[joneeboi]

Not to be thwarted by power supply complications, I gave it a bit more thought and I'm wondering what all your reactions are to putting two TI 5V-to-9V DC boosters DCP020509P's in series and dropping it to 15V with Fairchild's KA78R15CTU (datasheet: http://www.fairchildsemi.com/ds/KA%2FKA78R05C.pdf). It seems that if we want to use isolated boosters, you either have to pay through the nose for lots of power or use multiple chips either in series or parallel. rds is going with two DCP020515DP's in parallel because he wants to drop it to 12V with the TL750L12CLP. If I go with the DCP020509P, I get 222mA right off the bat, and putting two in series gives me a substantial 18V. I imagine the ripples will superimpose themselves over the DC signal which will double my output AC voltage, but the output capacitors and LDO can be configured so as to care of that (right?). I figured the LDO should have a higher current rating than the DC booster so as to not waste precious resources. I'll have to figure out how to filter the 800kHz switching noise, but that'll come after figuring out which devices I want to work with.

What do you guys think?

 

[millwood]

Quote:

but the output capacitors and LDO can be configured so as to care of that (right?).

the chips oscillate at 400khz (800khz/2) and the ripples will be in that territory as well.

most regulators and eletrolytic capacitors are useless at that kind of frequency.

if you really want to lower ripples, try a pai-filter or something like that for high current application, or a simple crc filter if you are sure that the current drop isn't going to be significant here.

don't take my words on that: TI has an application note on how to reduce ripple for those chips.

 

[rds]

Quote:

so is that enough to establish the fact that dcp01 (no B) is a bipolar process?

Yes, the DCP01 was a bipolar. It is not longer produced. All the chip types are now CMOS/DMOS. Referring to the DCP01B as DCP01 is perfectly acceptable as there is only one DCP01 now - that is the non bipolar one.

Quote:

the first one is the dcp01/02, using bipolar processes.

So this is wrong

Quote:

I would take a wild guess that Bi in BiCMOS means Bipolar?

Yes it refers to a bipolar bjt and cmos in one package. It does not refer to the chip being made using a bipolar process.

Quote:

nor the higher wattage? nor the 2% better regulation? ...

Read my post - I'm referring to the DCV. The DCV is 1W. The DCP02 is 2W and better regulated.

Quote:

nobody said it is. better reading would have helped here.

Read the last page, it is there (not from you, some else).

You are also incorrectly referring to the ripple at 50% when I was referring to it at 70%.

I'm not going to bother refuting you point by point anymore. You've done a great job of making this very confusing for anyone trying get some factual information from this thread.

I hope you can stop spamming this thread with nonsense.

 

[rds]

Quote:

If I go with the DCP020509P, I get 222mA right off the bat, and putting two in series gives me a substantial 18V

That sounds like a good idea. Just remember that the usb bus limits you to about 105mA at that voltage.

A pi filter between the dc-dc and ldo is the best way to get rid of ripple.
As I suggested earlier this choke with these ceramic caps on either side will make a very appropriate pi filter for this application.

This Murata DC-DC converter is a very good isolated alternative to the ti chips. This will allow you to have 140mA at 15V using a single chip.

 

[joneeboi]

Deep breaths all around.

Hm, I forgot about doing the power conversion. Where is that 105 coming from? I get,

P = V1*I1max = V2*I2max
(5V)(500mA) = (18V)*I2
I2max = 138.9mA

But I guess

2W = (18V)*I2
I2 = 111mA

and I shouldn't try to get to that limit. I knew I was missing something with the higher voltage. I looked at that Murata unit, but it has 5 stock and a 20-week factory lead time. It has the higher power rating and a nicer end voltage and price than the two chip method. I should probably keep myself busy for the next five months then, eh?

Ending up with a 10V or 12V regulated supply wouldn't be all bad.

I'm just going to continue thinking out loud so that people can correct me or give input where necessary.

Current needs:
55mA for BantamDAC
Suppose AD8397 as opamp choice. Max current draw is 30mA.

Max power:
P = (5V)(500mA) = 2.5W

Set voltage boost with (5V)(500mA) = V2*I2 to get max current output:
V2 = 15V allows I2 = 166mA
V2 = 12V allows I2 = 208mA

Leftover current for PS sections:
V2 = 15V leaves 81mA
V2 = 12V leaves 123mA

Keep in mind inefficiency and tolerances.

Some candidates for DC boosting:
15V

• Cincon EC3AE03M
  EC3AE03M
  Efficiency 61%, no load current draw 120mA (!!!)
• Murata NDTS0515C
  NDTS0515C

12V

• Murata NDY0512C
  NDY0512C
  Efficiency 73%, no load output current 33mA, input current full load 533mA (yikes),
• Murata NDTS0512C
  NDTS0512C
  Efficiency 76%, min output load current 63mA, max input current 769mA
• Murata HL02R05S12YC
  HL02R05S12YC
  Efficiency 69%, min load input current 70mA, max input current 580mA
• Murata HL02R05S12ZC
  HL02R05S12ZC
  Efficiency 69%, min load input current 70mA, max input current 580mA

Components for PI CLC filter:
C330C105K5R5TA
PLY10AN1121R8R2B

Humbug.

 

[millwood]

Quote:

Originally Posted by rds /img/forum/go_quote.gif Referring to the DCP01B as DCP01 is perfectly acceptable as there is only one DCP01 now - that is the non bipolar one.

I'm not going to bother refuting you point by point anymore. You've done a great job of making this very confusing for anyone trying get some factual information from this thread.

I hope you can stop spamming this thread with nonsense.

 

[millwood]

Quote:

Originally Posted by rds on what BiCMOS is /img/forum/go_quote.gif Yes it refers to a bipolar bjt and cmos in one package.

taking a class on semiconductor processes would great benefit your knowledge and save all of us more confusion.

"bipolar bjt"? as supposed to bipolar MOSFET, or CMOS bjt? what do you think that "b" in "bjt" stands for, rds?

no answer is needed,

.

 

[rds]

Quote:

Where is that 105 coming from?

You need to multiply by the efficiency of the dc-dc converter and the voltage it outputs (~.85, and 20 respectively):

(5V)(500mA)*0.85 = (20V)*I2

~105mA

Remember that is assuming you aren't drawing any power for the DAC.
So (5V)(450mA)*0.85 = (20V)*I2

~95mA

I really like that 15V Murata for the pimeta with lmh6321 since it draws about 60mA and the min current for the Murata is 50mA.
Also on USB 2.0 it allows for 130mA at minimum efficiency.
I may end up ordering one of those too

 

[millwood]

you may want to do the math backwards.

out of your usb, you get 5v 500ma max.

of that, the dc-dc converter takes 40ma (at 5v) per dcv010512D. so you have 5v 460ma (2.3w usable) to be converted.

the converter efficiency is about 50 - 80% in the 30% load - 100% load range so let's assume 75% efficiency. that means you have 2.3w*75%=1.7w available on the output.

at +/-12v output, that leaves you 1.7/24=70ma both rails.

math for the rest is simple.

obviously, it doesn't make sense to use such a high voltage for headphone amps but the math is similar.

btw, ad8397 idles at 7ma/amp typical and 8.5ma/amp max. where did you get the 30ma figure?

 

[millwood]

Quote:

Originally Posted by millwood /img/forum/go_quote.gif btw, ad8397 idles at 7ma/amp typical and 8.5ma/amp max. where did you get the 30ma figure?

I figured that out: i was looking at the wrong page,

.

the 30ma figure is correct for +/-12v.

 

[error401]

Quote:

Originally Posted by millwood /img/forum/go_quote.gif I have a very difficult time understanding how that could be the case. could you elaborate please?

Standard centre-tapped full-wave rectifier, as in the datasheet:

Standard centre-tapped half-wave doubler, as we can assume is being used:

Circuit above with the outputs connected together:

Not a dead short but, uhm, I wouldn't want to wire it up and apply power...

 

[joneeboi]

Come on, guys, deep breaths all around.

I probably won't go with 24V, but thanks for the math help, fellows. I knew I had to factor in the current draw of the booster circuit, but I was just mixed up about the pre-boosted and post-boosted currents.

*more thinking out loud*

The NDTS0515C outputs 50mA with no load, meaning it draws 197mA from the input at minimum.

I1*(5V)*0.76 = (15V)*(50mA)
I1 = 197mA
I(leftover) = 500mA - [I1 + I(DAC)] = 500mA - 197mA - 55mA
I(leftover) = 248mA
(248mA)*(5V)*0.76 = (12V)*I2max
I2max = 78mA

At 10V, the AD8397 draws 24mA in stereo configuration. This converts to 76mA at the input, so total current draw so far is 288mA from the USB. The TL750/751 datasheet doesn't mention anything about its quiescent current, but we still have plenty of headroom, it seems. It has a maximum output current of 150mA, so we should be quite safe there. But just to make sure,

I1*(5V)*0.76 = (15V)*(150mA)
I1 = 592mA
I(total) = 647mA

General equation for future use is

I1 = (3/0.76) * I2

I don't think we could get that kind of current with the amplifier, but now I'm wondering what happens when headphones are plugged in and unplugged from the jacks. What kind of current spike is happening there? Maybe I'm being paranoid. I haven't really seen a discussion about it yet except for in the PPAv2 "Parts Selection" section for the output transistors. Google hasn't really turned up much info either, so I'll take the general silence to mean that I don't really have to worry about it.

Hm, does that mean I have one possible configuration lined up? Must let information settle before continuing...

 

[millwood]

Quote:

Originally Posted by error401 /img/forum/go_quote.gif Not a dead short but, uhm, I wouldn't want to wire it up and apply power...

I thought we were talking about paralleling the chips / 2ndary windings of different chips, not paralleling 2ndary windings of the same chip.

 

[millwood]

I don't know if your interpretation of "minimum load current" is accurate. the way I read the datasheet, it is saying that converter has to have a minmum load of 50ma to work properly. that 50ma is going to be satisfied, likely, by your opamp idle current + other load, and if anything, you can parallel a resistor to generate the rest.

if you do that, you don't need to double count the minimum load current.

the datasheet doesn't, unfortunately, provide the idle current for the converter itself or an efficiency curve to help us estimate that.

so if you stay with the 15v side, you have a current draw of 50ma (which will include your opamp + dummy resistor load), maybe a few ma for your linear regulator and you are looking at maybe 60ma.

@ 15v, that's a power dissipation of 60*15=900mw.

at this kind of load levels, your efficiency will be much lower than 77% specified in the datasheet. it is likely in the 50% - 60% range. let's take 50% for now.

that means your usb side will have to provide 900mw/0.5=1.8w. At 5v, that means a current draw of 1.8w/5=360ma.

maybe your dc-dc converter needs to draw another 40ma at idle so you have 100ma (500ma-360ma-40ma) to spare.

the real numbers are likely in the ball park, I think.

Quote:

now I'm wondering what happens when headphones are plugged in and unplugged from the jacks. What kind of current spike is happening there?

a PROPERLY designed headphone amp will always have a resistor OUTSIDE of the feedback loop exactly for such occasions - however, I have seen lately plenty examples of poorly designed headphone amps in that regard,

.

the short taken place when the phone is being plugged into the amp is likely going to last a "long" while (a few ms to a few hundred ms?), they are more like a true DC short from the amp's perspective.

if your amp is not properly designed, the amp will try to maintain a voltage over a short. a few things may happen:

1) the amp's protection circuitry kicks in;
2) the amp's RE resistor, if it has one, took the bulk of the fall.
3) the output transistors get destroyed,
4) the rail voltage droops to a point that it saves the amp.
...

I would never use a headphone amp that does NOT have an output resistor OUTSIDE of its feedback loop.

 

[rds]

Joneeboi-

The maximum efficiency of a dc-dc converter is rated for its maximum output current.
Since the converter has a load independent current draw the proportion of output to input at low currents is very poor.
The nice thing about efficiency is that it is defined as the power output over the power input. So many of these other calculations are pointless. Once you're drawing about 40% of the maximum load the efficiency doesn't change a lot. So for a 77% efficient converter it is safe to assume about 73% efficiency at 50% load.