meek13

Hi,

I am a complete "noob" when it comes to electonics. I have a 2 part question that I would like to ask the forum. Here is the first part.

I am trying to control the output voltage by a fixed amount. I used a potentiometer for this.

I hooked the source voltage (input) to terminal 1. I grounded terminal 3 and used the wiper (terminal 2) as my output voltage. The source voltage is NOT fixed. It varies from 0V to 5V. The input voltage does not pulsate, it is user controlled. Since I discovered the pot is a voltage divider I could not achieve the desired reults.

eg.

Desired result of 0.25V differential is required on the output voltage. (Please note that this can be any figure between the min and max of input voltage).

IN V ---> OUT V
1 ---> 0.75
1.5 ---> 1.25
2 ---> 1.75
3 ---> 2.75
. ---> .
5 ---> 4.75

Actual result using a pot

IN V ---> OUT V
1 ---> 0.75
1.5 ---> 1.13
2 ---> 1.5
3 ---> 2.25
. ---> .
5 ---> 3.75

Is there anyway I can achieve output voltage by a fixed amount? Please treat me like a 10 year

Gaston

i would leave the input a five volts. ten make a divider out of multiple resistors and tap it a different points to get your voltage. i don't have any way to draw schematics and post them right now so its hard to explain. but keep in mind that the voltage is going to change with the current. thats what voltage regulators are for. to keep the voltage constant even though the current changes

Roff

Gaston is on the wrong track.
There are ways to do this. One way is a constant current through a resistor. This question is typical of a noob (not criticism, just an observation). If you would tell us, in as much detail as you can, exactly what you are trying to do, it would be much easier to help you. Give us the big picture. Where does the signal come from? Where is it going?

__________________
Ron

Gaston

sorry, i missunderstood the question.

meek13

Thanks for the reply.

No offence taken Roff

This is the picture of the source:

http://www.picotech.com/auto/wavefor...diesel_big.png

Here is a description of what is going on:

This waveform shows a test of the fuel system on a common rail diesel engine, using the fuel rail pressure sensor.

The PCM varies the rail pressure between about 280 bar at idle and 1600 bar at full speed and load. The sensor is the feedback component in a control loop, and informs the PCM what pressure is in the rail. The PCM can then tell the pump to increase or decrease output accordingly. The PCM controls the pressure regulator or “M-Prop” valve on the pump to control pump pressure. When you press the pedal, the PCM will immediately calculate how much fuel to give the engine based on speed, load, etc. and the internal calibration table. This fuelling table is specifically for that engine/vehicle combination. The sensor gives a continual feedback of rail pressure so that the PCM can make any pressure adjustments almost instantaneously.

We can analyse the performance of the system by graphing the output of the sensor against time, whilst we start, run, accelerate, hold at full speed, and return to idle. We finally switch off and wait for PCM power down (normally around 10 seconds after key-off). The scope is best set to a slow time base in chart recorder mode.

The waveform starts on the left just after key-on, where the voltage is 0.5 V, corresponding to a pressure of 0 bar. The sensor does this to provide a plausibility check: it should never normally read 0 V, so if it does, it has failed. When we start the engine, the voltage rises to about 1.3 V, which corresponds to about 280 bar, a common value at idle. We then put the pedal to the floor, and the PCM immediately adds a shot of fuel to accelerate the engine to red-line, where it is held by the speed governor. The voltage then settles back to a lower value, about 2.5 V, until we release the pedal back to idle, when it settles back to 1.3 V as at the start. We then key-off and the engine stops. Note how the signal drops slowly back to 0.5 V over about 10 seconds, before the PCM powers down near the right-hand end of the waveform. If the voltage drops very quickly to 0.5 V, then the residual pressure is leaking away too quickly, and may indicate a problem with the system — for example, a leaky injector, or a leak back through the pump. Remember that this test is done on an unloaded engine. On a fully loaded engine the centre section of the graph will rise well above 2.5 V. It won't go above 4.5 V, as this represents about 1600 bar. Again, this is a plausibility check on the sensor: if it goes to 5 V (the sensor supply voltage), there could be a fault with the sensor.

So as you can see the voltage input can vary depending on how much you press the accelerator. What I am trying to do is intercept this signal and adjust the voltage by a tiny amount and send to the ECU.
The amount of adjustment needs to be determined by trial & error.

Here's where it get's more complicated. The adjustment of the voltage should only start from 1.5V, any lower and it could interfer on how the car idles. Also, once the input voltage reaches 4.5V the adjusted voltage should be the same as the input voltage, 4.5V.

There will be alot of trial & error here to reach the optimum adjusted voltages, hence I need the ability to adjust the voltages as required. If I use fixed resistors then I would need to make several circuits for various voltage adjustments. By altering this voltages by a tiny bit, you get more torque and power out of the diesel engine without going outside the engines limits.

I hope that is enough information to get your teeth into it. Please let me know if you need any more info. Thanks is advance.

meek13

If we break this down into simple steps then lets assume the following:

Vin = source voltage (Ov to 5V)
diff = amount of voltage to be taken off (set between 0.1V to 0.9V)
Vout1 = (Vin - diff)

1. If Vin <= 1.3V then output voltage = Vin
2. If Vin > 1.3V then output voltage = Vout1
3. If Vin >= 4.5V then output voltage = Vin

I will have to experiment with the "diff" value to get the optimum figure. That basically is what I need in a nutshell. How can I achieve this please?

mneary

I would use an MCU.

Aren't you troubled by the discontinuities in the response? For 1.3V < Vin < (1.3V + diff) the output will be less than 1.3V. And when Vin passes through 4.5V, the output voltage suddenly jumps from (4.5V - diff) to 4.5.

Roff

I think a graph of Vout vs Vin would be peachy.

__________________
Ron

meek13

mneary

I am a total "noob" when it comes to electronics. I am assuming MCU stands for microcontrollers right?

How would I go about making this circuit using MCUs? I'm quite surprised that no one wants to take a stab at this...or maybe what I am asking for can't be done. Either way it would be nice if someone could help me out with this.

Roff

OK, if you won't draw a graph, then I will.
Is this really what you want? I might take a shot at it if I was sure this is what you want.

Attached Images

Vout vs Vin.PNG (16.9 KB, 18 views)

__________________
Ron

meek13

Sorry about the graph Roff. Yes that is what I am looking for although I would need the following options:

1. Ability to modify the kick-in voltage, let's call it V1. (Where I would like the circuit to start adjusting the input signal)

2. Ability to modify the drop-out voltage, let's call it V2. (Where I would like the circuit to stop adjusting the input signal)

3. Ability to modify input signal at any point between V1 and V2

Here is a image with your original plot alog with 2 other examples marked in red and green. Thank you for taking interest.

http://img292.imageshack.us/img292/3...utvsvinzs3.gif

Roff

The plots you drew look nothing like what you described originally. The one I drew (which is what you described) is easy. The other ones are very hard to do with an analog circuit. With a microcontroller you would probably need switch-selectable lookup tables, unless you can come up with an equation with adjustable parameters that will work for all those curves.

__________________
Ron

meek13

Hi Ron,

Yes I agree with what you are saying, that is why this was a 2 part question. Hey if you can help me build the plot you made that will be great.

Roff

I just got back from a mini-vacation. I'll try to get a schematic together.

__________________
Ron

Roff

OK, try this out. It looks complicated, but it only has 3 ICs.

Attached Images

window level shifter.PNG (49.7 KB, 13 views)

__________________
Ron