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Your diode needs to be a Schottky diode for its speed. Any normal diode (1N4001 etc)will be too slow, the energy in the inductor will not dissipate quick enough into the output capacitor and you will get a spike where the diode and inductor meet. Not good for noise and long life of the diode
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I am surprised no one has replied to this. You need an H Bridge circuit:
http://en.wikipedia.org/wiki/H-bridge
and apply your load across the centre. You can then set the polarity of the voltage to your load by the switching of the transistors -
back to the asylum on Monday....
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Connect the positive terminal of your ammeter to the output of your power supply and the negative end to your load. This will read the current going into the load
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get two pnp transistors. connect their emitters together and connect these to the power input. Drive the bases low via a 10k resistor. Take the outputs from each collector.
Once again, your posts do not make a lot of sense. If you used proper English, you might get more help.
At a guess, I think the above answers your question. -
If you posted more sensical questions, you will get more replies. I cannot understand what you are asking...
It sounds like it is 3.9k. 3.9k and 3,9k sound like the same thing -
... and you need to place the capacitors as close to the supply pins as possible...
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It depends on what type of battery you have. To charge batteries you need a constant current source, but the method of charge termination varies from chemistry to chemistry. Lead acid batteries charge with a constant current then need a float voltage top off. NiCads and NiMH need a change in voltage with time charge (or temperature). Lithiums need an extremely accurate charge termination (to 100mV accuracy).
Look at the MAX712/713 datasheet from Maxim for NiMH and NiCads. -
I appreciate inductors and capacitors have resistance and will dissipate heat. I was talking theoretically to make the point easier to understand.
Considering the wider picture, you need to look at switching times (passing through the linear region of the FET will dissipate heat). You will also need to consider track length with its impact on resistance and cross coupling
You should also consider the proximity of the switching components to the inductor to avoid radiation. You need to look at the value of any feedback resistors as making them too high may cause the inductor to induce fields into them causing instability.
You will also have radiated and conducted emissions from the inductor. There will be switching losses in the controller itself.
Then you have stability issues by picking the wrong output cap and inductor values. That is a whole different arguement...
I tried to keep it simple... ;D -
Have a look at ISD.com and their Chipcorder range. You can just record your own speech into it and play it back. You dont have to worry about triggering an MP3 player etc.
Not sure what you amp looks like and whether you can feed a line in signal into it, so connecting this circuit up could be a challenge -
switched mode power supplies rely on 2 priciples:
a switch does not dissipate any heat:
heat = volts x current
in a switch the current is zero when switch is open or voltage is zero when switch is closed. therefore heat dissipation is always zero (theoretically).
Also inductors and capacitors do not dissipate heat
If you want to smooth out the switched waveform (from the switched ac to dc), use inductors and capacitors.
Thus you can achieve power regulation with theoretically zero heat loss. Heat is energy from your input supply, so zero heat means zero power loss means longer battery life.
Bill Naylor
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Electronic Kits for Education and Fun
Does MAX232 support UART?
in Projects Q/A
Posted
The MAX232 is nothing more than a level shifter taking in 0-5V and outputting +/-10V