectifiers and a perfect 18V rms sinusoid you will get more than 25V d.c.  Your diodes will subtract a bit (~1.2V for a bridge on a single winding, half that for a center-tapped bridge with respect to each d.c. output) from this, but your transformer is probably rated for that 18V at the rated load, resistive.  They will usually be a fair amountl higher at lower currents. Factor in the lowest a.c. line voltage proportionally though, and remember that the tops and bottoms of the line voltage waveform can often already be a bit clipped due to lots of diode-cap loads in local equipment---both of these stealing some of your d.c. voltage.

Also, put the transformer as far away from the other circuitry as possible, and minimize loop areas of conductors, especially in the early stages of the preamp.

The rectifier diodes themselves can have a nasty very-high-frequency snap when they are just reverse-biased after conducting forward current.  This will radiate RF energy which can as well be partially rectified in low-level circuits with nonlinear elements, typically the base-emitter junctions of bipolars.  Diodes specially designed to minimize this transient exist, and as well noise suppression networks can be connected across the diods to damp the energy or at leasy shift it to a lower frequency that doesn't radiate as efficiently.

newbie choosing toroid
« Reply #7 on: March 14, 2005, 10:46:13 PM »
Quote from: "bcarso"
Kelly, as far as sizing for current, 0.5 amp will typically sound like overkill.  However, toroids especially have a tendency to saturate on the peak charging currents associated with the typical diode/cap-input filter, and when they do they tend to radiate transient magnetic fields.  The charging transients also have more high-frequency components and the toroid cores tend not to work as well at higher frequencies, at least when they have been optimized for mains frequencies. These transient mag fields can get into the low-level circuitry and produce some nasty buzzing. And magnetic shelding is tricky and expensive.

I like to put some resistance in series with the transformer leads and use a few more volts over what would otherwise be plenty of headroom for the regulators.  This limits the peak currents and reduces the magnetic radiation effect.  Also, the higher current rated transformer helps as well.  With your 18VAC windings you should have a little room to do that already---as PRR pointed out, with ideal rectifiers and a perfect 18V rms sinusoid you will get more than 25V d.c.  Your diodes will subtract a bit (~1.2V for a bridge on a single winding, half that for a center-tapped bridge with respect to each d.c. output) from this, but your transformer is probably rated for that 18V at the rated load, resistive.  They will usually be a fair amountl higher at lower currents. Factor in the lowest a.c. line voltage proportionally though, and remember that the tops and bottoms of the line voltage waveform can often already be a bit clipped due to lots of diode-cap loads in local equipment---both of these stealing some of your d.c. voltage.

Also, put the transformer as far away from the other circuitry as possible, and minimize loop areas of conductors, especially in the early stages of the preamp.

The rectifier diodes themselves can have a nasty very-high-frequency snap when they are just reverse-biased after conducting forward current.  This will radiate RF energy which can as well be partially rectified in low-level circuits with nonlinear elements, typically the base-emitter junctions of bipolars.  Diodes specially designed to minimize this transient exist, and as well noise suppression networks can be connected across the diods to damp the energy or at leasy shift it to a lower frequency that doesn't radiate as efficiently.


Wow. Thanks so much for such comprehensive advice. I now understand the RMS thing (or at least insomuch as I need to for the moment), which is good, 'cause I already got the regulators :green:.

Now it's time to start researching how to get phantom out of it. If I wanted, could simply tap the secondaries for a third supply rail, then voltage multiply it and regulate it to 48 v?

Again, thanks.

kelly

bcarso

newbie choosing toroid
« Reply #8 on: March 15, 2005, 12:01:40 AM »
Kelly: "Now it's time to start researching how to get phantom out of it. If I wanted, could simply tap the secondaries for a third supply rail, then voltage multiply it and regulate it to 48 v? "

Well, sorta.  Voltage multipliers are beastly unless currents are quite low. Hey, Cockcroft and Walton iirc used them for power supplies to first artificially split the atom, so they can't be all bad. Having said that, there are versions with nearly doubled numbers of parts that at least do a full-wave loading of the given winding.  More typical ones tend to be half-wave and have ripple at the mains frequency (and harmonics thereof) rather than at double the mains freq.  Putting a bit of d.c. into the secondary with a half-wave isn't the end of the world if the current is low enough.

But if you can manage it, I'd opt for an independent winding or transformer.

Family Hoof

newbie choosing toroid
« Reply #9 on: March 15, 2005, 12:27:18 AM »
Kelly,

I like the Amveco 7000 series torroids which are available through digikey for less than $20 in the smaller sizes. They're encapsulated, usually fit in a 1RU case, and are very easy to mount because of the PCB pins and single screw hole on the bottom. I know that you've built some SCA stuff so you should recognize these. Check out the part # 70054 for example.

As for phantom and voltage doublers using the same mains transformer, you'll see that this topic has been covered very many times if you check the power supplies Meta and Group DIY power supplies pages. Most people seem to use a Villard or Greinacher voltage doubler. I don't know anything about the benefits and disadvantages like bcarso has begun to describe, but just keep in mind that phantom requirements should never be more than 15mA per mic, and theoretically less that that much to keep the regulator working. Do a google search for "voltage doubler" or "voltage multiplier" to learn about the many different ways to arrange diodes and capacitors for rectification with a boost. :thumb:

Also worth checking out, Keith (SSL Tech) has a +/-18V w/phantom supply from a 2x20V mains transformer design that is very popular around here. http://www.beatbazar.com/guests/ssltech/kps-1/index.htm


PRR

newbie choosing toroid
« Reply #10 on: March 15, 2005, 02:24:17 AM »
> it would be very handy knowing how much current such a mic pre is going to draw. To do this, would I just take voltage/resistance at the opamp somehow and figure current requirements using ohm's law?

Sadly, no. Transistors in circuits work very different at 10V-40V than they do with the 1.5V (or 0.2V) from an ohm meter.

If it is entirely chips (power goes only to chips, or chips and high-value bias resistors), then you can estimate the idle current by looking up the chips. TL072 eats about 3mA, 5532 eats about 8mA. Ideally you find a specification for the same power voltage as you will use, but most chips eat nearly the same current at any useful voltage. Maybe rising 30% from 8V to 40V.

When an amp drives a load it takes more power. Load a 1,000 Watt amp with speakers, hit a power-chord, the lights dim. You can approximate this, for common chip topologies, for sine-wave test tones, as a resistor across the power rails with a value 6 times the load resistor. If a perfect (zero idle current) amp drives a 600Ω load to clipping, the total supply rails see something like a 3,600Ω power drain. If you happen to be using +/-18V rails (36V total), then it works out 36V/3,600Ω= 0.010 Amps or 10mA.

So a chip-only mike amp will eat 5 to 20mA idle, plus another 10mA if you do sine-wave tests with 600Ω load. (Speech/music power drain will be much less.)

Many mike amps have a pair of input transistors that eat more power. Usually 1mA-2mA each, up to 4mA total, but a few run richer. Usually the emitter resistor runs from a power rail to near ground (base is grounded and emitter is 0.6V offset). So if you find two 15K emitter resistors, and +/-18V supplies, figure 18V/15K= 1.2mA each, 2.4mA total for these transistors.

You may have LEDs and other frills. One design I looked at recently, the LED ate as much juice as everything else. Relays can be real power pigs.

Back to the power transformer. Just for illustration (and easy math), say it is 10V and 1A. As you know, the "10V" is the RMS of a bent (sine) wave, and our rectifier will charge up to the Peak voltage which is √2 or 1.4 times higher. So we get 14 volts. The rectifier loss is 0.6V (or 1.2V), so we are down to 13V (don't sweat small fractions). The filter capacitor is never big enough for perfectly steady DC: 120 times a second it charges to 13V, then between peaks it discharges. We typically pick a capacitor size to give about 1V of ripple. And what we really care about is how low it goes. So now we are down to 12VDC from a 10V AC winding.

But that "10V" will go up and down with utility-company voltage, and with load. Small transformers rise 20% when un-loaded. Hot summer days drag the wall-power down 10%. My power company (apparently) does not own a voltmeter. So that estimated 12V DC could be 14V or 10V, with 1V of 120Hz ripple.

If we need power cleaner and steadier, we regulate. The regulator can only waste excess voltage, it can't fix low voltage. The regulator itself needs 2V to 4V to stay alive. So we can get maybe 8V or 10V of solid regulated DC from a "10VAC" winding. Since some of those losses are fixed-voltage, and we can usually over-size the transformer (so it never runs at full rated load and sag) at low cost, we can expect to design for good regulated 18VDC from 18VAC windings.

There is no "free lunch" in that √2 or 1.4 times higher raw DC voltage from the rectifier and capacitor. The downside is that the peaky current wave heats and sags the transformer. A winding rated 1A AC can only give about 0.6A DC before heat or sag become real problems.

Worked example. Amplifier to drive 8Ω to 20V peak sine wave. Assume the amplifier wastes 4V on peaks, we need +/-24V DC rails. Assume that we are not too concerned with ripple: no regulator. But we do allow 1V of ripple-dip, 1V for rectifiers, we need 26V peak AC from the transformer winding. 26V peak is 18.4V AC RMS, we want 2*18V (or 36VCT) winding.

At full sine-wave power the amp will act like a resistor of 8Ω*6 or 48Ω. With +/-24V rails, 48V total supply, this is 48V/48Ω= 1 Amp DC. The transformer winding must be rated 1/0.6 or at least 1.6 times higher: we need 1.6A winding rating.

Most chip-based mike preamps eat more like 30mA. Rack four of those, 120mA DC. Times 1.6, 192mA or 200mA or 0.2A AC needed. So yes, a 0.5A winding is generally ample for a few mike preamps. And for total regulated DC voltage of 24V-48V with simple rectifiers (not doublers) we can rule-of-thumb say that VAC=VDC: we need 18VAC to rectify to 25VDC +/-20% that we can regulate to solid 18VDC. So the spec looks like 2*18VAC at 0.2A. This is 36VAC and 0.2A, 36V*0.2= 7.2VA. Transformers are priced by the VA, the maximum energy they can handle. And trannies smaller than 10VA are not cheaper (it costs more to make them too small), and 25VA trannies are not much more dollars than 10VA trannies. So for mike-amp type supplies, always super-size the transformer: you get more raw DC for better regulation, and it will run cooler, for only a buck more.

Rectifiers must be rated for 2.8 times the maximum transformer voltage. These days, 500V rects cost the same (in DIY quantity) as 50V, so buy a bag of 500V rects. The rectifier must be rated for the Amps of the transformer (not the load). The rect sees big spikes from the transformer to the capacitor, not the average DC out of the cap. Big transformer, big spikes. A 1N1006 rectifier rated 1A is barely big enough for a 1A winding. If you miss this point, it will work for a few hundred turn-ons and one day die at switch-on. So if you are close, always round-up the rectifier current rating to the next bigger size.

> rectifier diodes themselves can have a nasty very-high-frequency snap

bcarso is quite right. But if you are just starting to build power supplies, this is a frill. Excellent audio can be made with $0.07 1N1005 diodes. Build your first few supplies with generic parts. Tweak, listen, tweak. One of those tweaks will be $7 rects instead of $0.07 rects. But it is discouraging to make beginner mistakes with expensive parts. And you may find that, in some circuits and systems, fancy rects don't make a difference. Or if they do, trying the ordinary rect first gives you a comparison point.

> Keith (SSL Tech) has a +/-18V w/phantom supply from a 2x20V mains transformer design that is very popular around here. http://www.beatbazar.com/guests/ssltech/kps-1/index.htm

And whenever possible, plagiarize! There is nothing new under the sun, and no shame in following the paths of others, who followed paths worn by those before them. Be sure what you are stealing is appropriate to the job, and that you understand the basic design, and adapt as needed, but plagiarize! (Only: call it "research".)

newbie choosing toroid
« Reply #11 on: March 15, 2005, 03:29:44 PM »
Quote from: "PRR"
Sadly, no. Transistors in circuits work very different at 10V-40V than they do with the 1.5V (or 0.2V) from an ohm meter.


If it is entirely chips (power goes only to chips, or chips and high-value bias resistors), then you can estimate the idle current by looking up the chips. TL072 eats about 3mA, 5532 eats about 8mA. Ideally you find a specification for the same power voltage as you will use, but most chips eat nearly the same current at any useful voltage. Maybe rising 30% from 8V to 40V.

When an amp drives a load it takes more power. Load a 1,000 Watt amp with speakers, hit a power-chord, the lights dim. You can approximate this, for common chip topologies, for sine-wave test tones, as a resistor across the power rails with a value 6 times the load resistor. If a perfect (zero idle current) amp drives a 600? load to clipping, the total supply rails see something like a 3,600? power drain. If you happen to be using +/-18V rails (36V total), then it works out 36V/3,600?= 0.010 Amps or 10mA.

So a chip-only mike amp will eat 5 to 20mA idle, plus another 10mA if you do sine-wave tests with 600? load. (Speech/music power drain will be much less.)

Many mike amps have a pair of input transistors that eat more power. Usually 1mA-2mA each, up to 4mA total, but a few run richer. Usually the emitter resistor runs from a power rail to near ground (base is grounded and emitter is 0.6V offset). So if you find two 15K emitter resistors, and +/-18V supplies, figure 18V/15K= 1.2mA each, 2.4mA total for these transistors.

You may have LEDs and other frills. One design I looked at recently, the LED ate as much juice as everything else. Relays can be real power pigs.

Thanks so much for breaking this down PRR. Actually having a list of addititives is more than I could have hoped for!
Quote from: "PRR"

Back to the power transformer. Just for illustration (and easy math), say it is 10V and 1A. As you know, the "10V" is the RMS of a bent (sine) wave, and our rectifier will charge up to the Peak voltage which is ?2 or 1.4 times higher. So we get 14 volts. The rectifier loss is 0.6V (or 1.2V), so we are down to 13V (don't sweat small fractions). The filter capacitor is never big enough for perfectly steady DC: 120 times a second it charges to 13V, then between peaks it discharges. We typically pick a capacitor size to give about 1V of ripple. And what we really care about is how low it goes. So now we are down to 12VDC from a 10V AC winding.


Thanks to you and mcarso, now I understand WHY 60 hz full-wave rectified will get me 120 hz.

Quote from: "prr"



At full sine-wave power the amp will act like a resistor of 8?*6 or 48?. With +/-24V rails, 48V total supply, this is 48V/48?= 1 Amp DC. The transformer winding must be rated 1/0.6 or at least 1.6 times higher: we need 1.6A winding rating.

I understand ohm's law math for the 1 amp dc requirement, but where is the 8 ohms x 6 coming from? That one's got me!

Quote from: "prr"
Most chip-based mike preamps eat more like 30mA. Rack four of those, 120mA DC. Times 1.6, 192mA or 200mA or 0.2A AC needed. So yes, a 0.5A winding is generally ample for a few mike preamps. And for total regulated DC voltage of 24V-48V with simple rectifiers (not doublers) we can rule-of-thumb say that VAC=VDC: we need 18VAC to rectify to 25VDC +/-20% that we can regulate to solid 18VDC. So the spec looks like 2*18VAC at 0.2A...


Crazy. How much would it cost me to learn this electronics school? PRR and mcarso, if you send an address, I will send you beer/cheese/crackers. You guys are the best.

Kelly

PRR

newbie choosing toroid
« Reply #12 on: March 15, 2005, 03:59:04 PM »
> I understand ohm's law math for the 1 amp dc requirement, but where is the 8 ohms x 6 coming from?

In my third paragraph, I state (without derivation or proof) "You can approximate (load-power demand), for common chip topologies, for sine-wave test tones, as a resistor across the power rails with a value 6 times the load resistor".

newbie choosing toroid
« Reply #13 on: March 15, 2005, 05:12:29 PM »
Thank you kind sir.

kelly


 

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