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Hero999
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Posts posted by Hero999
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What did you use to power the transmitter?
Was it a cheap unregulated wall wart or a battery or regulated wall wart/mains supply?
I've tried powering an FM transmitter from a chap wall wart and it made the radio hum rather than transmit sound from the mic. You need to power it from a steady DC source such as a battery or regulated mains supply. -
The amount of current that actually flows depends on the resistance and the voltage of the supply.
I = V/R
Where I is the current, V is the voltage and R is the resistance.
The power supply will have a current rating which just specifies the maximum amount of current it can safely supply.
If the device is rated to 12V and the supply is 12V it will work fine, it will adjust its resistance to draw the amount of current it needs.
Don't listen to your friend, from the sounds of it he doesn't know any more than you, he just thinks he does. -
This sort of question gets asked pretty often, there should be a sticky somewhere.
http://www.electronics-lab.com/forum/index.php?topic=18917.msg87206#msg87206 -
What sort of sequence do you wand?
Red, yellow then green?
Or:
Red, red + yellow, green, yellow, green?
I can think of a simple way of doing it using the LM339, the CD4017 will also work.
The comparator idea has the advantage of using less active parts but needs more passives and is only limited to red, yellow and green.
The CMOS idea requires two ICs but has the advantages of being able to do either sequence and requires less passives. -
Yes, the device will take as much current as it needs, research Ohm's law.
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Yes you do have to find another solutaion for your video transmitter.
You're right, that you can alter the frequency of an FM transmitter by changing the value of the capacitor and inductor but when the frequency approaches the transistor's transition frequency you need a faster transistor. -
With the values shown, it could switch at up to ~48KHz... your not going to drive that MOSFET with a 555 directly at that frequency.
Why not? The 555 can work all the way up to 500kHz.A 555 can only sink ~200mA whereas the 2N2222 is good for 1A.
0.8A.
Would using a push-pull emitter follower configuration be any better? -
It's not very good is it?
If you can get hold of the 2SA16 then good for you but if you can replace it with a cheaper equivalent then all the better. -
The lack of a schematic was a greater barrier to getting help than your poor English.
I couldn't find much information about 1A16 on the Internet. I imagine that any PNP transistor rated to 1.5A will do. -
Unfortunately, the turn on will be slower due to the 330R.
The 2N2222 can sink a lot more current than the 555, so the "turn off" will be faster... and thats a good thing.
I doubt there's a significant difference between driving it directly from the 555 and via the 2N2222A. -
Unfortunately protection diodes can shorten the life of the relay because the current decays more slowly when it's turned off causing the contacts to open more slowly which increases arcing across the contacts. A resistor in series with the diode (about the same value as the coil resistance as a general rule of thumb) can increase the rate of current decay in the coild.
A diode in parallel with an inductor, like many protection diodes, prevents adverse voltages generated from the inductor from shortening the life of the connected devices. -
Now we getting somewhere !!
Yes, you'll nevery get any help until you give enough information.
Yes, it will work although it's a very crude and inefficient solution, a switching regulator is better idea.In my mind i had a theory that if i wanna use for example a car battery as feed and
it would be great if i could draw 5V - 3-5A out of this circuit.
I'm sure that it is better solutions for this task out there but i got curious about
that A16 thing !!! -
Yes, it's being used as a current limiter.
The 2N2955 is a booster transistor, when the voltage across the 50R resistor (or is it 60R? the schematic is too fuzzy) exceeds about 0.7V it turns on, allowing current to bypass the regulator.
A16 is acting as a current limiter, when the across the 0.1R resistor exceeds 0.7V it turns on which pulls the base of the 2N2955 closer to the emitter, causing it to turn off.
What output voltage do you want to get out of this?
With a 9V transformer, the maximum output voltage is 5V.
How much current do you want to draw?
The transformer is rated to 2A but with the values shown, the current limit will be about 7A. -
I don't see what the 2N2222A is adding to the circuit, if anything it's making it worse not better, I'd get rid of it and drive the MOSFET directly from the 555.
MOSFETs can be paralleled but why not use a single better MOSFET such as the IRL540, rather than loads the obsolete IRF510. -
This will work, when the output voltage is set to under 12V the power is drawn from the centre tap cutting the power dissipated by U1.
It's designed for a 15-0-15V transformer, for a 9-0-9V change the D2 to 6.2V.
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It could be used as a switch, amplifier or (as you've said) part of a current limiting circuit.
Bipolar transistors have so many applications that it's impossible to tell you what it's doing from what you've said. For example if I asked you "I've seen a flat head screw; what's it holding together?", how could you know the answer?
If you post a picture I might be able to tell you but even then the only way is to post a schematic. -
Use the heaviest copper board you can and cover all high current carrying traces with a thick layer of solder.
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In a SMPS, the output transistor is either on or off.
In a linear power supply, the output transistor always operates in its linear region, hence the name linear power supply. -
This can be solved by using a centre tapped transformer and taking the power from the tap when the output voltage is set below a certain level. I'll post a schematic if you're interested.
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It does, for a start 1.5A is only guaranteed at (VIN-VOUT) ≤ 15V, see page 5.
And there's a graph showing the safe area protection, see page 6.
http://www.national.com/ds/LM/LM117.pdf -
Hello, an idea struck me the other day when i was using my laptop.
Ive noticed that with many people, if the battery is left in whilst the power plug is in (which is how allot of people tend to leave their laptops) the battery will reach 100% and then the amount of power recharge cycles will increase from that point onwards. so for instance, it will use the battery down to 95% and then recharge again. A li-ion battery only has a certain amount of recharge cycles.
This always reduces maximum battery life. If it is constantly left in when the user is at home, the battery life will decrease and ever make the battery un unusable.
I've heard of this sort of thing before but never seen any real evidence.
It's certainly true that laptop batteries have a limited number of charge and discharge cycles but the rating is in full cycles, not partial cycles. Discharging a battery to 95% capacity then recharging to 100% does not cause as much wear as discharging it to 5% and recharging it to 100%. Li-on batteries also don't like being fully discharged, in fact most batteries don't and will wear more if fully discharged, then completely recharged many times. If your proposed device ensures that the battery is completely discharge before it's recharged, it would actually shorten the life of the batteries not lengthen it.
The grain of truth in what you've said probably comes from the fact that, it's bad to store Lion batteries in a fully charged state, the recommended storage charge is 40%. The laptop or battery will probably also have a digital fuel gauge which might only be calibrated when the battery is completely discharged and recharged. This might be important if the laptop shuts down when the battery is low; if it's not calibrated then it might shut down when there's plenty of charge left.
If you want maximum life then don't discharge past 30% or leave your laptop battery, for long periods, >40% charged. A more sensible option is to remove the battery from the laptop (again 40% charged if possible) when you're running it off mains power. Heat is the biggest battery killer so store the battery in a cool and dry place.
http://www.batteryuniversity.com/parttwo-34.htm -
A lot.
Don't even try to do it, it's a very stupid, dangerous and illegal idea. -
It probably also uses a voltage doubler circuit to generate both positive an negative voltages.
Making an inverter to get 12VAC from 12VDC is possible but not simple.
First you need to convert 12VDC to 17VDC, then use an h-bridge and oscillator to convert it to AC. -
In the modified circuit, the gain can be varied between 1 and 11 not 10.
You're not haveing a good day are you audioguru? ;D
Incidentally, there would be nothing wrong with being able to set the gain below 1 on an inverting amplifier.
What the value of Zout???
in Theory articles
Posted
Yes, the output impedance will be equal to the impedance of the inductor and capacitor in parallel.
The impedance of an ideal inductor and capacitor in parallel is infinite, in reality is won't be infinite but it'll be very high and depend on the Q of the tank. Any load connected to the output will reduce the amplitude and stability of the oscillator because it will affect the Q of the resonant circuit.