Who can explain this page in more detail to me?

Thread Starter

gte

Joined Sep 18, 2009
357
Ok,

vcc = 5vdc

r6 = 24k

r5 = 1k

r4 = 1k

r3 = 1k

r2 = 1k

r1 = 12k


Thank you :)


You had a schematic for the previous example. Schematics are the language of electronics, they have no ambiguity. So lets start from this example:



You tell us the values of Vcc and R1-6. We'll show you how to solve for them.
 

Wendy

Joined Mar 24, 2008
23,415
1st step, Total resistance.

R1+R2+R3+R4+R5+R6 = 40KΩ

Side recommendation, do the math yourself, people make mistakes all the time.



2nd step, Calculate current

V/R = I, 5V/40KΩ = 125µA (or 0.000125A)



3rd step, Calculate individual voltages.

You now have resistance and current, Ohm's Law will tell you voltages.

Remember though, Ohm's law deals with voltage across a resistance. It doesn't care about ground.

So if you want the voltage across R4 say, plug the numbers in.

V=IR, 0.000125A * 1KΩ = 0.125V


If you want the voltage from V4 to ground, you have to consider all the resistances in between, but Ohm's Law still applies.

Resistance from V4 to ground is R1+R2+R3+R4, so 12KΩ+1KΩ+1KΩ+1KΩ = 15KΩ

V=IR, so 0.000125A * 15KΩ = 1.875V


If the resistors are ±5% then the voltage will be too. So you will measure something between 1.78V and 1.97V (±0.09375V).
 

Thread Starter

gte

Joined Sep 18, 2009
357
How do you know the 1mA is sufficient? Is this a general rule of thumb? I'm not powering anything, but I am sending a signal to an automotive ECU if that makes a difference? Is my 40k ohms incorrect, even though it yields the proper voltages per resistor where I want it to?



Let's go back to your:


First, let's decide on a range of current that you need in this scenario. Let's just say you simply needed the voltages present, and you weren't going to power anything. Somewhere around 1mA should be enough.

So the question now is, how much resistance is needed to limit current to 1mA with a 5v source?
R = E/I, or Resistance(Ohms) = Voltage/Current(Amperes)
Since E=5v, and I=1mA = 0.001 Amperes...
R = 5,000 Ohms.

So to get 1.5v, we know that R = E/I, or Resistance(Ohms) = Voltage / Current(Amperes).
Since E=1.5v and I = 1mA, R= 1.5/1mA = 1.5/0.001 = 1,500 Ohms (1.5k)

Now you need 1.75v - but that's just 0.25v higher than the 1.5v you already have.
So to go 0.25 higher than the 1.5v to get the 1.75v, you'll use R=E/I again.
R = 0.25v/1mA = 250 Ohms.

We're up to 1.75v, and have to drop the rest of the 5v. 5v-1.75v=3.25v.
R = 3.25v/1mA = 3250 Ohms.

Another way to look at it would be since 5v/5,000 Ohms = 1mA, and 1mV/1 Ohm = 1mA, for each Ohm resistance there is 1mV dropped in this particular circuit that has a total resistance of 5,000 Ohms.
 

Thread Starter

gte

Joined Sep 18, 2009
357
Ok, perfect, this is how my real world testing came out

And if I wanted higher resolution (that may not be the proper term) I could substitute (2) 500ohm resistors in place of one of the 1k resistors, correct?

Also, how do you determine what wattage resistors are needed?


1st step, Total resistance.

R1+R2+R3+R4+R5+R6 = 40KΩ

Side recommendation, do the math yourself, people make mistakes all the time.



2nd step, Calculate current

V/R = I, 5V/40KΩ = 125µA (or 0.000125A)



3rd step, Calculate individual voltages.

You now have resistance and current, Ohm's Law will tell you voltages.

Remember though, Ohm's law deals with voltage across a resistance. It doesn't care about ground.

So if you want the voltage across R4 say, plug the numbers in.

V=IR, 0.000125A * 1KΩ = 0.125V


If you want the voltage from V4 to ground, you have to consider all the resistances in between, but Ohm's Law still applies.

Resistance from V4 to ground is R1+R2+R3+R4, so 12KΩ+1KΩ+1KΩ+1KΩ = 15KΩ

V=IR, so 0.000125A * 15KΩ = 1.875V


If the resistors are ±5% then the voltage will be too. So you will measure something between 1.78V and 1.97V (±0.09375V).
 

SgtWookie

Joined Jul 17, 2007
22,230
How do you know the 1mA is sufficient? Is this a general rule of thumb?
Since you did not specify any kind of a load, I made the assumption that this was simply an experiment, and the only "load" on the circuit would be that of a multimeter or DMM. The old-style multimeters with a D'Arsonval movement generally have an impedance of 10k Ohms per volt, and this must be taken into account when making voltage measurements. Digital multimeters have a far greater impedance, thus have a very small effect on the circuit being measured.

The 1mA circuit current was chosen as a reasonable trade-off between low power consumption and the effect of a meter's impedance on the circuit.

I'm not powering anything, but I am sending a signal to an automotive ECU if that makes a difference? Is my 40k ohms incorrect, even though it yields the proper voltages per resistor where I want it to?
Now you're heading into unknown (and likely expensive) territory. Giving your ECU an incorrect input will at best result in poor performance, poor economy, and high emissions. At worst, it will lead to destruction of the engine due to excessively lean operation, causing detonation (pinging) and resulting severe damage to pistons, wrist pins, connecting rod and crankshaft bearings, and/or destruction of the ECU itself.

Automotive environments are brutal and unforgiving. Temperatures can range from sub-zero to hundreds of degrees F in the span of a few minutes. There is a LOT of electrical noise under the hood, which can wreak havoc on digital logic and low-level analog circuits.

I suggest that it's a bad idea for someone at a beginner level to attempt modifications to their ECU's control circuits.
 

Wendy

Joined Mar 24, 2008
23,415
Wattage is a formula, just like Ohm's Law.

P = VI = V²/R

The second formula is derived from Ohm's Law.

You have current. You can figure out the voltage across each resistor, and since we are talking wattage you need to work on an individual resistor.
 

Thread Starter

gte

Joined Sep 18, 2009
357
Well the good news is that I have this circuit inside the cabin and that I have lots of automotive experience, even though almost no electronics experience.

Here is my dilemma, I need to simulate the voltage readings in this video. When an electric pump turns on, the voltage input goes from ~1.61vdc at atmosphere, to ~1.83vdc under 1 psi of pressure, up to ~2.04vdc under 3-3.5psi of pressure and then down to 1.57vdc.

At first I built a circuit that has a SPDT relay and when the pump was off, the voltage was 1.6vdc (first via a potentiometer and now via resistors) and then 1.85vdc when the pump was on. This worked, except for the check sum that is done at the end where it spikes up the pressure and then goes into a very slight vacuum. I'm not sure how to simulate that, any ideas?

The other strange thing is that when the pressure sensor is not plugged in, the sensor wire has 4.92vdc on it, but when I plug the sensor in it drops to ~1.6. Where this creates problems is that when I add my circuit to the sensor wire, with the sensor NOT plugged in, it changes the voltage output of my circuit? Somehow the 4.92v is adding .10 to .15vdc to my circuits output?


http://www.youtube.com/watch?v=OB6EN6tdc-Y - gauge and multimeter

http://www.youtube.com/watch?v=UfjOp3295gc - multimeter only



Since you did not specify any kind of a load, I made the assumption that this was simply an experiment, and the only "load" on the circuit would be that of a multimeter or DMM. The old-style multimeters with a D'Arsonval movement generally have an impedance of 10k Ohms per volt, and this must be taken into account when making voltage measurements. Digital multimeters have a far greater impedance, thus have a very small effect on the circuit being measured.

The 1mA circuit current was chosen as a reasonable trade-off between low power consumption and the effect of a meter's impedance on the circuit.



Now you're heading into unknown (and likely expensive) territory. Giving your ECU an incorrect input will at best result in poor performance, poor economy, and high emissions. At worst, it will lead to destruction of the engine due to excessively lean operation, causing detonation (pinging) and resulting severe damage to pistons, wrist pins, connecting rod and crankshaft bearings, and/or destruction of the ECU itself.

Automotive environments are brutal and unforgiving. Temperatures can range from sub-zero to hundreds of degrees F in the span of a few minutes. There is a LOT of electrical noise under the hood, which can wreak havoc on digital logic and low-level analog circuits.

I suggest that it's a bad idea for someone at a beginner level to attempt modifications to their ECU's control circuits.
 
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