NEW: Learning electronics? Ask your questions on the new Electronics Questions & Answers site hosted by CircuitLab.
Basic Electronics » Farads and Capacitors
February 17, 2011 by devinsbusiness |
I am trying to build a charge to drive a solenoid. I am thinking of using a capacitor to hold the charge and building up the voltage I need with a DC-DC convertor. The charge the solenoid needs is 13V, 1A for 22ms. I found a formula on Wikipedia that says F=(A * s)/V. So if I plug in my values, F=(1 * .022)/13 = .022/13 = .001692 = 1692microFarad. So if I use a 2200 microF capacitor, that should be plenty? Then the next part of this is how long I need to give the convertor to charge the capacitor before I let the H bridge send the charge to the solenoid. Since the convertor I found ( http://www.mouser.com/ProductDetail/Texas-Instruments/TPS61093DSKT/?qs=Mw8JyaFIW4tgepQ0OpElBQ%3d%3d ) has an enable pin, in order to save juice in my batteries I will only activate the convertor when the system needs the charge. Somewhere I thought I read that the same formula applies to the time required to charge up the capacitor. The convertor I found allegedly puts out 15V @ 50mA. So, using the same formula .0022=(.05 * s)/ 15 = .033=.05s = s=.033/.05 = .66 or 660ms. So, just to be safe, give it 1000ms. Hopefully I have all of this right. If not, somebody please let me know if this is completely out to lunch. |
---|---|
February 18, 2011 by Ralphxyz |
Won't you need some sort of latching solenoid or do you just need a pulse? Once the charge in the capacitor is used up won't the solenoid return to it's original position? Now as far as your question, I haven't the foggiest idea but your theory sounds right. I'd sure like to see some discussion about that TPS61093DSKT Converter, like what exactly is it doing? Is it actually boosting the output or just switching it on/off? That TI web server is really slow I went on to see if I could get some free samples (which they have) and it is taking 10 minutes to show a page. Maybe they are real busy processing sample request. Ralph |
February 18, 2011 by devinsbusiness |
Hey Ralph, If you look at the schematic on page 1 of the Datasheet for the TPS61093DSKT, it gives you a pretty good idea of what is going on. I found a pretty interesting article onHow Stuff Works about camera flashes. This integrated circuit is essentially doing the same thing only much less voltage. To answer your question directly, yes. It is boosting the voltage from an input range of 1.6V-6V up to an output range of up to 17V. There are also discrete components that have to be paired up with this IC to make it work properly. If you look at page 10 of the datasheet, you will see that you can program the output voltage based on the values of R1 and R2 and they give you the mathematical formula to figure it out. Further, if you look at page 4 of the datasheet, pin 5 is an enable pin. So we can use an output pin on our MCU to set pin 5 on the convertor high, thus turning it on. When pin 5 is again pulled low, the converter becomes inactive and only consumes < 1 microAmp. To say exactly how all this happens, I don't think I am going to try to explain because I don't really have a strong enough of a grasp and don't want to put out bad information. The How Stuff Works article does a pretty good job describing the concept. You mentioned free samples :-) . Please tell more. I have spent hundreds and hundreds of dollars on this project purchasing components. Some of them, to simply find out they don't fit my application and have to try something else. Any free samples would be great help. Thanks, Devin |
February 21, 2011 by devinsbusiness |
Hey Ralph, I want to thank you for the idea of free samples. I just got off of TI's website and ordered some samples of the converter I mentioned and also got some digital temp sensors that may go well in my project. I had no idea that free samples even existed. Thank You, Devin |
February 22, 2011 by devinsbusiness |
Just an update for anyone interested in this subject. I just tested the idea described in the first post in this thread. I used a wall wart as a power supply with resistors to simulate the output that my DC-DC converter will have to charge up a 2200uF capacitor. I then removed the power supply to see if just the charge in the capacitor would be enough, without the help of the power supply. Then with the help of a momentary switch, I applied the charge from the capacitor to my solenoid, and it worked.@ Ralph in his first post in this thread. Yes , I am using a latching solenoid. I will be using an L293 quadruple half h driver to drive the solenoid. Also, I just got an email from TI and they have already shipped the samples I ordered just yesterday. So far, so good. |
February 22, 2011 by Ralphxyz |
I just got my samples from TI today, could you post some of your code and maybe a schematic to show how you are using the booster? Apparently most IC manufactures have free samples. I got some of the DS3232 RTC (Real Time Clock) from Maxim free samples. Ralph |
February 22, 2011 by devinsbusiness |
Honestly, I have not yet written any code to address the converter. Or done any schematics. My plans are to set one of the available pins on my MCU as output. Tie that pin to the enable pin on the converter (pin 5). Then when whatever condition is meet for a need for the greater voltage, use an if statement to set the output pin on my MCU high, thus setting the enable pin on the convertor high. Then using the delay_ms function to control the needed delays in the sequence of things I want/need to happen. For example, I need about a 1000 ms delay between when the convertor kicks on and when the H-Bridge sends the voltage to the solenoid in order to allow the capacitor to charge up, so I will use delay_ms(1000) between the command to enable the convertor and the command to enable the H-Bridge. !?/#!? As I am writing this I just refered to the datasheet for the L293 and realized that min Vcc is 4.5V. I want to operate on 3V. Back to the drawing board. If anyone has any suggestions I would be eternally grateful. Anyway, I am sure there is an easier,more efficient way to do all of what I just described, but it has worked for me in the past. I hope that was of some help to you Ralph. |
February 23, 2011 by Ralphxyz |
Excellant, keep us posted. Ralph |
Please log in to post a reply.
Did you know that essentially all power supplies' voltages drop when current is drawn from them? Learn more...
|