265V PS Needed For Braille Display 10
Harald Klein writes: "We are developing a cheap braille display for visually impaired people. A big problem is the power supply for the piezo elements. They need 265V DC, 0.5 mA max. The logic needs 5V DC, 15 mA. It is absolutely necessary to provide logic power before providing the high voltage. Even so the high voltage should be removed before logic power and signals. Any power supplies in mind that meet the specs?"
Power supply design (Score:3)
Shouldn't be too hard to make one. Start with a 5V supply capable of giving 1A or so. Have a delayed signal from the "logic voltage" start up a 50Hz (or so) sine wave oscillator (simple 555 circuit + small IC-based audio power amp would do), feeding a mains transformer in reverse. Put the output from the transformer through a suitably-rated full-wave bridge rectifier and smoothing circuit. The input to the transformer will have to be capable of driving at least 0.5*265/x mA, where x is the rms voltage output of the amp.
You'll have to work out the conversion factors to ensure you can get to 265V output based on oscillator amplitude (make it adjustable), transformer ratio (depends what mains voltage the "primary" (which you are using as a secondary!) is designed for), rectifier ratios and losses. You might need to start with more than 5V to be able to get up there: say 15V perhaps?
BE CAREFUL: despite the low current capability, 265V can do you serious harm.
ac.uk
Piezos do braille ?? (Score:2)
It seems you would have to leverage it up with a lever ratio of at least 5.
In any case, the piezo will draw very little current. I'll check on our power supplies in a few hours when I hit work. We use piezos for somatosensory stimulators (at about 100 microns of amplitude).
Re:Delayed switching of high voltage - SIMPLIFIED (Score:1)
Too complicated. Comparators are non-linear op-amps. Support requires at least a chain of resistors to set thresholds, or the RC constant. And it's a chip with current overheads, cost overheads, size overheads and reliability overheads - that you can do completely without.
First off, I've solved the problem of the power supply itself; just use a backlighting inverter for a notebook computer. (See the earlier thread.)
Now, it seems to me that the current required was 0.5mA in the 265V range... That's under 0.15 watts. Even assuming a really inefficient inverter circuit, I'll bet that the operating power consumed by the inverter and the piezos will be less than 0.25W. But, for calculation sake - and because I design failsafe military equipment which gets you into the habit of building brick outhouses - let's assume that the RMS operating current is 0.5W.
Now, we know the power consumed based on estimated efficiency, and we're in the ballpark. 0.5W at 5V means that we're drawing 100mA. So, add that to the logic circuit requirements when designing the main power supply.
As for the power-up sequencing, comparators and stuff are cool; relays would be more elegant, but they're big, expensive and unreliable. So, how do we do it as easily as possible?
Between the 5V power supply and in series with the logic, put in a forward-biased garden variety diode. 1N4001 or similar, or if the current required by the logic cicuits are low, you might even get away with a 1N34 or similar germanium diode with a low forward voltage drop.
Now, across the logic, isolated from the power supply by the diode, put in a good size filter capacitor. Calculate the value as a compromise between keepalive time for the logic, power supply output impedance, and inrush current as the capacitor charges.
To make sure that the logic goes on before the high voltage supply, put a reverse-biased zener diode in series between the inverter and the logic. The zener diode will remain non-conducting until the power supply voltage has risen to the zener voltage, at which point power will be applied. Calculate the zener voltage once you know the load for sure, and you've got a handle on whether the logic is CMOS or TTL, and therefore how forgiving it is. The direct use of a zener diode here is only practical because the inverter's current rating will be so low. Do *not* add any capacitance to this side of the circuit, BTW.
In operation, the power supply will turn on and the voltage will ramp up. As it rises, the logic circuits will receive power through the diode. At the same time, the capacitor will charge. At some point, the logic will be fully operational and it will internally generate a reset vector and the system will come up.
Meanwhile, the capacitor charging in the logic circuit will load the power supply enough that the voltage won't ramp up to nominal levels as quickly. Once it does, the capacitor will also help stabilize the logic when the inverter is turned on by the zener diode (mini brownout).
When the power supply reaches the zener voltage, the logic will already be on and the zener diode will apply power to the inverter. The power supply voltage might fluctuate a little as this happens, making the inverter stumble a bit as it starts - which will slow the charging of the secondary capacitors in the inverter. Either way, the logic remains protected by its own capacitor.
Power down is the opposite. The power supply will fall below the zener voltage and the inverter will cut out. As the power supply voltage continues to fall, the capacitor associated with the logic circuit is protected from being discharged thanks to the diode. The logic continues to run off this capacitor for a few moments, providing the necessary shutdown sequence.
Parts list?
One zener diode (calculate voltage) reverse biased and in series with the inverter.
One el-cheapo silicon rectifier diode, or, if current requirements are light, a 1N34 germanium diode, forward biased and in series with the logic.
One electrolytic capacitor, value to be calculated, across the power input to the logic circuit, and "downstream" from the diode.
Even in the electronics 7-Eleven that is Radio Shack, these parts should be readily available and won't cost more than $6 or $7. (At Newark, Mouser, DigiKey or MCM or something, I'd be surprised if the bill for these parts broke $2.00)
Next lesson, build a tesla coil and amuse your friends! :)
GetMeAGreenCard@spam.really.sucks.yahoo.com
Look to tube audio (Score:1)
Not too difficult... (Score:2)
Assuming you are going to be getting you main power from the AC line, get a low-amperage step-up transformer. It will convert your 120VAC to 240VAC. Using a rectifier and filtering will get you about 300VDC at no load. If you load it correctly you'll end up with about 260-270 VDC, which should be well within spec for your piezo elements.
If you need this to be battery operated, or you need the above to be exactly 265VDC, then you need to get into switching power supplies. At that point, unless you know what you are doing, you should get someone else to design it for you. High voltage power supplies are something of an art, since insulators don't act like you expect them to. It's kindof like trying to charm rattlesnakes...
-Adam
All power corrupts - but we need electricity!
Re:Piezos do braille ?? (Score:2)
I used to work with lasers, where long-throw piezos were commonplace as mirror adjusters etc.
Some use levers, but that's awkward to make and less accurate in use. Maybe they'd work for this, where it's a fairly simple "up or down" output.
The usual method was to stack piezos, so that they were mechanically in series, but electrically in parallel. This gave you long throws (a few mm), but with voltages in the hundreds of V, not the thousands.
Re:Power supply design (Score:1)
Low current PSU design is dead easy these days. I'm not certain that they'll reach a few hundred V, but there are very many DC-DC converters out there which are based on single-chip chipsets. Most are capacitor based, where the inductors are only used for filtering purposes, not even as transformers.
Any decent electronic component vendor (rswww.com) will list these and you can just grab the application note. I've built these things for the 5V->50V region starting from scratch in an evening and it's no big deal. Some even do developer kits with a ready-etched PCB.
There's certainly no need to use 555s any more - the chipsets are as easy, and more sophisticated. Many also have a simple "output disable" input that would solve your other problem.
Re:Delayed switching of high voltage (Score:1)
build your own (Score:1)
If you need some help with it, ask me and I will spend some time prototyping a supply up for you.
cya, Andrew...
Re:Power supply design (Score:1)
Yeah, that's exactly right, this is by far the best way to do it.
However, I'd suggest that you change the frequency of the inverter. The higher the frequency, the smaller and lighter the step-up transformer can be. It's for this reason that, when you open up a computer's power supply, the transformer that couples the mains voltage to the secondary side of the circuit tends to be pretty small despite a 200W+ rating.
In order to achieve 200W current rating from a 50/60Hz transformer, it has to be pretty large, heavy and expensive. The reason is that in order to have a given reactance (effective resistance at a given AC frequency) so as to not overload your low-voltage supply, at a lower frequency you will need more windings. The math for this is:
Xl = 2*pi*f*l
Xl = reactance of the inductor (transformer) winding in ohms
pi = 3.14159...
f = operating frequency in hertz
l = inductance of winding in henries. (Inductance is tough to calculate since it depends on gauge of windings, shape and size of the transformer core, etc.)
Copper is expensive and heavy, not to mention the iron laminates that make the best transformer cores that this lower frequency.As the frequency moves up, you can cut your needs down to even just one or two turns of winding if the math behind it makes sense. With small and cheap powdered iron (ferrite) core, you're well ahead.
Most switching power supplies in computers tend to run somewhere in the range of 50kHz. That allows for nice small transformers that can be easily mounted to a compact printed circuit board. If you try to center your inverter to run around 50kHz, you'll be able to use readily-available switching power supply transformers. Kick it into the circuit backwards and you'll step up your voltage nicely.
Designing and building a small switching supply to do this with any great degree of efficiency will require an electrical engineer. And, the large component count that will be required to do this may seem daunting. But if you're producing more than a few hundred of these things, the cost savings on being able to use a cheaper and far smaller transformer will more than make up for it.
And finally, here's an excellent idea. Take a look at a notebook computer. Most of their displays are backlit by either tiny fluorescent tubes or electroluminescent sheets. Either one of these backlight solutions requires somewhere in the range of 100-300v to run. Notebook display inverters tend to be tiny, cheap to make, lightweight and efficient, all for obvious reasons. As for suppliers of these things, I'm sure Aztec Power Supplies made the inverter module in my old Compaq 386SX.
Best of luck! :)
lwade@i.hate.all.spammers.the-wire.com