I have a voltage supply of 12-14.5 DC with a 7805 Regulator, what would be the capacitor values before and after the regulator to "smooth" the 5 volts outpur of the regulator?
Also....
How would I calculalte these valuesfor future reference?
I have a voltage supply of 12-14.5 DC with a 7805 Regulator, what would be the capacitor values before and after the regulator to "smooth" the 5 volts outpur of the regulator?
Also....
How would I calculalte these valuesfor future reference?
I have a voltage supply of 12-14.5 DC with a 7805 Regulator, what would be the capacitor values before and after the regulator to "smooth" the 5 volts outpur of the regulator?
Also....
How would I calculalte these values for future reference?
I have a voltage supply of 12-14.5 DC with a 7805 Regulator, what would be the capacitor values before and after the regulator to "smooth" the 5 volts output of the regulator?
Also....
How would I calculalte these values for future reference?
C = (I/Vpp)T
where C is the capacitance in Farads I is the load current Vpp is the maximum (allowable) peak to peak ripple voltage T is the period in seconds
So, as an example, for a 60 Hz full-wave supply at 5 amps with an allowable ripple of 1 Vpp, the capacitor should be about 40,000 uF.
The above will guide you in picking the capacitor before the VR. The one on the output side is typically 10 to 100 uF.
The output of the 7805 doesn't need any smoothing, it's already 'smoothed' by the 7805 ! A small amount of capacitance at the regulator terminals helps stability though ( see manufacturer's data sheet / application note ). I normally use a 10uF electrolytic in parallel with a 47n plastic film or ceramic type. That's in addition to any load decoupling you also need of course. If the regulator isn't near the bulk storage cap you may also want to add some local decoupling on the input too.
As for the input.... I assume you're using a full wave rectifier at ac line frequency ?
You need to consider ripple current ( an approx rule of thumb is 1.6x DC load current ). You'll find ripple current ratings on the capacitor manufacturers data sheet. 'Derating' ripple current ratings - being more 'conservative' - makes the cap last longer btw ( also consider operating temperature for lifetime ) !
You also need to the consider ripple voltage that the storage cap will result in.
Pk-pk ripple voltage is ~ Iload/Cstorage*( tdisch) where tdisch is ~
7.5 ms for 50Hz supplies and ~ 6ms for 60Hz supplies.Eg 4700uF - 1 A load and 50 Hz > ( 1/4700*10^-6 ) * 7.5*10^-3
= 1.6 V ( quite an acceptable figure )
Graham
The only thing you need on the output of the regulator is a 0.1µF capacitor and, if your DC supply never goes below 7.5V when the 7805 is supplying maximum current into the load, a 0.33µF cap across the input, and they should both be as close to the 7805 as possible.
-- John Fields Professional Circuit Designer
Maybe I am thinkng this all wrong, with so many different answers, Whic one is roght?
This is a DC input from a 12-14.5 volt vehicle battery.
what is happenning, with a circuit a made to flash lights for certain gears, is that my gear sensor out put varies a lot with acceleration, maybe the cap or whatever, should be on the gear input side????
Why didn't you say that before ? That's somewhat important. You don't 'smooth' a car battery with a cap any more than a person can stop an elephant !
Sheesh. I went to all the trouble of typing out how to select a reservoir cap for an ac supply and all along you were just pulling my leg !
10uF on the in and out should be fine.I haven't a clue what you're rabbitting on about. What 'gear side' ? What sensor ? What does this have to do with a 7805 ? Why does acceleration affect it ?
Would you like to start again at the beginning by explaining ( as ever ) what you *really* want to know ( with full accomapanying details ) instead of just hinting at it ?
Graham
Yikes....it said in my original post that it was DC!!!
Anyone who knows anything about a car, knows that with acceleration and deccelration, the battery voltage can rise and fall.
I have an output from a sensor on the transmission that that puts out 1
- 5 VOLTS DC, deoending on which gear you are in.
What I am finding is that if I am in first, I would like the output from the transmission to my circuit board to stay around 1.5 volts dc, not 1.4 to 1.6 depending on acceleration and deccelration. The fluctuaion is affecting my circuit. I would like to see if I can get the variance closer than 1.4 - 1.6.
This voltage changes for every gear but it still flucuates whithin it's voltage. around 15%.
Or is this something that I could accomplish with another component, say a series resistor 10K or so.....I don't know.
Hopefully this is clear enough.
Is the sensor (a potentiometer that varies with gear position) powered by your 5 volt regulator, or by a connection directly to the battery (or through a dropping resistor to the battery)?
Is your regulated 5 volts varying with engine speed, or only the battery voltage?
snipped-for-privacy@persona.ca wrote: :>I have a voltage supply of 12-14.5 DC with a 7805 Regulator
: Yikes....it said in my original post that it was DC!!!
: Anyone who knows anything about a car, knows that with acceleration and : deccelration, the battery voltage can rise and fall.
: I have an output from a sensor on the transmission that that puts out 1 : - 5 VOLTS DC, deoending on which gear you are in.
: What I am finding is that if I am in first, I would like the output : from the transmission to my circuit board to stay around 1.5 volts dc, : not 1.4 to 1.6 depending on acceleration and deccelration. The : fluctuaion is affecting my circuit. I would like to see if I can get : the variance closer than 1.4 - 1.6.
: This voltage changes for every gear but it still flucuates whithin it's : voltage. around 15%.
: Or is this something that I could accomplish with another component, : say a series resistor 10K or so.....I don't know.
: Hopefully this is clear enough.
No cap on the input of the regulator is necessary in your situation.
The information that you need to give me to answer your other question is how much the CURRENT of the load (your gear sensor) of the 7805 changes and how fast (actually, we could assume the worst case, that the change is instantaneous.) If the answer is that your gear sensor draws a constant current from the regulator, no cap on the output of the regulator is necessary. Otherwise, I believe a previous poster gave an equation for the cap. value vs. the peak-allowable ripple on the supply. Use that.
Joe
The gear postition sensor is part of the bike, not my circuit, my circuit just takes this information and campares it to voltages on my voltage ladder and kicks on the proper comparator to give an indication light.
Even with no circuit attached to the bike, I read the outpout of the sensor and this is what I get(min to max.....manual stated value)
1st - 1.68 to 1.76....1.782 2nd - 2.13 to 2.21....2.242 3rd - 2.83 to 2.98....2.960 4th - 3.51 to 3.63....3.63 5th - 4.23 to 4.33....4.31 6th - 4.56 to 4.66....4.66 neu - 4.85 to 5.15....5.02The 5v from the I am not sure if it flucuates with vehicle accel and decel, however it shouldn't matter should it? If my 7805 output goes down, say due to decel, my voltage ladder cutouts will drop also. And I would think that everything should be equal.
I am still awaiting my PIC programmer, so it will eliminate all this stuff, however, I beleive in the last week , I have learned a lot about this stuff, and would still like to solve my initial problem.
Post back if you need any more info.
As I understand it, the position sensor is a variable resistor (potentiometer) connected to some supply voltage. If that voltage is regulated, the engine speed (alternator voltage) should not affect the output voltage. If the potentiometer is connected in series with a fixed resistor to the positive side of the battery, then engine speed will vary the total voltage across the potentiometer, and also the output at each of the gear positions. You need to have your A/D converter have the same voltage as its reference(full scale voltage) as the potentiometer has, either by duplicating the way the potentiometer develops its voltage (a second voltage divider across the battery, instead of using the 5 volt regulator) or you need to change the supply to the pot to power it from your 5 volt regulator. This is the only way they will agree in spite of the battery voltage variations.
This looks like there are still clear divisions between these ranges to make an unambiguous decision, as long as you can filter out any noise. For example, taking the average of the highest the lower voltage can be and the lowest the next highest level can be I get these decision points:
1-2, 1.95 volts 2-3, 2.52 3-4, 3.25 4-5, 3.93 5-6, 4.45 6-n, 4.76You can also do digital filtering, like I described earlier, to prevent flicker on an occasional spike.
It is the full scale reference for your A/D converter, so if it bounces around, it bounces your measurement around. 50% of full scale output on the A/D occurs at half of the 5 volt regulator voltage. So you may need a little high frequency capacitance on the regulator input and output, and, perhaps a bit of resistor or inductor between its input and the battery line.
I thought we were talking about the microprocessor doing the job. Are we back to the string of comparators?
If so, yes, the ladder string will vary, but is it varying exactly the same way the potentiometer supply is varying? If the potentiometer uses a series resistor to drop the battery to approximately 5 volts full scale, then your resistor string also needs a series resistor, not a 5 volt regulator to drop the battery voltage to a range of taps between 0 and about 5 volts. The two processes have to track each other.
Okay, I think I understand, now. If you use 5 volts as the full scale reference for the A/D, don't be too surprised if the problem is exactly the same as it is now.
Most ppl's DC comes from an ac supply of some sort.
You asked about *smoothing*. A battery doesn't need smoothing at all.
Actually the battery voltage rises and falls with the *load*. Vehicle speed has no direct effect at all. Besides you didn't even mention a car in your original post !
That's going to depends on what your *output* is ! Nothing to do with any
7805 !
Without knowing what your circuit is - no - not at all. How can I possibly answer a question about a circuit if you won't tell me anything about it ? Do you think I'm psychic ?
You can't even explain yourself properly. I suspect you haven't a clue what your're doing. Seems common these days.
What has education come to ?
Graham
You reckon ?
Never heard of ( trace ) inductance ? Instability ? It's the blind leading the blind here !
Graham
Oh, sorry Pooh Bear.....me stupid....me know nothing.......you know everything.
I have no education.......just figured with everything else that is done online, why not get "brilliant, witty, biting" people like you to do my work for me.
The battery voltage rises and falls with rise and fall of the engine speed not speed of the vehicle.
For your ultimate clarity.....
With the rise and fall of engine speed, the output of the gear position sensor goes up down also. This is with no circuit hooked up.
When I hook it up to the circuit, (pin 3 on U2 and U3) and put the bike in first gear, and rev the engine, the second light flickers. This does not happen in every gear, but about 3 out of six of them.
Maybe all I need is a series resistor, to bring the voltage down a bit.
But then I have no education and just want someone to do my work for me.
So if anyone feels like helping they can if not....go to the Pooh Bear side. Some people like to share knowledge, and others like to point out people's lack of it.
:> No cap on the input of the regulator is necessary in your :> situation.
: You reckon ?
: Never heard of ( trace ) inductance ? Instability ? It's the blind leading the : blind here !
: Graham
Jackass. Why does this matter on the INPUT of the regulator.
In case you didn't hear me: Jackass.
You might have an argument for a compensating cap on the OUTPUT of the regulator, but I'm not familiar with the inner workings of a 7805, and assumed that any good designer would compensate his regulator so that it was stable over the specified (in the datasheet) operating conditions (i.e. load, frequency.)
Once more: You are a jackass.
Joe
: snipped-for-privacy@ccwf.cc.utexas.edu wrote:
:> No cap on the input of the regulator is necessary in your :> situation.
: You reckon ?
: Never heard of ( trace ) inductance ? Instability ? It's the blind leading the : blind here !
: Graham
One more time: JACKASS.
I forgot to put this in my last post: For a DC INPUT, why do you care about trace inductance? I'll make it easy for you with a multiple choice question:
Joe
the
In case the bulk cap is some distance away.
I suggest you read up on the app notes.
It shows.
You're mistaken. A typical beginner's error.
And you're a monkey who talks out of your ass. Read up on the parts before opening your idiot mouth again will you. No shortage of know-nothings talking garbage here sadly.
Graham
See Fig 1 of
There's 0.33 uf on the regulator input.
National's LM340 ( replaces 78xx ) series datasheet
the
And you're just plain simply *WRONG*.
I've seen ppl do what you suggest and their supplies are sometimes unstable ( high frequency ripple superimposed on the DC output ) .
Graham
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