Need to shut down half my pcb... - Electronic Circuits Projects Diagrams Free

iso9001 · 28th March 2006, 03:53 AM

Need to shut down half my pcb to save power. Didn't really want to add another 5V regualtor since I already have one and its pretty large.

I have a car battery feed (10-15V) coming in and need to knock it all down to 4.75 - 5.10V. Then being able to source upto 250mA. My plan was to use my 5V regulator and then a mosfet to turn half the board off, but I forgot that fets cause a small voltage drop that would put my at about 4.3V.

ops:

I've been looking for the best way to do this. Maybe a transistor setup or an opamp a voltage follower mode ?

I'm not opposed to buying more components as long as they are small, cheap, and will help save power. Just not sure what the best approach is,

Any ideas ???

DigiTan · 28th March 2006, 04:05 AM

I don't know if this works all the time, but Ron H showed me you can get a 5V regulator to output ~5.7V by connecting a Si diode between the ground pin and the actual ground. That should just about make up for the FET drop assuming the diode current is right.

iso9001 · 28th March 2006, 04:48 AM

Overclock the regulator huh ? I really don't understand what would be going on there.

Sounds like it would lessen the life of my 5V regulator.

I'de buy a 5.2V but I need parts mouser has in stock NOW. And the regulator needs to able to handle automotive loads. So I'm pretty much stuck with 5.0.

I ruled out an opamp, its just not going to supply the current I need.

Maybe some sort of transistor deal where I can shut it off with an input from the micro,

DigiTan · 28th March 2006, 05:03 AM

Well, your 5.0V regulator places its output voltage relative to what it thinks is ground. If its ground pin is really at GND+0.7V due to a diode, it will place 5.7V on the output. More or less. All it needs is a series diode between the GND pin and ground.

iso9001 · 28th March 2006, 05:20 AM

Ah..... I see.

So then, if that works in my application... I'll need to pass my secondary circuit into a transistor, I get a ~.7V drop, but then I will also need to pass my primary circuit into some junction like a diode to get rid of that now pesky .7V.

If I put the micro to sleep and the total primary draw is 300uA... That diode is going to waste a good chunk of power no ?

I wonder if it is going to change voltages with temperature more then before the diode at the regulator?

I like the idea, I'm not sure its going to work tho. I need to get this whole thing down to as little power as possible.

DigiTan · 28th March 2006, 06:54 AM

Yes, that's correct. The actual volt drop on the diode will vary with both current and temperature, so your line and load regulation will suffer slightly.

The diode voltage drop will vary exponentially with the current exiting the GND pin. Additionally, its volt drop will fall about 2mV for every 1°C increase in temperature. So considering the extereme temperatures you can get in a car, that might be significant. What kind of 5V regulator did you have in mind?

iso9001 · 28th March 2006, 07:34 AM

I don't mind the temp drop as much as I do the increased used of power at the diode knocking the 5.7V down for the primary circuit.

I've have 1mA to work within. IF i can shut my circuit down to bare minimum I am looking at 900uA.

I'm thinking that .7 I'm going to be throwing away in the diode is going to screw up my 900uA no ???

The regulator is a ST-KF50, its not perfect but it meets all the criteria. Being instock at mouser is one of those requirments. I was hoping to order tommorow!

http://www.mouser.com/index.cfm?hand..._pcodeid=51138

Someone Electro · 28th March 2006, 02:30 PM

A mosfet has no foward voltage.It has a chanel resistance.This is the resistance betwen the pins when its saturated.I have mosfets that have this resistance of 0,008 Ohm and this is almost a wire.But you dont need that low. meaby like half a ohm.The less the better

Roff · 28th March 2006, 04:08 PM

Quote:

Originally Posted by Someone Electro

A mosfet has no foward voltage.It has a chanel resistance.This is the resistance betwen the pins when its saturated.I have mosfets that have this resistance of 0,008 ohm and this is almost a wire.But you dont need that low. meaby like half a ohm.The less the better

Adding to that: You need a P-channel MOSFET with Rds(on) specified at 4.5V. As Someone Electro said, there are many to choose from. Try irf.com or fairchildsemi.com, or some of the other MOSFET mfrs.

iso9001 · 28th March 2006, 07:36 PM

So...... If I have my secondary cicuit being driven @5V through a mosfet with .5ohm drop, and I draw 300mA from it, my voltage drop will be ???

I'm crappy at figuring this out but, its low.... like .015V no ?

If thats the case, that'll do!

So to calc that its just the original voltage - (resistance * amperage) ?

edit:
I read this data sheet:
http://www.fairchildsemi.com/ds/FD/FDC6304P.pdf

I'm curious as to why they list all the values as negitive. As if everything is in respect to the source and not the drain ? But really the drain is there the feed is coming into the fet so why is it listed in the sources view ?

This transistor has an Ron of 1.1Max @ 4.5/.5A... It seems as voltage and amperage go up the Ron goes down. Is this true for Voltage AND Amperage or just Voltage ? Is there a way to determine the value @ 5V/.3A ? I looked at the graphs, but it didn't exactly tell me, looks like ~1ohm.

Roff · 28th March 2006, 07:53 PM

Quote:

Originally Posted by iso9001

So...... If I have my secondary cicuit being driven @5V through a mosfet with .5ohm drop, and I draw 300mA from it, my voltage drop will be ???

I'm crappy at figuring this out but, its low.... like .015V no ?

If thats the case, that'll do!

So to calc that its just the original voltage - (resistance * amperage) ?

Yeah, except 0.5ohms *0.3A=150mV. Not a problem, though. It's easy to find a PMOS with Ron<0.05ohms.

iso9001 · 28th March 2006, 09:15 PM

Oh, yea my mistake. .150V Still fine.

So was I right about MOSFETs being measured to thier Source and thats why those values are almost all negitive ?

I included a quick diagram but I'm thinking its layed out of N-Channel all of a sudden. I've got 4 other project I'm working on and I know I've got to be making mistakes somwhere

I need to check and see if this is correct or not, also I need to figure out how I should bias this.... Or if I even need to ?

Roff · 28th March 2006, 09:24 PM

Close, but no cigar. See below. No biasing needed, if your logic input swings from zero to +5V.
And it does have to be P-channel.

iso9001 · 28th March 2006, 09:37 PM

Edit: I'm dumb just noiticed I used G instead of S.... sorry, hit the wrong key apparently.

Ok, I'll make the change.

I'm a little confused about the biasing tho. The last time I accidentely made a touch switch, but that was an N-Channel

But I understand getting a few stray electons to trigger then N-ch is probably easier then finding a close place to sink enough to fake a P-ch out,

I'll trust you and leave the biasing out for now. Just need to make sure I cut the power before I make the micro sleep.

Thanks Again Everyone!

Roff · 28th March 2006, 10:28 PM

Quote:

Originally Posted by iso9001

Edit: I'm dumb just noiticed I used G instead of S.... sorry, hit the wrong key apparently.

Ok, I'll make the change.

I'm a little confused about the biasing tho. The last time I accidentely made a touch switch, but that was an N-Channel

But I understand getting a few stray electons to trigger then N-ch is probably easier then finding a close place to sink enough to fake a P-ch out,

I'll trust you and leave the biasing out for now. Just need to make sure I cut the power before I make the micro sleep.

Thanks Again Everyone!

I think you know this, but be aware that the PMOS is ON with the gate at zero volts (GND), and OFF with the gate at the same voltage as the source (+5V). As I pointed out above, be sure you use one that has Rds(on) specified at -4.5V or less (actually, I guess -2.7V is MORE than -4.5V, but I think you know what I mean).

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28th March 2006, 03:53 AM	(permalink)
iso9001	Need to shut down half my pcb... Need to shut down half my pcb to save power. Didn't really want to add another 5V regualtor since I already have one and its pretty large. I have a car battery feed (10-15V) coming in and need to knock it all down to 4.75 - 5.10V. Then being able to source upto 250mA. My plan was to use my 5V regulator and then a mosfet to turn half the board off, but I forgot that fets cause a small voltage drop that would put my at about 4.3V. ops: I've been looking for the best way to do this. Maybe a transistor setup or an opamp a voltage follower mode ? I'm not opposed to buying more components as long as they are small, cheap, and will help save power. Just not sure what the best approach is, Any ideas ???

28th March 2006, 04:05 AM	(permalink)
DigiTan	I don't know if this works all the time, but Ron H showed me you can get a 5V regulator to output ~5.7V by connecting a Si diode between the ground pin and the actual ground. That should just about make up for the FET drop assuming the diode current is right. __________________ The most-hated young engineer in America. * AVR Projects * Robot War: TI * About

28th March 2006, 04:48 AM	(permalink)
iso9001	Overclock the regulator huh ? I really don't understand what would be going on there. Sounds like it would lessen the life of my 5V regulator. I'de buy a 5.2V but I need parts mouser has in stock NOW. And the regulator needs to able to handle automotive loads. So I'm pretty much stuck with 5.0. I ruled out an opamp, its just not going to supply the current I need. Maybe some sort of transistor deal where I can shut it off with an input from the micro,

28th March 2006, 05:03 AM	(permalink)
DigiTan	Well, your 5.0V regulator places its output voltage relative to what it thinks is ground. If its ground pin is really at GND+0.7V due to a diode, it will place 5.7V on the output. More or less. All it needs is a series diode between the GND pin and ground. __________________ The most-hated young engineer in America. * AVR Projects * Robot War: TI * About

28th March 2006, 05:20 AM	(permalink)
iso9001	Ah..... I see. So then, if that works in my application... I'll need to pass my secondary circuit into a transistor, I get a ~.7V drop, but then I will also need to pass my primary circuit into some junction like a diode to get rid of that now pesky .7V. If I put the micro to sleep and the total primary draw is 300uA... That diode is going to waste a good chunk of power no ? I wonder if it is going to change voltages with temperature more then before the diode at the regulator? I like the idea, I'm not sure its going to work tho. I need to get this whole thing down to as little power as possible.

28th March 2006, 06:54 AM	(permalink)
DigiTan	Yes, that's correct. The actual volt drop on the diode will vary with both current and temperature, so your line and load regulation will suffer slightly. The diode voltage drop will vary exponentially with the current exiting the GND pin. Additionally, its volt drop will fall about 2mV for every 1°C increase in temperature. So considering the extereme temperatures you can get in a car, that might be significant. What kind of 5V regulator did you have in mind? __________________ The most-hated young engineer in America. * AVR Projects * Robot War: TI * About

28th March 2006, 07:34 AM	(permalink)
iso9001	I don't mind the temp drop as much as I do the increased used of power at the diode knocking the 5.7V down for the primary circuit. I've have 1mA to work within. IF i can shut my circuit down to bare minimum I am looking at 900uA. I'm thinking that .7 I'm going to be throwing away in the diode is going to screw up my 900uA no ??? The regulator is a ST-KF50, its not perfect but it meets all the criteria. Being instock at mouser is one of those requirments. I was hoping to order tommorow! http://www.mouser.com/index.cfm?hand..._pcodeid=51138

28th March 2006, 02:30 PM	(permalink)
Someone Electro	A mosfet has no foward voltage.It has a chanel resistance.This is the resistance betwen the pins when its saturated.I have mosfets that have this resistance of 0,008 Ohm and this is almost a wire.But you dont need that low. meaby like half a ohm.The less the better __________________ Il give you shocking experience.

28th March 2006, 07:36 PM	(permalink)
iso9001	So...... If I have my secondary cicuit being driven @5V through a mosfet with .5ohm drop, and I draw 300mA from it, my voltage drop will be ??? I'm crappy at figuring this out but, its low.... like .015V no ? If thats the case, that'll do! So to calc that its just the original voltage - (resistance * amperage) ? edit: I read this data sheet: http://www.fairchildsemi.com/ds/FD/FDC6304P.pdf I'm curious as to why they list all the values as negitive. As if everything is in respect to the source and not the drain ? But really the drain is there the feed is coming into the fet so why is it listed in the sources view ? This transistor has an Ron of 1.1Max @ 4.5/.5A... It seems as voltage and amperage go up the Ron goes down. Is this true for Voltage AND Amperage or just Voltage ? Is there a way to determine the value @ 5V/.3A ? I looked at the graphs, but it didn't exactly tell me, looks like ~1ohm.

28th March 2006, 09:37 PM	(permalink)
iso9001	Edit: I'm dumb just noiticed I used G instead of S.... sorry, hit the wrong key apparently. Ok, I'll make the change. I'm a little confused about the biasing tho. The last time I accidentely made a touch switch, but that was an N-Channel But I understand getting a few stray electons to trigger then N-ch is probably easier then finding a close place to sink enough to fake a P-ch out, I'll trust you and leave the biasing out for now. Just need to make sure I cut the power before I make the micro sleep. Thanks Again Everyone!