i need the said inverter chip to control four BDX53/4s that for max power output need a base amperage of 10.6 mA (8A/750) or would it be better to use transistors to do it. or...
using another four small transistors make the 2 BDX53s and BDX54s into super darlintongs by adding the third transistor as Vbe is not a problem

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Hi Thunderchild,
A 74HC04 (high speed Cmos) hex inverters can source and sink more than 25mA (its max output rating) with a 5V supply. It will overheat if the output swing is slower than about 1ms or if it operates as a linear amp.
You can use the 74HC14 Schmitt trigger hex inverters to get fast output swings with slow input signals. Its output is also more than 25mA. :lol:

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Uncle $crooge

 

ok sinking fine but is that for sourcing as well ? the frequenzy will be like a few KHz so the 1ms pulse is ok and actually the frequenzy depends on the frequenzy that a standard diode (for say 10A) will work well at so perhaps u can tip me on that. i'd rather not use expensive skotchy diodes especially at that amperage and will resolve with larger condensers instead that will not be a big issue anyhow as it will be 1 KHz or more and not 50 Hz so that helps a lot. this will all be working on 12 volts is that ok on the gates (will check for the data sheet now)

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The expected minimum sourcing current of 74HCxx depends on supply voltage and is about 15mA with a 4.5V supply and about 25mA with a 6V supply. Typically it is way more than 40mA but the rating has a max output current of 25mA.

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Uncle $crooge

 

so will this chip work on 12-14 volts like in a car ? i had some trouble trying to find a useful datasheet and only came up with something that did not realy tell me much (and that was great fun when surfing at 2.5 KB/s)

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Hi Thunderchild,
We need complete details of your circuit to determine a suitable darlington transistor and driver for it. The gain of your BDX53 is rated at only 3A. It isn't even rated at 8A but the curves show its gain dropping quickly above 3A.
How much saturation voltage can your load tolerate? At 3A, the BDX53 has a max saturation voltage of 2V with a base current of 12mA. Its saturation voltage will be much higher at 8A where it might need a base current of 140mA!

The max recommended supply and input voltages of the 74HCxx series is 6V, so might not be suitable for your circuit. :lol:

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Uncle $crooge

 

thanks for saving me from disaster. i will explain.
i will have an oscilator in the form of a 555 timer and this via the gates that will buffer separate and where necesary invert the pulses to drive a bridge of transistors so that i have 12 ac from 12 dc so i will be driving four transistors 2 at a time to reverse the polarity as the clock signal ticks away. perhaps u can sugest a good set of transistors and inverters and i will be needing at least twelve amps in the end as i was only intending to experiment on the bdx53/54s as i have them already but 2volts vce means i will have only 8 volts left on 3A if i have understood right ? and maybe nothing at 8A.
bottom line what the hec can u do with a transistor that has a Vce of 2 volts ?

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Maybe it is time to use nice modern (but expensive) power Mosfets. An IRF2804 has a max on-resistance of only 2.3 milli-ohms. At 8A it will have 0.018V or less across it. It is an extreme example since it is rated for over 1kA current pulses and has an extremely high input capacitance.
A cheap old IRF540 power Mosfet will have 0.35V or less across it.
A cheaper old IRF530 will have 0.7V or less across it. :lol:

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Uncle $crooge

 

i see but how do i control a mosfet

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For an N-channel Mosfet, 10V to 20V to its gate from a fairly low-impedance source (to quickly charge its high gate capacitance and prevent oscillation) turns it on, and 0V to 2V turns it off.

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so i can just control the mosfet directly with the outputs of a 555 take a look at this diagram if my damn ffffing isp will let me upload it. the switch will be replaced by the mosfets and 555 (or maybe i'll use a good old relay hehe).

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so if i use one N and one P mosfet how do i control the P one 0 volts to turn it on and VCC to turn it off ?

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That's correct. Just connect its source to the positive supply.
You should have a part of your circuit to provide some "dead-time" when both transistors switch. Or else when they are switching then both will be turned on at the same time shorting the supply. :lol:

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Uncle $crooge

 

ok so if i use a 555 won't that do the trick as it is the same output so if it is high then it can't be low or is there a risk of partial turning on that can still be dagerous. how can i create a delay ?if i use say some not gates all in cascade this will also delay the turn off so causing the same problem.
is there a P equivalent for the IRF540.
by the way i did a small scale prototype using a manual switch and it worked only thing is if i just left it it would gradually discharge to 12 volts and stay there so it was not the tester that was discharging it so fast and i assume that one of the diodes was reverse conducting. it did this on both so i doubt one diode was faulty. they were 1N4007's so they should of had no problems

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Hi Thunderchild,
Even an ordinary 555 has about a 400mA power supply current spike when its outputs switch because both output transistors are on at the same time. If it drives a pair of Mosfets then they would make a nice toaster.

It sounds like you are making a charge-pump voltage doubler. Power supply ICs are available with an oscillator, two separately wired output transistors, variable dead time control, error amplifiers etc. like the TL494 that might do the job best. :lol:

A P-channel equivalent to an IRF540 is an IRF5210 that has a slightly higher on-resistance and costs about 3 times more.

A 1N4007 rectifier diode is very slow and is designed for 50Hz-60Hz operation. For higher frequencies a very fast-switching Schottky diode is used. :lol:

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