I'm having trouble finding a replacement power supply that has the same specs as the original empeg power supply (15V @ 1A).

Could anyone tell me if they have succesfully operated the player on a power supply producing 13.8V @ 1.5A, or whether this would theoretically be safe and sufficient?

FAQ entry.

Edit: Wow, I actually managed to FAQ him first. Is this the Tonybot's night off?

Yep, saw that. Doesn't answer my question, though.

Doesn't answer my question, though.

From the FAQ: The voltage can be in the range of about 12v-16v DC, the exact voltage doesn't matter much.

I run my on 12V, @ 3amps on a regular basis with no ill affects. Your specs should be fine, worst case scenario is the drive(s) don't have enough power to spinup and you have to find a more powerful supply. Just make sure it is center positive so you don't fry anything.

-Mike

I don't know much about electronics. I've always wondered why the exact voltage doesn't matter much. Those words were paraphrased from either Hugo or Rob if I recall correctly, but I never understood why. Anyone care to explain? In layman's terms?

Ok, that sounds good. If you're running fine on 3 amps, 1.5 must be safe.

Actually, the worst case scenario is the one involving blue smoke, which does get a mention in the FAQ entry

Oh, I see. You were worried about it being too much amperage. Actually, if I understand it correctly, what you should be worried about is too little amperage, not too much amperage. Re-reading the FAQ entry I see that it doesn't make that point very clear. I will update it.

Thanks.

It's funny how a FAQ entry can be 50 lines long, and still not answer the question for everybody...
I guess a FAQ maintainer's work is never done.

Ok, that sounds good. If you're running fine on 3 amps, 1.5 must be safe.


No, no, no!

The amperage rating of the power supply is irrelevant as long as it is equal to or greater than the amount of power the player needs in order to operate.

You could have a 13 volt power supply capable of outputting 2,000 amps -- and the player would only draw about 1 amp from that power supply.

Where you run into problems is taking a dual drive player which requires perhaps 1.2 amps to run it, and hooking it up to a power supply which can only deliver 8/10 of an amp.

tanstaafl.

Anyone care to explain? In layman's terms?

Well, my laymans understanding is that the power must be regulated within the player. A car is going to put out a wide range of power (probably anywhere from 9V to 16V) while being started, using headlights, accessories, etc. which the player must compensate for. That same regulator gives us leeway on which power supplies we can use and as tanstafl mentioned the amperage is irrelevant as long as it is sufficient.

-Mike

Yep. What tanstaafl said.

Just make sure the supply outputs DC and not AC and that the center connector in the plug is positive.

But if the player is running at 12v, and the wire supplying it the power is running at 14v, then isn't some of that energy being converted to heat by the voltage regulators?

Or is it like amperage, in that it only draws what it needs?

I admit, I still never understood the difference and relationship between voltage and amperage, no matter how many times I looked at the water-faucet-and-tank metaphor pictures.

And I think once I briefly grasped the concept of torque vs. horsepower, but I think that's now lost again.

Yes. The voltage regulators inside the player will drop whatever excess voltage there is and output 5V/12V/whatever that the circuitry inside needs. That's why there is a limit on how high the voltage can go. Any higher and either heat output will exceed what the voltage regulator is specified for or it will exceed what the rated maximum voltage is. It has the same effect which is to burn out the regulator and you'll end up with a dead empeg.

Higher voltage = More the regulator has to drop off

Current is would be equivalent to how much water I can get out of the pipe in a second.
Voltage is how much pressure that water is coming out at.

The two are linked but aren't the same thing.

Say I was thirsty, I can drink exactly how much I want from the end of a pipe.
I can cope with large amounts easily, it'll be like drinking out of a river but I can't cope with high pressures. That would be like drinking out of the end of a firehose.

I'm having trouble finding a replacement power supply that has the same specs as the original empeg power supply (15V @ 1A).

I have two original empeg indoor power supplies here and neither of them have those specs on them. They both say output is DC 12V 1A. That shouldn't be very hard to find.

Really? I threw mine out after it broke, but I'm almost certain it was 15V...

I have been surviving on 12V @ 1A, which is easy to find, as you say. With the two disks in mine, that is sufficient to do uploads to the unit, but during playback it regularly has short periods of no sound output, which I assume is each new disk access.

Maybe mine is an earlier unit, and the power requirement dropped slightly with later HDDs?

I don't know much about electronics. I've always wondered why the exact voltage doesn't matter much. Those words were paraphrased from either Hugo or Rob if I recall correctly, but I never understood why. Anyone care to explain? In layman's terms

I'll try.

Because very little in the empeg actually runs directly off of 12V. All the digital circuits (and hard drives) are 5V. I suppose there might be some 3.3V stuff too (I haven't really looked for it, and don't recollect seeing it).

Anyhoo, the 5V supply is created from the '12V' line by a regulator. Basically any modern (*) power regulator is a specialised form of integrated circuit - ie it has a lot of internal components to make it do what it does. One of the things that a modern regulator does is accept a wide _range_ of voltage inputs and provide a stated output with a stated precision. Truthfully, that is its entire purpose in life, and a 5V regulator that could _only_ provide exactly 5V from _exactly_ 12V would be useless - any noise (ripples, spikes, notches, glitches etc) on the 12V input would be passed through to the 5V output (by definition).
So the manufacturers specify and design regulators to provide a stated output with a stated maximum noise figure from a stated range of input. For example the 78M05 linear regulator (common as muck regulator chosen as example) is spec'd to have a maximum of 50mV ripple on it's output as long as the input it at least 8V. This is useful - many digital circuits will only operate correctly with a supply voltage of 4.75-5.25V, ie less than 500mV ripple on exactly 5V, so we know that a 78M05 would be a good candidate for such circuits.
It will actually operate down to 7V, but that maximum ripple number doubles to 100mV. Still good enough. The 2V difference between the minimum input and the 5V output is what the circuitry internal to the regulator needs as overhead to operate.
The 78M05 will operate with an input range of 7-35V. I have no idea where the 35V figure comes from, possibly an internal component breakdown voltage. I've never been particularly bothered about finding out.

But the 5V circuits are not the whole story in the empeg. The display uses 60V to drive the VFD, and that has to be created from the 12V too. We have some analog electronics that I haven't really looked at. And don't forget that the 12V is switched for the Remote Amp line too. Not only that, but the empeg's 12V input has a transient suppressor to eliminate ignition spikes that are common in automotive systems. Any of these elements could be the cause(s) of the 12-16V range quoted by the guys.

I'm speculating that the Remote Amp line is the reason for the 12V figure (who knows how various amps would respond to less than 12V?), and that the transient suppression is the reason for the 16V number.

(*) Modern in this context meaning approximately post 1980.

I've always wondered why the exact voltage doesn't matter much.

To answer in a completely different manner...

Why should it?

If I asked you for a 12V power supply, I bet you could find one.
How about if I asked you for a 12.0V power supply?
12.00V ?
12.000V ?

I have this wonderful Tektronix bench power supply that I can set to be 12.00V. It has a great 4 digit LED display to tell me so. It's guarranteed to provide line regulation to under 0.03%. Its list price is ~$900 (Gotta love ebay). It's a great piece of kit. But since its display accuracy is limited to 0.5% of reading, that could really be 11.94V or it could be 12.06V, and I wouldn't know the difference. Until I got out my 0.2% accuracy mutimeter, in which case I would then know that it was somewhere between 11.976V and 12.024V.
Understand? Even with some reasonably decent electronics gear, I cannot accurately quote any better than 12.0V.

With that in mind, can you imagine the havoc that would result if consumer electronics needed an accurate 12.0V supply? How much more would the power supplies cost to produce? (Each one would require calibration). How many people would get frustrated due to voltage drifting that occurs with temperature, load and age? How much would the tech support cost? Frankly, requiring such accuracy from an external supply would be incredibly bad design.

Fortunately, most consumer electronic equipment is easily tolerant of 5% voltage variations in their rated supply, if not more, so has no problems with your dirt cheap 12V supply that may really be 11.4V or 12.6V.

I'll admit that 12-16V is a large range to see quoted. Remember that sometimes things get simplified in the marketing and documentation stages (Can you imagine Jane NoBrain asking in RatShack for a 12-16V adapter? Can you imagine the employee's response? "Dur...dunno what that is")
That 12-16V figure did come directly from engineers.

I'll admit that 12-16V is a large range to see quoted

Well... considering a lot of stuff use "universal" power supplies that take 110-240V these days, 12-16 isn't a very wide range.

There has been a large-scale migration from linear power supplies to switched ones, and switched regulators have an easier time coping with a wide input range.

In efect, a linear regulator burns off the excess voltage as heat. A switcher is in effect a variable transformer, using just the energy that is needed (modulo switching losses).

In both cases, the device/supply will only draw as much current as it needs - a good thing, considering your car battery can supply hundreds of amps.

A linear regulator will essentially dump the excess voltage as heat. So, for example, if you take a 1A 5V linear regulator (ie an LM78CT05), and wire it across a 12V line, you are getting out (at full load) 1A @ 5V as usable voltage (5 watts), but the regulator is taking 1A @ 12V (12 watts) to produce it. The excess 7 watts is dissipated as heat, so one needs to provide for facilities to get this heat away from the circuitry before something horrible happens, ie a heatsink. The same regulator run from a 16V power supply would dissipate 9 watts, and from 6V only 1 watt. You get the idea.

So, to minimise the wastage (or conversely increase the efficiency), ideally one wants to run the regulator input as close to it's output voltage as possible. The problem is that there is a certain minimum amount of headroom required to allow the regulator itself to work, usually a couple of volts. Some more modern low-dropout regulators can be run down to only about 0.2V above the output, and are normally short-circuit and overtemperature protected as well. Most linear regulators will work at up to a 32V differential between input and output.

Now, in the case of the empeg main power supply, a switchmode design was used, which is much more efficient. I won' t get overly involved in the technical aspects, but essentially this works by means of a resonant LC circuit which stores energy at one voltage and releases it ad a different one, which can be either higher or lower than the input level. The higher frequency that the circuit uses the more efficient it gets, and the smaller the inductors can be. The chip used in the empeg is the MAX1631, which runs at 300kHz and gives typically 95%+ efficiency. It will actually accept voltages up to 30V, but there are other components in the player that are limited to about 18V maximum, hence the 16V limit (with an overhead for safety). It will also work down to about 8V, although external amps generally won't, which sets the lower limit.

The SMPSU produces the +3.3V and +5V rails, which power the majority of the circuitry. The CPU, memory, etc, all run off 3.3V, while the drives and a few other things run from 5V. The power supply can produce about 3A on the 5V line and 2A on the 3.3V one.

Although switchmode power supplies are very efficient, they tend to be electrically noisy due to the high frequency and heavy current switching going on in the inductors, so they're not ideal for analog equipment. The audio sections of the player are thus powered by a 5V and 10V linear regulator from the main 12V input, but at very low current so efficiency and heat aren't major problems.

The display board 60V supply is produced by yet another SMPSU (MAX770), which is configured as a step up (or boost) controller. This runs from the main 12V input, and produces a nominal 60V, which is wound down to about 40V to dim the display.

Boost SMPSU chips are also very efficient, although usually somewhat less so than step down (or buck) converters. The 770 runs at about 82% efficiency. I designed a +5 and +12V psu board for my UAV project using some of the things, which runs from a single cell. It produces about half an amp on each line, and can do amusing things like spinning up a hard drive from one small lithium cell. The cell doesn't last very long, of course!

Does this help?

pca

Okay, somebody please grab PCA's response here and stuff it into the hardware section on riocar.org!!

Thanks Patrick!

Does this help?

Yes, except for one question.

You said:

A linear regulator will essentially dump the excess voltage as heat.

But then proceeded to explain that the empeg used a switch mode power supply. Does a switch mode power supply also dump the excess voltage as heat?

I've noticed that the player will run hot or cool depending on which AC power supply I've got it plugged into, and sometimes it will also run hot in the car. Is that because sometimes the car or the AC adapter is supplying more than 12-13 volts?

Current is would be equivalent to how much water I can get out of the pipe in a second. Voltage is how much pressure that water is coming out at.

See, that's the problem I've always had with the water metaphor. Those two are exactly the same to me. If I open a faucet, it's either going to deliver a lot of water (high pressure) or a little water (low pressure). I don't see how that relates to voltage and amperage.

The river versus firehose metaphor makes a bit more sense, but I still don't understand it. There is either a certain amount of water or there isn't.

Maybe the difference between volts and amps can be better explained to a more literal person like me in terms of exactly what the electrons themselves are doing. Does more volts mean more electrons or what?

To answer in a completely different manner... Why should it?

Ah, that's a very good explanation. Thanks!

It's the size of the pipe that is the important bit. The cross sectional area is equivalent to the resistance.
Your faucet is actually restricting the flow. And by turning the knob you're affecting that flow by widening or closing that restriction.

Say I want to fill a bucket. I can fill it with a big pipe that gives me 1 litre a second (current) but at 1 psi (voltage). Or I can fill it was a small pipe that also gives me 1 litre a second (current) but it comes out at 10 psi (voltage).

Basically what I'm trying to say is that I can get the same amount of water out of something with a big pipe but low pressure or a smaller pipe but higher pressure. The amount I get out the end is exactly the same but how fast it comes out is related to how big the pipe is.

Vast quantities of water going past is okay I just let it go past. But I can't cope with huge pressures.

If you've ever seen them use a firehose you'll notice that it comes out a high pressure but once it's on the ground then it drains away slowly and it's spread out in a stream. The amount of water is the same. It's just how fast it's going past is different.

It's like 6am so if this is incoherent then I apologise

I can get the same amount of water out of something with a big pipe but low pressure or a smaller pipe but higher pressure.

So which one is voltage and which one is amperage? Is the pipe amperage or is the pressure amperage?

And how far can I carry this metaphor? How does it relate to real-world electronics? Can I compare filling that bucket to lighting an LED? For example, too much voltage will burn out an LED, so how does that relate to the water metaphor?

For example, too much voltage will burn out an LED, so how does that relate to the water metaphor?

The pressure from the water hose would do a number on you, just as voltage does bad things to the LED; but it's actually the overcurrent that follows from the overvoltage that does the job, same as the pressure in a fire hose feeding through a pinhole won't harm you, but a 2" firehose fully open will knock you over... With high enough pressure/voltage, even small flows/currents will be dangerous though - think waterjet cutting machine...

But LEDs are a bit odd - they're better controlled by controlling the current run through them than by controlling the voltage fed to them. This is because small differences in voltage lead to large differences in current...

But take a normal lightbulb; it can handle a given voltage - like a hose that can handle a given pressure. It will pass a given current at the rated voltage, as the hose will allow a certain water flow at the given pressure.

/Michael

I've noticed that the player will run hot or cool depending on which AC power supply I've got it plugged into, and sometimes it will also run hot in the car. Is that because sometimes the car or the AC adapter is supplying more than 12-13 volts?

A switch mode PSU is still not 100% efficient, so there is a small amount of excess heat produced. The larger the differential between the input and output voltages, the more heat, hence the difference you notice with different mains adaptors. In addition, as I said, there are a couple of linear regulators which also produce some waste heat. Also, I suspect that the SMPSU chip runs more or less efficiently depending on the ripple and noise on the input supply, which would produce different amounts of heat with different, nominally identical output mains adaptors.

If the main regulators in the 3.3V and 5V supplies for the empeg were linear ones, it would get MUCH hotter, as it would be disippating around 16-20W of heat when running in the car. Also, don't forget, ultimately pretty much all the power put into electronics comes out as heat in the end, except for that radiated as light, RF, magnetism, or sound. Which all end up as heat in the long term anyway, just somewhere else. Such is entropy.

pca

>Does a switch mode power supply also dump the excess voltage as heat?

Some of it. The "efficiency" rating tells you how much. I think PCA said the switching supply was around 95% efficient, meaning that 5% of the power drawn would not go to power the empeg, but would be lost as heat instead.

For the 5V supply, that could mean (13V x 1A = 13W x 5% = small number.

Cheers

So which one is voltage and which one is amperage? Is the pipe amperage or is the pressure amperage?

The pipe's restriction is equivalent to resistance. The pressure is equivalent to voltage and the _flow_ is equivalent to current.

You can take this analogy surprisingly far. The same math that is often used to describe and model linear electronics is also used to describe and model a lot of other linear physical systems - it's simply a case of finding the appropriate models for various elements. The water analogy is a fairly simple one.

Here's a simple kitchen experiment that may help you to really get a grasp;
Get an old soda bottle or clear juice bottle, about 2L is ideal.
Find a nail, screw or screwdriver, and put a small hole in the bottle about an inch from the bottom.
Cover the hole with your finger and fill the bottle with water.
Now place the bottle near the sink and take your finger off, and observe what happens.

At the beginning, water spurts from the hole and shoots out a fair distance. This distance is related to how high the water level is in the bottle, which in itself is analogous to voltage. Now watch the water level for a while and get an idea of how fast it is dropping. This speed is proportional to the amount of water flowing out of the hole, ie current.
When the water level is nearer the bottom, you will notice that the spurt is much shorter (lower voltage), but more importantly that the water level is dropping much slower, ie the flow (current) is much lower. The hole (resistance) hasn't changed.
So here you can see I=V/R. As the height of the water level(V) drops, the rate at which water flows out of the bottle(I) drops too. The hole (resistance) is constant.

When your happy that you have a handle on how fast this happens, enlarge the hole and repeat the experiment. You should notice;
1) The spurt lengths are about the same as before.
2) The water level drops faster, ie the average water flow (current) is increased. Note that the I=V/R relationship we saw earlier is still there - the water flow will still be higher at the start than at the end, since V (height) is still decreasing with time. Since the height of the bottle hasn't changed, the range of voltages is still the same. If you marked a line somewhere on the bottle and labelled it x Volts, then it is still x Volts. So the only way that the water flow (I) can have increased is for R to have decreased - and that is exactly what we did when we enlarged the hole.

Play with it a bit and hopefully you'll get it.

Play with it a bit and hopefully you'll get it.

...and please send us the photos Tony