hello folks,
I am an electrical engineering stufdent working on a senior engineering project, and I am unsure how to implement something I need. I am using 2 or 3 1.5 volt AA batteries to run my heart rate/blood pressure/temperature monitor. I need to figure out how to step the 1.5 volts d.c up to 5 volts dc. I am not familiar yet with power sources, as I have not had that sequence of classes yet. I have been told to hook up the batteries to a set of loops connected to a toroid on the positive side, and then to hook that into a bridge rectifier. It then loops into a pulse width modulator that will detect if the five volts is lacking due to a large load. It is in turn hooked to a transistor that acts as a switch, running into the negative side of the battery pack. THere will be a set of terminal running off the side of this gizmo. This is what I was tols by another engineering student. Frankly, I am a little confused. If anyone can help me by providing a little clearer instructions/help on it I would be ecstatic. Also, if you know of a small chip that already is designed for this purpose, please refer me to it. It will be running CMOS technology, so the amperage requirements are going to be rather low. If you have any other suggestions on how to power it, I am open to any ideas. THank you for your time.
>>If you have any other suggestions on how to power it, I am open to any ideas.<<
Use a 6V Lithium photoflash battery or several of them in parallel. It is much easier to step down than it is to step up.
Hi raistlinrandall and welcome to the board.
If you have several 1.5 V batteries in series you have 3, 4.5 or 6 Volts input. If it is 6 then coming down to 5 is quite easy. If you have a lower input voltage and need to raise to 5V, there are many ways to skin this cat. Assuming you do not need ground isolation the simplest way is probably this simple switching circuit which is used in most 12V adaptors for laptop computers. I have always found this simple solution very elegant and it is the same principle used in water ram pumps which pump water to a higher level using only the energy contained in the water, no electricity or other source of energy is needed.
A transistor T is used as a switch and it briefly shorts to Gnd the inductance L. When T stops conducting the energy stored in L flows through diode D and charges capacitor C.
2.5 - 6V ----\/\/\/\/\ -------->|----------- 5V out
DC L | D |
| |
/ ---
-| T --- C
\ |
| |
| |
Gnd ----------------------------------
I will let you work out the details. You can have a feedback loop which adjusts the duty cycle % of T as a function of output voltage but there are simpler ways of doing it, especially for small power. You can just set the duty cycle to a convenient constant if the output requirement is pretty constant or just use a zenner diode or other means to adjust the output.
This is a very simple and pretty standard circuit and you can find plenty of examples.
How close to 5 v. do you have to be? The usual method for getting a higher voltage with dry cells is to use two or more of them in series. Is there some special reason that you need to use such a complex scheme as “stepping 1.5 vdc up to 5 vdc?” This would require converting the 1.5 v to ac with some kind of chopper, stepping it up with a transformer and then rectifying it to get back to dc.
Four 1.5 v cells give 6 v, no load. If that is too high and your load (that is the current taken by your medical monitor) is relatively constant you can use a resistor in series with the monitor to lower the voltage on it to 5 volts.
I wouldn’t connect batteries directly in parallel to get a higher current capacity. (If AA’s won’t supply the needed current for long enough, then use A’s, or B’s.) Batteries seldom have exactly the same voltage under load and they do not run down at the same rate. This leads to circulating currents in the batteries, because of the voltage difference, and consequent heating. Batteries are modeled as a “black box” containing voltage source in series with a small internal resistance with two terminals on the outside of the box. If you hook two of these “boxes” in parallel and the voltages aren’t precisely the same, a current will flow through the internal resistances from one voltage source to the other, i.e. will merely circulate around throught the two “boxes” and result in heat being generated in the internal resistances.
What exactly do you need the 5V for? If it’s for powering logic devices, then you may not need all 5V. It depends on the logic family (technology) you’re using.
For example, if you’re using TTL parts, I believe 4.5V (3 batteries) would be plenty.
I think Sailor has you going the right way. We had to do the exact same thing for my design class. There are chips out there that will solve the problem of switching for you. Check out something like the chip LM2577 to get an idea of what it would look like. Its made by National so just go to their web site and get the datasheet. You might want a chip with a lower input voltage if you are going to use only two double AAs. A chip like this will give you a regulated voltage over the range of your batteries.
If you need under 1A LM2621 would work too and is cheaper.
If you need under 1A LM2621 would work and is cheaper.
Man, that National Semi-Conductor site is great. Its got some pretty useful tools on it.
thank you all for the help. THat schematic gave me an idea of how that type of circuit worked. I am gratefull. ALso, that LM2621 was EXACTLY what I needed. It will work great for what I am doing. It gives me the 1.5 volt to 5 volt jump I need, and at .5 amp. THank you guys a bunch!
:smack: Also thanks for the information on not putting the batteries in paralle. I would NEVER have thought of that. The technology is CMOS by the way.THanks again!
>>Also thanks for the information on not putting the batteries in parallel. I would NEVER have thought of that.<<
I think a couple of blocking diodes per battery will solve that problem but sailor has the solution. In fact, every solution he has ever presented has been elegant. So sailor…what is the best schematic cad program for XP?
Question: What if you set it up so that the batteries weren’t actually connected to each other until the device is turned on? Say when you throw the switch to start the unit, it completes the connection between the 2 positve terminals. Once the device is on won’t the current flow along the path of least resistance, thereby sending most of the current thru the device instead of circulating thru the batteries?
Chimming in late…
Sailor has the right idea. This month’s Nuts-n-Volts magazine has a circuit that will do basically what you want – it uses a 555 astable multivibrator to pulse width modulate a current through an inductor using a power transistor. Keep in mind, however, that these switching “boost regulators” can be touchy, and may require extra bench time to get them to work correctly.
Whenever the batteries are in parallel there is the likelihood of circulating currents. In your scheme, whenever the switch is turned on such current will flow. Yes, most of the current will go to the load (probably) but the batteries are a constant voltage source and the current that circulates in the batteries because they are in parallel will be relatively unaffected by any other load on them.
Connecting batteries in parallel isn’t recommended. If you need more current use a bigger battery. If you’re already using the biggest available, then break up the load so it divides between the two, or more, batteries.