Re: Pull-up resistor question / data lines
Lol yeah it's very tiny I have microscope I use for this. Helps tons.
Thanks for the tidbit about the bubble symbol, never noticed it before.
iPhone schematics are confusing, they seem vague compared to most other schematics im used to.
The previous model used reset in similar function for multiple perifperals, low to reset, high active. This model has a reset and active/ready lines so makes things more complicated to understand.
Pull-up resistor question / data lines
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Re: Pull-up resistor question / data lines
01005 resistors are smaller than a pepper spec. You got guts measuring this!
Notice your schematic net labels tell if a signal is an input or output or bi-directional.
The body outline has an arrow: <- OUT, -> IN, <> BI for bidirectional lines. Bubble symbol means, example on RESET and IRQ, pulled low to assert.
A MCU input can have a pullup or pulldown resistor or nothing- this is assigned in firmware.
Pullup resistors are used on open-drain (driven) signals like RESET or IRQ because several IC's can activate (pull low) the line. It is a "wired-OR" setup. This IC or that IC can pull RESET low, for example.
A MCU output can be push-pull or open-drain.
For that iPhone, I think those are open-drain MCU signals (RESET and IRQ), so they need pullup resistors to +1V8.
They should stay high unless pulled low by anybody connected to cause RESET. BUT during off/sleep I think the +1V8 is switched off so it is confusing.Leave a comment:
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Re: Pull-up resistor question / data lines
Just took some measurements, there was 0v when the phone was off and 1.8v when it was on.
So looks like it's in reset mode when that line goes low. That's why schematic is confusing it appears to be the opposite like you saidLeave a comment:
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Re: Pull-up resistor question / data lines
Those pins are connected to other circuits which you should find so you can see how they function.
The example above as you can see, the 'reset' has line on top of the word reset, which mean it will go into reset when that pin goes low, normal operation the pin will be high.
Your IC 'RESET' pin is active high, so high = reset, so in normal operation that line should be low, but it has * after the RESET so you need to find out what the * is for.
Another sample:
https://cdn.badcaps-static.com/pdfs/...e24167a17b.pdfLast edited by budm; 03-22-2017, 09:45 AM.Leave a comment:
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Re: Pull-up resistor question / data lines
So if they're pull-ups that means those reset/system alive lines would always be pulled to 1.8v unless some other circuit pulls them low?Leave a comment:
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Re: Pull-up resistor question / data lines
They are pull-up resistors.
You can also look up the spec sheet of that IC to see what the requirement are.
http://www.electronics-tutorials.ws/...-resistor.htmlLast edited by budm; 03-22-2017, 09:31 AM.Leave a comment:
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Re: Pull-up resistor question / data lines
Thanks for those links.
Is there a way to know via a schematic if a resistor is blatantly being used as a pull-up/down?
For example the schematic attached, are r2260 and r2261 providing 1.8v to the reset and system alive lines, or is 1.8v being sent from those lines to the resistors? This confuses me at timesLast edited by caphair; 03-22-2017, 09:01 AM.Leave a comment:
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Re: Pull-up resistor question / data lines
http://www.ni.com/white-paper/4453/en/
http://pcbheaven.com/wikipages/How_Key_Matrices_Works/
Switching can be done with active high, active low, Voltage level sensing, etc.Leave a comment:
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Re: Pull-up resistor question / data lines
Ok I ask because my curiosity of how digital power buttons work. How shorting a line momentarily enables something to turn on.
Also how data lines are constantly being grounded to create pulsesLeave a comment:
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Re: Pull-up resistor question / data lines
Ten 1K resistors connected in parallel to the 10V power supply that can handle 1A (1000mA) and you will be drawing only 100mA so the power supply can easily maintain 10Vdc, if you use the power supply that can only handle 10mA then the power supply is not going to be able to handle 100mA load, it has to do with the output impedance of the power supply.
Study here:
https://cdn.badcaps-static.com/pdfs/...5fc335bf1a.pdf
http://powersupply.blogs.keysight.co...edance-of.html
Last edited by budm; 03-21-2017, 09:11 AM.Leave a comment:
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Re: Pull-up resistor question / data lines
So no unless one of those resistors was just a straight short to ground then the whole rail would be pulled down right?
So circuits need to be designed to handle the minimum current (in this example 10ma) in case all resistors were connected to ground?Leave a comment:
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Re: Pull-up resistor question / data lines
"does it create a low condition for all other circuits attached to that rail via other resistors " Nope.
Lets say you have 10Vdc rail (that can handle lets say 1A) and you have Ten 1K resistors connected to that 10Vdc, and another ends of those resistors are not connected to anything else yet.
So if you connect one end of one of those Ten resistor to GND, will you see the 10Vdc drops down? How about all TEN resistors have other ends all connected to GND, will the 10Vdc drop down?
Notes: 10V and 1K load = 10mA.Leave a comment:
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Re: Pull-up resistor question / data lines
When there's multiple pull-ups on the same rail, does it create a low condition for all other circuits attached to that rail via other resistors when only one is pulled low or it has no affect?Leave a comment:
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Re: Pull-up resistor question / data lines
Ah ok that helps.
So for example, a preceding chip could ground a circuit thus completing the electrical path flowing in it and this produces an active high on it's output? Am I understanding that right?Leave a comment:
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Re: Pull-up resistor question / data lines
Looks like this is partially due to terminology. When a chip is said to have an ACTIVE low or high, it refers to the output, not the input. In this case the term means that the chip's output can source or sink current. The ideal device would have a very high input Z (won't load the source) and a very low output Z (can drive a load). When CMOS logic came along, you got pretty close to that ideal. Previous logic types like TTL, not so much.
If a chip has a pull-up resistor on one of its pins, and no other connections, that would be called a "static" state. But in that case, no resistor would be required because the pin would never have to be pulled LOW.Leave a comment:
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Re: Pull-up resistor question / data lines
Ok thanks I guess my confusion stemmed from thinking of a chip has an active high due to a pull-up then there's no need for it to become an active low from a preceding chip pulling it to ground.
In other words I was looking at it in a way that if a chip is using a voltage due to a pull-up then pulling it low turns that chip off and I couldn't see why it'd need to be pulled lowLeave a comment:
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Re: Pull-up resistor question / data lines
So, yes, the preceeding logic chip might (for example) be capable of 2 output states: open and active ground. When that chip is in an "open" or disconnected state, then the following chip has to provide its own input logic state (voltage) via the pull-up. When the preceeding chip changes to a LOW, or ground state, the following chip's input is forced to be LOW. Keep in mind that any particular logic gate is constantly receiving input information, processing it, and providing output to the next logical device.
Not all logic inputs need pull-up or pull-down resistors. Typically you will need them with TTL devices, because their design provides an ACTIVE LOW output.Leave a comment:
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Re: Pull-up resistor question / data lines
Active HI/LO:
Open Collector output:
http://www.evilmadscientist.com/2012...ector-outputs/
http://www.electronics-tutorials.ws/...-resistor.htmlLast edited by budm; 02-25-2017, 04:46 PM.Leave a comment:
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Re: Pull-up resistor question / data lines
Threre are 2 reasons why you would use a pull-up (or down) on a logic chip or any other kind of chip. The first is that a particular input pin is not active in your design, but needs to be a solid High or Low. Without a pull-up or pull-down resistor in place, a logic input voltage can wander around aimlessly and cause lots of trouble.
The second one is that the preceeding chip might be active in only one direction. In other words, whatever is driving an input (with a pullup) has an open collector output for example, and can only produce an active low. (low impedance path to ground)
How does the circuit know when to create a pulse? That's up to the circuit designer. Logic chips do not create data. They respond to their inputs.Leave a comment:
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Re: Pull-up resistor question / data lines
Threre are 2 reasons why you would use a pull-up (or down) on a logic chip or any other kind of chip. The first is that a particular input pin is not active in your design, but needs to be a solid High or Low. Without a pull-up or pull-down resistor in place, a logic input voltage can wander around aimlessly and cause lots of trouble.
The second one is that the preceeding chip might be active in only one direction. In other words, whatever is driving an input (with a pullup) has an open collector output for example, and can only produce an active low. (low impedance path to ground)
How does the circuit know when to create a pulse? That's up to the circuit designer. Logic chips do not create data. They respond to their inputs.Last edited by Longbow; 02-25-2017, 09:46 AM.Leave a comment:
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by mikey5791Hi to all forum members,
Got this faulty 350 watt atx psu dirt cheap from a local online seller as i am keen to use it as a learning tool to repair faulty power supply.
The faulty part is quite obvious from the burn area near the 16pin KA7500 BD pwm. The fuse is still intact with good continuity.
Upon power on ac, there is 5vdc standby power on cathode of diode D14 and about 10vdc on cathode of diode D13.(This should be the aux power to generate the 12v and 5v rail.)
Initially i replace the two burned resistors near my marked red arrow to 200 ohm resistor... - Loading...
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