1n5408 diodes are 3 amp devices in my op would do a better job as the resistors will change value with current drawn through them over time where there is heat there is loss…
resistors have a wider tolerance rage than the diodes
the diodes would be much more stable given there tolerance is in a different ball park.
thus a different device…they are meant to cope with voltage regulation as resistors are more meant to oppose current … semiconductors do both in general… https://www.jaycar.com.au/1n5408-3a-1000v-diode/p/ZR1018
ohms law states that voltage and current are proportional therefore resistance is the opposition to current flow.
diodes and transistors alike are more like an electronic resistor that you can vary .
if you really are so concerned with heat just use mosfets…the next best thing to valves to come out of the electronic era…size is an issue but where stepping into different devices now…i think we have come full circle if we delve into fets but to-92 devices may be an alternative due to space that is available for the op…i agree i think the op needs to decide with his engineers as to which is the better method for his purpose …i`d tend to go with the diodes over time i think they would stand up to the test much better…
I thought it would have been glaringly clear by now that if you reduce the voltage using a linear (resistor / diode) approach there will be a power loss which will manifest itself as heat. Just how much could build up in a confined space I have no idea. May not be a problem. The boffins would no doubt be able to calculate this but I am not one of those. I just know it happens. You have a walk up start in that the maximum time the bulb would be on is about 60 sec. I imagine the duty cycle would be quite low leaving time for temperature recovery.
To be sure of exactly what is happening the temperature build up under operating conditions should be measured. Either by using the stickers I mentioned earlier or thermocouple or whichever means is convenient. You have the device so you are the only one who can do this. You can then make a decision whether or not this is going to be a problem. It may well not be.
Resistors or diodes. Both have their pros and cons and behave quite differently.
Now I don’t know much about the characteristics of a xenon bulb but I am assuming similar to any incandescent device. As the voltage supply lowers the current lowers in proportion. Now with a resistor the voltage drop ACROSS it will also lower allowing more of the supply voltage to be available for the bulb which will use more current etc and an equilibrium is continually established. Bit if a balancing act really. Keep in mind the resistance of the bulb will change as the bulb gets dimmer, this also goes into the mix.
Now with diodes the voltage drop across each diode will be a fairly constant 0.6V over a wide current range. Which means the voltage at the bulb will be pretty much constantly about 1.2V (for 2 diodes) below the battery voltage at any one time. This will continue until the diodes switch off theoretically but the lithium battery protection should activate before then and shut down.
Choice of diodes? any power diode in that 1N400X series would be OK. as they are all rated at 1A continuous and are common and quite rugged.
Diodes or resistors? Up to you. Both will work. Diodes will provide a voltage at the bulb which will follow the battery voltage up and down. A Resistor may tend to self regulate to a certain extent in that the voltage drop across it and thus the volts at the bulb will vary with current through it. Try both and measure temperature at the same time. The solution may be quite simple. If temperature is a problem you will then have to look for alternative methods.
I think those xenon bulbs get pretty hot themselves don’t they.
Cheers Bob
You can calculate this based on the efficiency of each, but that only provides theoretical maxima, you’ll need to do some real-world testing to be able to determine how similar they are depending on the context and ambient conditions that they’re exposed to in your exact setup. Although, as Robert said, nothing is 100% efficient, so you’re going to have heat generated in either set up, and you’ll likely need to add some kind of cooling from a fan or heatsink if the ventilation isn’t sufficient to maintain a suitable temperature.
Yes, if you’ve got a noisy power supply that would be the way to go, as you can also quite easily overcompensate a little in order to reduce the risk of overvoltage while still maintaining a flowing current.
Kenneth’s description indicates that everything is in a container the same size as 2 C cells. That is 18650 lithium battery, Arduino Nano and whatever voltage dropping device decided on. So it will be a fairly confined space. That is why I keep bringing up the need to measure in operating conditions.
Having said all that though, electronics can get pretty hot and still be OK. If you can put your finger on it it’s not hot enough yet. To determine what sort of temperature can be tolerated safely you need to consult the data sheets for all components in there, including the battery as this will probably have a derating figure depending on temperature. I haven’t looked but I would be surprised if this is not the case.
Cheers Bob
One thing I haven’t seen anyone mention is actual battery voltage. All the calculations have been done as if the battery voltage is actually 3.7v - but it’s 4.2 at full charge, and 3.8-3.9 for most of its life. If you’re planning on using a resistor to prevent bursting your bulb you need to plan for max system voltage, not minimum voltage.
I’d suggest ditching the 18650. Go for an RCR123 to give yourself more room, and you’ll still get well over an hour of runtime while also powering your microcontroller. For this application you really don’t need the capacity of an 18650.
As for the voltage step down, a resistor is very simple and efficacious on space, but is converting 100% of the unused energy to heat.
A step-down converter will be enormously more efficient and only produce about 10% the heat of using a resistor if you can afford the extra space.
If you want it to be rechargeable by plugging in power, you’ll also need a charge controller in there somewhere - something like this is probably your best bet:
Take off the JST PH connector and solder it straight to your 3.7v lithium.
Excellent discussion, covering all of the options and a very in-depth about trade-offs!
.
(In regards to modeling the case the biggest factors would be the conductivity, emissivity, convective coefficients of the enclosure, and the ambient temperature outside, an interesting dive into multi-domain systems)
How have you been going with your project Kenneth?
I tried the diode idea but needed five to drop the battery voltage from 3.7 → 2.4v. Each diode only dropped the voltage 0.28V or so. The diodes were really big so there wasn’t room for that many in the case.
Instead I went with the voltage divider idea after all. I didn’t have any 2.2 ohm resistors so used two 4.7ohm in parallel which came to 2.3. It clearly isn’t a long term solution. But works for now
Next up I need to run the nano sense via the battery and the voltage regulator at the same time.
Finally (the hard part) I need to transmit the bluetooth signal representing a three-axis gyroscope and acceleration data from one ~60s ‘event’ through a natural hole in the base of the aluminium handle. this hole is perhaps 8mm in diameter. I really have no idea if that will work.
I’m fairly sure I can write some smart algorithms to filter/ parse this raw data into something useful.
I can keep you all in touch if you like. PS see the attached image: A 18650 inside a sleeve the size of two C batteries actually gives a fair bit of space. Think of the possibilities!
Hi Kenneth
What diodes are they? They could be schottky diodes which have a far lower forward voltage drop than the “ordinary” variety. The ones I mentioned, the IN400X series are about 10mm long and 2,5 - 3mm dia with axial leads. The voltage drop needs to be measured under load by the way.
I could be wrong but I don’t think Bluetooth will work too well in an aluminium can, hole or no hole. Once again a trial will tell if it is successful.
That enclosure looks good. We can get a much clearer picture of what you have now.
Yes please keep us posted. Looks interesting.
Cheers Bob
Just a comment on the the 18650 Lithium battery.
Some of these battery have built in over-charge under-charge protection, some don’t.
The pic looks like one from Jaycar; the product page states it is an unprotected battery.
A similar product from Battery World states it has built in protection.
I have used the Jaycar batteries successfully, but am aware of the situation and make sure it is never fully discharged and use a quality charger so as not to overcharge.
Just a long as you are aware of this limitation.
Regards
Jim
For those curious about how dangerous that can actually be Don’t try this at home! And always assume lithium batteries are dangerous, even with the appropriate protection circuits.
I was involved in a HAZOP study with CSIRO here in Newcastle a few years back on lithium batteries. They were developing a test rig for Tesla Power walls and similar that would fire a tow-bar into the battery bank with same energy as a 2 tonne vehicle travelling at 10km/h.
The nastiest thing to come out of them is actually the Hydrogen Fluoride - it’ll dissolve your bones very effectively if you get enough of it.
Just a quick reply to thank everyone for the advice and direction. The Bluetooth does seem to be able to transmit through the 8mm diameter hole in the handle!
I bought both the battery and charger from Core Electronics. I’ll let you know if they explode. Alternatively you can read about it in the news.
I bought this product, which is the same as above. I didn’t measure the voltage drop under load…
1000V, 3A. A bit of an overkill. Spec says 1.2V forward voltage drop @ 3A. Don’t know anything about these so forward drop @ 600mA is unknown. I DO know however that the drop of a 1N400X series diode will be very close to 0.6V hence the suggestion to use these. You didn’t see fit to do so, so you will just have to investigate.
To measure the voltage drop you must pass some current through the diode. Under load as it were. Measuring open circuit just passes the minute current the 10MΩ of the meter allows (0.5µA) and means nothing.
Cheers Bob
PS By 1N400X series I mean:
1N4001 100V
1N4002 200V
1N4004 400V
1N4007 1000V
All rated 1A continuous. All voltages quoted are reverse voltage withstanding capability. Any would do your job
An add on.
You should be able to check the forward voltage drop of the diodes with the diode test function of a digital multimeter. This should pass a known current through the diode and display the actual voltage drop.
Cheers Bob
you should be able to put a voltmeter across the two diodes positive to anode and negative to cathode and treating the 2 diodes as a single device whilst the bulb is lit and the circuit is powered up and that will tell you the true voltage that the diodes are dropping .
thus measuring in real terms how much drop is occurring.
Don’t try this at home! And always assume lithium batteries are dangerous, even with the appropriate protection circuits.
