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« on: October 05, 2012, 09:32:16 pm » |
What do you think would be the easiest, si plest way to turn a 5v signalinto a 12v signal? Im thinking either comparator with a 12 supply, or two npn transistors tho this doessnt seem as fast or energy effewcient any other ways im not thinking of? It will basically be usedfor powering a mosfet so I need atleast 10-11 of my 12v supply, and its actually more important it turns off faster than it does on, so im thinking I may want to drive it low but idk which is best(btw its about 1000nc gate charge)
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Massachusetts, USA
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« Reply #1 on: October 06, 2012, 12:02:16 am » |
An open-collector output and a pull-up resistor should convert 5V signal to 12V. You can use an NPN transistor (and base resistor) to change your output to an open-collector output.
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« Reply #2 on: October 06, 2012, 06:19:46 am » |
Not sure what you meant by "amplify", but if you meant linear amplification, a cheapo LM386 will do.
For digital, get a npn.
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« Reply #3 on: October 06, 2012, 09:01:52 am » |
Depending on the MOSFET being used, what it's driving, and the rise/fall time requirements (how fast it turns on and off), it may be possible to drive it directly with the 5V signal. One example: https://www.sparkfun.com/products/10618
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« Reply #4 on: October 06, 2012, 02:59:14 pm » |
Its a high voltge, high power type so definetly need 11 volts id say, what values would you use to achieve it? Because I want to drive it a little, but im not sure how strong a pull up down I an get without throwing off my gate voltage
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« Reply #5 on: October 06, 2012, 04:36:11 pm » |
Have a part number?
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« Reply #6 on: October 06, 2012, 05:18:53 pm » |
It's pretty easy to use an NPN to switch the gate voltage.
See attached.
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« Reply #7 on: October 06, 2012, 05:33:43 pm » |
If you need a "high-side" switch, and have an NPN and a PNP...  With a 2N3904 and a 2N3906, it's good for at least 100mA, where V out dips to 11.8V
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« Reply #8 on: October 06, 2012, 05:39:02 pm » |
Im using this http://rohmfs.rohm.com/en/products/databook/datasheet/discrete/transistor/mosfet/r6010anx.pdf and that last circuit I guess would work but wouldn't the vout be 11.3ish? Since the. 7v drop of the transistor? But now alos what would I do drive it low? A pulldown would waste energy and steal from the gate voltage as well Turning it on idk if it takes 1ms to fully turn on but turning off id like to make as fast as possible basically it will be receiving the output from a 5v comparator deciding whether it is on or off, id like to drive the mosfet in one step if possible but with the smallest footprint
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« Reply #9 on: October 06, 2012, 05:52:53 pm » |
that last circuit I guess would work but wouldn't the vout be 11.3ish? Since the. 7v drop of the transistor? But now alos what would I do drive it low? A pulldown would waste energy and steal from the gate voltage as well Turning it on idk if it takes 1ms to fully turn on but turning off id like to make as fast as possible basically it will be receiving the output from a 5v comparator deciding whether it is on or off, id like to drive the mosfet in one step if possible but with the smallest footprint
There's no 0.7V drop because it's all about V sat. As I stated, at 100mA, that gets up to 0.2V. Drawing less current results even less V sat. It's a tested circuit! The PNP opens and closes the path to +V (12V), so its output is +V (less V sat) when it's on or it's "open circuit" when it's off. It's pretty fast. The MOSFET switch time will have more to do with your Gate pull-down (and Gate capacitance). A high from the Arduino output turns the NPN on, which takes the PNP base low, turning the PNP on, and enabling the final output. The first circuit is the "easiest". Its only drawback is that it needs a high to the NPN to turn the FET off.
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« Last Edit: October 06, 2012, 05:55:12 pm by Runaway Pancake »
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« Reply #10 on: October 06, 2012, 05:57:52 pm » |
Its only drawback is that it needs an active HI to the NPN to turn the FET off. If that's indeed a drawback, tie the 4.7k to 5v and put the arduino's otuput pin to the base of the npn. Turning arduino's pin as input turn on the npn. If you had pre-written a 0 to the pin, turning it into output turns off the npn.
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« Reply #11 on: October 06, 2012, 06:02:43 pm » |
basically it will be receiving the output from a 5v comparator deciding whether it is on or off, id like to drive the mosfet in one step if possible but with the smallest footprint
It's a comparator output, eh? If it's one with an open-collector output then you won't need the NPN transistor/s.
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« Reply #12 on: October 06, 2012, 09:42:46 pm » |
Not an open collector output, however now imthinking about it I may go with another comparator to drive it because the ttotal part count is lower(1vs5), and while im pretty good at smd soldering I don't think I can make 5 of anysize components fit where a smd comparator could, anyone have any good suggestions for a comparator that can handle a 12v supply and also drive a mosfet (maybe directly?) Or another simpler circuit?
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« Reply #13 on: October 06, 2012, 10:49:14 pm » |
Both the 3904 and 6 are good as switches to at least 50 Mhz... The Alpha cutoff is in excess of 150 MHz, I've used them as RF amplifiers in Commercial american business band radios (151 -159 MHz) with good success, A little noisy though at 150 MHz. I shouldn't think you will have any speed issues.
Bob
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« Reply #14 on: October 07, 2012, 08:32:37 am » |
anyone have any good suggestions for a comparator that can handle a 12v supply and also drive a mosfet (maybe directly?) Or another simpler circuit?
Doesn't an op-amp with no feedback resistor act as a comparator? The gain is high so if the + input is higher than the - input the output goes full high and vice-versa. It shouldn't be hard to find a 6-pin op-amp package you can run at 12V.
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