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Thomas Blake
Joined: 18 Jan 2004
Posts: 22
Location: Burbank CA
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 Posted: Wed Jun 22, 2005 7:10 pm |
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I agree that a packaged DC-DC converter would be the cleanest solution, as long as it's truly packaged. If you make it out of parts it's more fun but you may have to do a lot more development to ensure stability.
www.digikey.com has really a lot of different versions of converter. |
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Yashu
Joined: 08 Oct 2003
Posts: 26
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| Re: General circuit power question from 12V source |
| Posted: Wed Jun 22, 2005 8:04 pm |
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| jepherz wrote: | First off, let me start by saying that I am just getting into this whole circuit design, so if it sounds like any of these are nooB questions, well, they are.
Alright, so I made a circuit for automotive use, so my supply is going to be 12V. Currently I am using a 7805 regulator and running it with a 18V wall adaptor. I know when I hook it up to 12-14V, the regulator should be less inneficient, but at this point the regulator gets really hot. I can hold my finger on it for MAYBE 2 seconds. There is also no heat sink on it currently either.
With the regulator, I am powering a PIC, two 7-seg displays and 7447 drivers, a sensor, a 10k potentiometer, and an output LED.
My question is if I am powering this circuit correctly or if I am sucking too much current from a regulator such as the 7805. Should I keep this setup and just add a heat sink no matter how hot it is getting, or do I have another, better option in terms of regulator efficiency?
Thanks for your help,
Jeff |
Use a heat sink
Reference the national data sheet and AAVID heatsink supplier to determine appropriate requirements you need for junction-to-ambient thermal drop.
7805 TO-220 is a bear and good for 125C junction temp (i think)
2 sec finger hold with no heatsink in no big deal, but, use a heatsink |
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asmallri
Joined: 12 Aug 2004
Posts: 1635
Location: Perth, Australia
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| Re: General circuit power question from 12V source |
| Posted: Wed Jun 22, 2005 8:45 pm |
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| Yashu wrote: | | 2 sec finger hold with no heatsink in no big deal, but, use a heatsink |
Two seconds with how many segments active? I estimate the current consumption of the entire circuit to be of the order of 180mA when all segments are lit. This translates to 2.3 watts being dissipated by the regulator and 2 seconds would see burnt skin left on the T0220. :-)
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Regards, Andrew
http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!! |
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jepherz
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| Posted: Wed Jun 22, 2005 11:10 pm |
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Thomas Blake, would you care to elaborate on the "multiplexing" of the display? Maybe some code or something so that I actually have some idea what you are talking about :-)
What exactly does a "low dropout" voltage reguator mean? how is it different than the 7805 linear regulator?
I was thinking that it's also possible to put 2 regulators in parallel and therefore cut the current running through each in half. Is this do-able? |
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asmallri
Joined: 12 Aug 2004
Posts: 1635
Location: Perth, Australia
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| Posted: Wed Jun 22, 2005 11:20 pm |
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| Quote: | | What exactly does a "low dropout" voltage reguator mean? how is it different than the 7805 linear regulator? |
This is the minimum voltage that must be applied to the input to achieve the desired voltage at the output. Say for example you wanted to have a 5 volt output and the regulators drop out voltage was 2.5 volts. This means you would need a minimum of 7.5 volts at the input to deliver 5 volt output. A low dropout reguator is one that requires less input voltage to achieve the same output voltage. If my dropout votage was one volt for the same example above then the input can drop to 6 volt without affecting the output voltage.
The 7805 is a generic voltage regulator, different manufactures have different dropout voltages for the same part number.
| Quote: | | I was thinking that it's also possible to put 2 regulators in parallel and therefore cut the current running through each in half. Is this do-able? |
Do-able yes - practical no.
_________________
Regards, Andrew
http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!! |
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Thomas Blake
Joined: 18 Jan 2004
Posts: 22
Location: Burbank CA
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| Posted: Thu Jun 23, 2005 9:35 am |
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| jepherz wrote: | Thomas Blake, would you care to elaborate on the "multiplexing" of the display? Maybe some code or something so that I actually have some idea what you are talking about :-)
What exactly does a "low dropout" voltage reguator mean? how is it different than the 7805 linear regulator?
I was thinking that it's also possible to put 2 regulators in parallel and therefore cut the current running through each in half. Is this do-able? |
Point 1: I do have some code but the MCU had so few pins available that I had to use an external multiplexer (CD4511) and discrete transistors (2N3904) for enable, driven by a 2-to-4 decoder from 2 MCU pins for 4 characters. Therefore, it will probably not match your schematic at all. But here's the algorithm:
1. Disable all the digits.
2. Assert the 7-segment data for the desired digit.
3. Turn that digit on.
4. Turn that digit off.
5. Repeat n times for n digits.
You disable the display between pulses (pulse = steps 3 & 4) to prevent "ghost" characters from momentarily flickering on the wrong digit. Vary the repetition rate as desired; usually 100 - 2000 pulses per second is used. If you go too high with the rate the power consumption can increase again because transitions draw a lot of current. I usually use a very low duty cycle (each digit is on for only a fraction of the repetition rate) but some people like to go as high as 50% (on for 1/2 of the rate). You would change this by varying the width of the pulses (i.e., put some other code between steps 3 and 4 to stretch them out. It could be useful code or just time-wasting code.).
You really should get acquainted with multiplexing; it is universally used. The only time you might want to use linear drive is in noise-sensitive microvolt-level circuits such as microphone preamps and other low-level transducer circuits. But you're using an automotive electrical system, so display noise is probably the least of your worries. Have a look at the CD4511 data sheet, which admittedly is not a model of clarity.
Point 2 was addressed above. These are linear regulators for use where the raw voltage is less than two volts above the regulated voltage. LVDO regulators are not necessarily more efficient that ordinary linears. To increase efficiency you would go to a switching supply (DC-DC converter).
Point 3: You certainly can parallel regulators, but you may trade a headache for an upset stomach, so to speak. I have seen manufacturers do this, but the regulators are hand-matched for operating current. If they are not matched, the greedier one will hog most of the current as the temperature goes up and you have your same old problem over again.
I'd like to stress that paralleling regulators is something you'd do if an emergency circuit change happened after production starts. It's not something you'd do on purpose, at least not these days when there are so many power choices. I'll also stress what has been said before, that the 78XX series regulators can run awfully hot in normal usage. If they are actually overloaded they shut down pretty reliably; they hardly ever catch fire, in my experience at least.
Hope this helps!
tcb |
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Thomas Blake
Joined: 18 Jan 2004
Posts: 22
Location: Burbank CA
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| Posted: Thu Jun 23, 2005 9:44 am |
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Sorry, forgot to mention one thing in my previous post. The best way to time your pulses is with the timer 0 interrupt (RTCC). That way, the display doesn't momentarily flicker when other interrupts or serial communications are going on. Be careful interrupting during software-based RS232 communications, though. (You're on a F873, right? You have plenty of comms hardware there, so fewer problems.)
tcb |
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Yashu
Joined: 08 Oct 2003
Posts: 26
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| Re: General circuit power question from 12V source |
| Posted: Thu Jun 23, 2005 9:55 pm |
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| asmallri wrote: | | Yashu wrote: | | 2 sec finger hold with no heatsink in no big deal, but, use a heatsink |
Two seconds with how many segments active? I estimate the current consumption of the entire circuit to be of the order of 180mA when all segments are lit. This translates to 2.3 watts being dissipated by the regulator and 2 seconds would see burnt skin left on the T0220. :-) |
I've got National 7805's with a 0.75 inch high TO-220 heat sink sourcing 200ma from a vehicle battery source (14V typ) in over 1000 units for over 7 years... no problems what-so-ever. Ambient temp of typically 100 deg F. Static air (no fan). |
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jepherz
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| Posted: Fri Jun 24, 2005 1:49 pm |
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Thanks for all of your input guys, I should have enough info here to figure out a solution.
Onto my next question, I am using the sensor shown here:
http://www.jameco.com/wcsstore/Jameco/Products/ProdDS/321234MOT.pdf
It says operating voltage is 5.1V +- .25V, and the output characteristic takes the input voltage into account I believe, but I'm not totally sure. Do any of you know if it does?
If so, I guess I will just have to have a line coming from the input voltage of the sensor into my pic, figure out that voltage, and calculate the pressure based on that? |
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SherpaDoug
Joined: 07 Sep 2003
Posts: 1640
Location: Cape Cod Mass USA
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| Posted: Fri Jun 24, 2005 4:48 pm |
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That is a ratiometric sensor, meaning the output is proportional to the input voltage. The best way to deal with that is to use the supply voltage for the sensor as the A/D reference voltage (or vice versa). That way any drift in supply voltage is automatically compensated. It also means the voltage does not have to be very acurate. It should be filtered for noise but it can drift without causing problems.
_________________
The search for better is endless. Instead simply find very good and get the job done. |
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jepherz
Guest
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| Posted: Fri Jun 24, 2005 6:52 pm |
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| How exactly, with one A/D port left, can I set the reference voltage? What voltage is normally used on a pic? I am using the 16F819, so if the powering voltage is 5.1 volts, does that mean any analog input voltage on any of the A/D inputs are compared to that voltage? thus a 5.1 volt input would read 1024 in my code? If this is true, I guess I will just use the same line to power the sensor as is powering the PIC. |
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kender
Joined: 09 Aug 2004
Posts: 768
Location: Silicon Valley
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| Posted: Sun Oct 16, 2005 11:49 pm |
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| jepherz wrote: | | I am using the 16F819, so if the powering voltage is 5.1 volts, does that mean any analog input voltage on any of the A/D inputs are compared to that voltage? thus a 5.1 volt input would read 1024 in my code? |
Yes, but it will be 1023 actually 
| jepherz wrote: | | How exactly, with one A/D port left, can I set the reference voltage? |
It depends on what port do you have left. If it RA2 on pin1, you are in luck (page 2 of the datasheet). Pin1 can be configured as a reference input (see setup_adc() function). |
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