each push buttons control an output (represented by an led) individually. The switch, on the top, would simulate all buttons pushed.
In practice, am using 8 outputs, and I am driving 4 ~ 5 v, 0.2A on each output ( I am not actually using LEDs )
Currently the drop voltage on 8 diodes ( 1N4148 ? ) is enough no to turn a LED on.
These are mosfets someone taught me how to use and I have always used them for many applications. But I confess I don't know much about other options (npn transistor) for achieving same effect.
bontempos:
In practice, am using 8 outputs, and I am driving 4 ~ 5 v, 0.2A on each output ( I am not actually using LEDs )
Currently the drop voltage on 8 diodes ( 1N4148 ? ) is enough no to turn a LED on.
The 2SK4017 isn't a logic level MOSFET. Try the IRLD120PBF, instead -- or any Logic level MOSFET that can drive 1A or more. Then you can use simple 1N4148 or 1N914.
bontempos:
But I confess I don't know much about other options (npn transistor) for achieving same effect.
I would stick with MOSFETs. They require very little drive current [only enough to charge the Gate capacitance, and the Gate pull down resistor -- which can be much higher in value than what you're using -- like 100k or more, rather than your 1k resistors -- unless there's some need to switch that MOSFET really fast], and because the Channel is resistive, they can drive to much closer to 0V than a BiPolar transistor ever can.
Ignore larryd's comment -- he missed that the LEDs indicated in your schematic are redherrings
I went to a local shop to look for a logic level set but I think I could not make myself clear. A logic level fet means, in short, the driving voltage is something btw 3.3 and 5V, right? 4 to 5V in the case of IRLD120PBF.
I am replacing the press of buttons by a decade counter (4017) and my output voltage drops dramatically to around 2V on the output pins of the IC, having 5V at its Vcc pin. Still enough to drive each load (represented by LED) individually, but it does not work for the IC output pin which passes by the parallel diodes and attempt to switch all mosfets on.
bontempos:
I went to a local shop to look for a logic level set but I think I could not make myself clear. A logic level fet means, in short, the driving voltage is something btw 3.3 and 5V, right? 4 to 5V in the case of IRLD120PBF.
It's a little more involved than that. A MOSFET is an analog device -- so there is a lot of gray-area analog stuff going on. When a MOSFET is used as a switch, there are still analog considerations, so saying, "driving voltage", doesn't really encompass the full scope. It's more like "Gate-to-Source voltage that will induce a certain resistance in the channel, that will, in turn, interact with voltages across the Drain & Source. Whether the "switch" is "on" or not has to do with impedances in the circuit the MOSFET is interacting with. So, Logic Level, essentially, means, Logic Levels [0 to 5V] on the Gate [Gate to Source], have a good chance of making the MOSFET act like a switch. If you are using 3.3V logic, it gets even more nebulous, since "Logic Level" tends to refer to 5V logic.
Find yourself a good tutorial.
bontempos:
I am replacing the press of buttons by a decade counter (4017) and my output voltage drops dramatically to around 2V on the output pins of the IC, having 5V at its Vcc pin. Still enough to drive each load (represented by LED) individually, but it does not work for the IC output pin which passes by the parallel diodes and attempt to switch all mosfets on.
The 4017 has very little drive capability. So, if you're still using 1k resistors [from Gate to Source], then that's probably why your voltage levels are so low. Try replacing those 1k resistors with, say 10k [or even 100k].
Also, because you haven't really divulged the end goal, it's hard to give solid advice. For instance, I could be leading you astray if high switching speed is a factor.
Hi,
I think this layout will help you understand your circuit.
I have added the series resistors needed for the LEDs, but I have left the diodes as you have them, so you can see how they are causing problems.
I don't see the point of the shottky's between the gates of the MOSFETs and Vcc... as shown they'll be forward biased anyway , so the gates will be at about 0.4v less than Vcc, whatever the switch states.
Just omit them.
Or, if you want to simply turn LED's ( or anything else) on and off, just use a suitable switch - don't bother with any electronics.
ReverseEMF:
The 2SK4017 isn't a logic level MOSFET. Try the IRLD120PBF, instead -- or any Logic level MOSFET that can drive 1A or more. Then you can use simple 1N4148 or 1N914.
The 2SK4017 definitely is a logic level MOSFET - read the datasheet.
ReverseEMF:
The 4017 has very little drive capability .
It think it would be good to explain here that the CD4017 CMOS decade counter chip (which indeed
has very low output drive) is nothing at all to do with the 2SK4017 power MOSFET, the numbers being
the same is a complete accident. So be careful to give the full part number to avoid confusion.
larryd:
In post #6, I assume the element near the red + sign was a N.C. switch.
ass u me ???
Looking at the Ops picture, yes, that switch is used to diode switch the three MOSFETs on together.
I'll post an updated circuit when I get home tonight.
Tom..
Thanks for the patience and taking the time to reply with useful information!
I was reading about the logic level in a mosfet, but it seems I need more practice or examples to absorb it as it goes. For now, I understand we are referring to a clearer way to distinguishing the Gate-to-Source voltage in order to switch a Mosfet. This, being a bit too low value ? when working with 2sk4014.
I am still looking for a logic level mosfet in the same package while still fixing some other issues: I realize I might need another set of diodes outcoming from each IC pin ( D9, in the picture ) so to avoid some reverse current when pin 9 is HIGH and switching all Mosfets on. (my pcb was originally 3.5 x 2.5cm)
The goal is to control a GND output on different sockets, each one with a load of 5V and (updated) around 0.2A (some peaks of 0.5A). Basically turrets, with 2 mini servos and a 5V light each.
At initialization, each socket will be powered one at a time by 4017 IC. (each mosfet will be powered individually). The IC pin 9 will remain stationary (all mosfets on) during normal operation.
The reason for this initialization powering each socket individually is that each device carries a serial number IC which identify if device is present on the system. I need to associate each serial number to a unique socket.
EDIT* - I haven't had seen the new messages!
I am using diodes to make it possible for one pin on the 4017 activate all Mosfets. I am now using 1N4148s. Please disconsider the LEDs on the first diagram, I meant to represent a load (explained above in more detail).
In post #6, I assume the element near the red + sign was a N.C. switch.
You are right! Sorry for very unclear diagram! When I posted I was more concern about the use of diodes for this application (one switch to activate multi switches).
I realize I might need another set of diodes outcoming from each IC pin ( D9, in the picture ) so to avoid some reverse current when pin 9 is HIGH and switching all Mosfets on
I am measuring ~5V on IC 4017 Vcc and Gnd, as reference.
I have added diodes for each outputting pin on IC 4017 (except pin 9) and mosfets are outputting ~5V individually, and I can read ~5V from all mosfets at same time when IC pin 9 is HIGH. Currently everything seems to work.
Now, if I replace any diode (say, D9 from IC pin 3) by wire, mosfet representing that pin (Q1) outputs ~5V only when switched on individually, but seems to shortcut when IC pin 9 is HIGH, and all mosfets outputs 0V.
allanhurst:
Why not do it all in software? Then you don't need diodes at all....
Allan
Originally, that was the question I did, in another post. I agree it would be easier,but seems like the answer for that is to have each socket associated with an OneWire object, thus connected to different arduino pins. Maybe this is not very scalable for this experiment.
