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Electronics Forum / Basics / May 2008optocoupler trouble
I'm getting strange results from an optocoupler (aka optoisolator, and this is the kind with transistor output). In brief: nothing I do seems to make the transistor "conduct", except very feebly. Here's my setup: Diode anode: 0V/1.07V, 0.4mA Diode cathode: ground Transistor collector: connected to a 330 Ohm resistor which is connected to 4.92V. Transistor emitter: ground Transistor base: open When the diode anode is at 0V, the transistor collector is, as one would expect, at 4.92V. But when I give the diode anode 1.07V, the transistor collector drops to just 4.88V. I want it to go down to 0V and I don't understand why that's not happening. > I'm getting strange results from an optocoupler (aka optoisolator, and > this is the kind with transistor output). In brief: nothing I do [quoted text clipped - 15 lines] > to just 4.88V. I want it to go down to 0V and I don't understand why > that's not happening. How much current to the diode? Applying a fixed voltage to a diode will get you a wide range of currents. 0.4mA in the diode is a recipe for minuscule currents in the transistor. What does the optocoupler data sheet say? You need about 15mA to pull the collector down close to zero (you won't get it all the way), that's a lot for an optocoupler with a plain transistor.  SignatureTim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html Thanks. I'm going to try answering your question about the datasheet, though I'm not quite sure what I'm looking for. Is it the "DC Current Transfer Ratio"? For that it says 20%, at I_F=10mA and V_CE=10V. > Thanks. I'm going to try answering your question about the datasheet, > though I'm not quite sure what I'm looking for. Is it the "DC Current > Transfer Ratio"? For that it says 20%, at I_F=10mA and V_CE=10V. Yes. That says that the collector current will be 1/5 of the diode current. You should check the maximum diode current, too -- that 10mA is certainly suggestive of what you ought to be able to do. Figure 10mA in, 2mA out, means you'll need to use at least a 2500 ohm resistor to pull the collector all the way down. SignatureTim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html > > Thanks. I'm going to try answering your question about the datasheet, > > though I'm not quite sure what I'm looking for. Is it the "DC Current [quoted text clipped - 6 lines] > in, 2mA out, means you'll need to use at least a 2500 ohm resistor to > pull the collector all the way down. With 10mA LED current NOT 400uA. Graham > > You should check the maximum diode current, too -- that 10mA is > > certainly suggestive of what you ought to be able to do. Figure 10mA > > in, 2mA out, means you'll need to use at least a 2500 ohm resistor to > > pull the collector all the way down. Thanks, and would you please explain why 2500 ohm is the correct resistance, indeed why the resistance at that point makes any difference to what the voltage is going to be at the collector pin? (I was under the impression that when the transistor is "on", it has no internal resistance...) >> > You should check the maximum diode current, too -- that 10mA is >> > certainly suggestive of what you ought to be able to do. Figure 10mA [quoted text clipped - 6 lines] >(I was under the impression that when the transistor is "on", it has >no internal resistance...) It has a *pretty low* internal resistance, as long as the current flowing through it is less than some value. That value depends on (for a normal transistor) the voltage/current on its base, or (for a photo-transistor) the amount of light hitting it. "On" and "off" are just shorthand. "On" is a short way of saying "there's plenty of base current (or light), which means that the transistor will allow as much current as I'm likely to give it". "Off" is a short way of saying "there's not much base current (or light), so the transistor will allow so little current through that I can ignore it." In between "on" and "off", the transistor is in what's called the "active" region, and it acts like a controllable current-limiter. That's what the "current transfer ratio" on the optocoupler data sheet is talking about: it's the ratio of input current (through the LED) to output current (through the output transistor). If you put 1 mA through the LED, it emits X amount of light; part of that light hits the transistor; as a result the transitor allows Y amount of current --- the ratio 1mA:Y is the current transfer ratio. The 2500 ohms Tim Wescott mentions is because there's a limit to how much current the LED in the optocoupler will take: if you max it out at 10 mA (which is an educated guess at what the limit is; the data sheet will say for sure), then the transistor half will be allowing 2 mA (that's 10 mA multiplied by the current transfer ratio). If you can only draw 2mA of current, the resistor has to be at least 2500 ohms in order to swing 5V (2500 ohms * 2 mA = 5V). SignatureWim Lewis <wiml@hhhh.org>, Seattle, WA, USA. PGP keyID 27F772C1 > " In between "on" and "off", the transistor is in > what's called the "active" region, and it acts like a controllable > current-limiter. Like a 'transfer resistor' in fact from which the word trans_istor is derived. Transistors were not originally designed as switches but as linear devices. Graham > > > You should check the maximum diode current, too -- that 10mA is > > > certainly suggestive of what you ought to be able to do. Figure 10mA [quoted text clipped - 6 lines] > (I was under the impression that when the transistor is "on", it has > no internal resistance...) No, it is NOT 'ON' or 'OFF' as in infinite or zero ohms. A transistor will conduct a CURRENT determined by its input drive (in this case the LED current). This is fundamental basics and you clearly need to read up about how transistors operate and how to use them. Graham >>> You should check the maximum diode current, too -- that 10mA is >>> certainly suggestive of what you ought to be able to do. Figure 10mA [quoted text clipped - 6 lines] > (I was under the impression that when the transistor is "on", it has > no internal resistance...) Because with a 0.2 current transfer ratio and 10mA to the diode, the transistor will only sink 2mA. If you want to drop 5V, you need at least 2500 ohms of resistance to do it. As pointed out elsewhere, transistors are never really "on" in every sense. They are only fully on in the sense that they are voltage limited, which depends on the circuit they're in. In this state any more drive to the transistor will not result in any more voltage drop at it's output -- but in your case that means you can't ask the transistor to sink more than 2mA. SignatureTim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html > Thanks. I'm going to try answering your question about the datasheet, > though I'm not quite sure what I'm looking for. Is it the "DC Current > Transfer Ratio"? For that it says 20%, at I_F=10mA and V_CE=10V. It means if you put 10mA into the LED the transistor will conduct 2mA with its collector-emitter voltage held at 10V. It also means (assuming the CTR hasn't wildly dropped off) if you put 400uA into the LED the transistor will conduct 80uA. Why are you using such a low CTR ? Graham > I'm getting strange results from an optocoupler (aka optoisolator, and > this is the kind with transistor output). In brief: nothing I do [quoted text clipped - 3 lines] > > Diode anode: 0V/1.07V, 0.4mA You're only putting 400uA through it ? There's your problem. Try something more like 10-20 mA depending on the precise device. Graham > Transistor collector: connected to a 330 Ohm resistor which is > connected to 4.92V. 330 ohms ? You're hoping for high speed operation are you ? If you don't need that use 4k7. Graham > I'm getting strange results from an optocoupler (aka optoisolator, and > this is the kind with transistor output). In brief: nothing I do [quoted text clipped - 15 lines] > to just 4.88V. I want it to go down to 0V and I don't understand why > that's not happening. The diode in an optocoupler is an Infrared LED. It takes more than 1.07 volts to turn on. You can see this with the feeble current of 400uA you measure. Run the diode on 5 to 10 mA. The voltage should be nearly two volts. If this thing has to run fairly fast, place a 10 K resistor from the base of the transistor to ground. The open base will slow operation appreciably. |
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