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DELETED (Inactive)
3 Posts |
 Posted - Feb 19 2003 : 7:27:02 PM
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I'm attempting to build an air/fuel indicator using the oxygen sensor. I've got the thing to work using very little amperage. I can turn on the LED by using my body to conduct current flow. My problem is that I also need to turn on the LED with minimal voltage, perferably no more than 100mV. It takes 2V or so to light my LED using two transistors to reduce the current flow from my source (the line that I want to hook to my O2 sensor). Is there a type of transistor that will allow a larger amount of voltage to pass(collector) with real low voltage to the base or is that how all transistors work? Thanks for any help.
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Aaron Cake
Administrator
Canada
6718 Posts |
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DELETED (Inactive)
3 Posts |
Posted - Feb 20 2003 : 6:50:43 PM
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Thanks for the info. I've been driving myself crazy trying to build this thing. How does the light strip driver thingy work? I'm new to alot of this electronics stuff. I like to make things but I also want to understand how it works. Also, is there a transistor that can switch more voltage than is on the base or that just the way they work? Thanks again for your help.
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Aaron Cake
Administrator
Canada
6718 Posts |
Posted - Feb 21 2003 : 10:12:49 AM
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The O2 sensors outputs a voltage depending on the oxygen content of the exhause. The standard A/F guage is simply a voltmeter with fancy lights. You can get the same effect by connecting a high impedance voltmeter set on the 1V scale to the sensor. As you drive, you will be able to observe the voltage output of the O2 sensor and thus the mixture ratio of your engine. During cruise, the voltage will rapidly bounce up and down while the computer adjusts the mixture close to stoich for best milage. During accelleration, you will see the voltage rise as the engine runs rich. Under decel, it should fall very low as the engine goes full lean.
Long story short, the A/F guage is very similar to the sound level meter on my circuits page. In fact, it's nearly the same circuit. The chip just contains a bunch of op-amps, configured as comparators. An electronics text will explain what that is in detail. Basically, a compatator looks at a voltage input and compares it to a reference. If the input is higher (or lower, depending on the config) the comparator's output goes high (or low, depending on the config). If you connect an LED to the output, you have a LED that lights depending on the input voltage. A bunch of these things, alll with different references, will cause a bunch of different LEDs to light at a bunch of different voltage levels. Configure so that each LED lights with a slightly higher voltage, and there you have it.
For info on transistors specifically, check out http://www.aaroncake.net/electronics/schem.htm
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DELETED (Inactive)
3 Posts |
Posted - Feb 23 2003 : 12:40:47 AM
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Thanks Aaron. The op-amp is what I was looking for. I was wanting to know how to turn on an LED that requires 2V with a very low voltage. I had a 450mV bias voltage and wanted to make the LED come on at anything above that. 500 mV for example. Obviously 50 mV isn't enough to run a 2V LED. I'll have to do some more looking at my local radio shack. They don't stock much. As a mechanic I've had training on basic electronics but nothing in detail. They figure we don't need to know how to build stuff, just test and replace them. I'm trying to learn more about electronics. Your web sight has been very informative, thanks.
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DELETED (Inactive)
21 Posts |
Posted - Mar 25 2003 : 12:34:56 AM
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Okay heres the run down. All you need is a 741 op - amp sold at any radio shack for a few dollars. You're going to have to use your battery voltage to run the op amp, you can divide it down to a smaller voltage, but its not necessary. A 741 op-amp is a bi-polar device meaning it can handle both positive and negative voltages (the 741 must be hooked up differently to handle bipolar voltages) Okay, you're probably not familiar how to configure it, but you want a "Non-Inverting configuration" meaning that the op amp will only make your signal larger, it will not invert it. I can't remember what the voltage was but I'm going to assume it was 100mV for this application, and then I will calculate out some standard resistor values for you that you can buy. First off you want a gain of 20 I found that by Vo/Vi where Vo = 2v and Vi = 100mV. I can't find a list of standard resistor sizes, but find 2 resistors that have a multiple of 19 between them. IE 1 is 19 times larger than the other. Here is the configuration, so bare with me while I explain it. On the back of the op - amp box there will be a pin out -- Connect the 100mV signal that you want to be amplified to the V+ terminal input. Connect the resistor that is 19 times smaller to V- and ground the other end of it. and then connect the Larger resistor to the V- terminal and Vout (which I think is pin 6, but I'm doing this from memory). That configuration will give you a 2V out put given that the input is 100mV
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BEatonNo1
Nobel Prize Winner
USA
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transistor question for air/fuel indicator
