Ok. I tried my tutorial and I got all the answers wrong (except 1a). Worse still I can't attend the tutorial session.I have posted my WRONG answers to show that I have done my work. Can anyone pls correct my conceptual problems. Q1a. max current = 50mW/2V = 25mA b. I(0.5Led) = 25mA * 0.5 = 12.5mA // half of LED max current R(c2) = (Vcc - V Led - V cesat) / 12.5mA = ( 12 - 2 - 0.75 ) / 12.5 = 0.74Ω WRONG! c. Ic = βIb for minimum Ib, β must be 50. 25*10^-3 = 50Ib Ib = 0.5mA (WRONG!) d. Value of R b2 = (Vbb - V besat)/Ib = (5-0.75)/598μA = 7107Ω (WRONG!) e. ?? Q6a. It is to amplify current Ib????? (right or wrong?) b. It is to prevent leakage current. (right or wrong?) BTW are there similar problems on any website? I am totally lost. Thanks for the help.
Hi hitmen, i havent got time to look at all the questions but the first one just jumped out at me as a simple miscalculation. 50mW/2v = 25mA Hope this helps ;-) I wouldnt be entirely sure about this but, for variable b, you should assume worst case scenario, i.e. b = 20.
To follow up on mentaaal's comment, you have propagated the erroneous value calculated in 5.a into 5.b. On top of that, your calculation in 5.b has introduced an additional error. hgmjr
I have edited the answers but as usual my answers are still wrong. I have already spent hours on it. Pls help. Thanks.
You appear to be neglecting to consider the units of your terms. You should a take another look where the decimal point is located. You are treating milliamps as amps. There is a factor of 1000 between the two units of measure. hgmjr
It isnt the decimal places that are wrong. My entire concepts are wrong. I wouldnt be posting questions if they are mathematical errors. I just need someone to post how you get the correct answers.
5a. 25ma, correct. 5b. 25ma/2 = 12ma (aprox), (12V-2V)/.012 = 833.3Ω, or 820Ω (820 is a conventional value, see attached; ohms law, V/I=R) Reversing this, 10V / 820Ω = 12.2ma (again, ohms law, V/R=I) Try working it from there.
Umm... I dont understand why you are using approximated it to 12ma and not some other values. Also, what are the conditions in which Vcesat or Vbesat must be substracted? I dunno when the BJT saturate. I have tried doing part c using the formula Ibβ = Ic but whether β is 20 or 50, I compute Ib = 12.2mA / 20 and 12.2mA /50 and I still dont get the answer of 598μA. Thanks for the help.
When a transistor is saturated it is like a switch that is turned on, it drops no voltage between collector and emitter. The tutorial clearly stated "operating current ILED to be about half the maximum current allowed (or slightly less)", so I rounded the value off low. Since the answer was 12.5ma this would 12ma rounded. Why did I pick the resistor I did? I attached the chart of standard resistor values, which was the calculated value take to the charts nearest value. A quick hint for the next problem, calculate both values for beta (high and low), then pick the one that would meet both conditions.
Hmmm... I dont understand what you mean by 2 conditions. I have tried using the equation βIb = Ic for both beta 20 and 50. I assume that Ic = I Led, am I right? I got 0.625mA and 0.25mA for Ib which are not the answers. I dont get it.
The transistor is a current amplifier. A gain (beta) of 20 will give you one answer, 50 will give you another. Have you figured out 5b yet?
Just looking quickly, I'd say you have the concepts down correctly. There are a couple of obvious typos in your post, but I think you are basically correct. 820 ohms is not correct. (12V-2V-0.2V)/.0125=784 Ohms your other approaches seem right too, but I just looked quickly so I can't be sure.
Also, unless the question specifically says to use standard values with 5% tolerance, I don't think it is correct to change the value to match standard values. Now, we typically use 1% tolerance resistors without even thinking about it.
Most folks don't have access to 1% (including me), but standard 5% resistors are common. Like I quoted from the test, it states equal to or slightly less than half the calculated max current.There was a time 10% and 20% resistors were the norm, for that matter. I didn't see the answers until after I calculated it, but I got the first try. The .2V for the transistor saturation is an appoximation at best, I've see a LOT lower, so I didn't add it. Power transistors can handle a lot of current when on and not heat up. Another hint for OP, look up the formula for beta, you're not in the ballpark with the values you quoted.
Bill, I'm curious why you and most people you know don't have access to 1% resistors. They are commonly available and have been the standard for the two companies I worked at in the last 10 years. Honestly, they are available and cheap for either surface mount or thru-hole. I know many electrical engineers, and none of us design with resistor tolerance greater than 1% anymore, at least for low power resistors such as 1/8 and 1/4 W as we are discussing here. I live in the USA, so maybe it's a location thing? I'm not sure, but really am curious about it. By the way, I agree that 0.2 V saturation voltage is only approximate, and that it was a good idea for you to ignore it for a worst case calculation such as the one presented here. Surely, the value 0.75 V used in the original post is too high. Steve
Hitmen, It just occurred to me that you may have used Vsat=0.75 V as the saturation voltage because you confused it with the base emitter voltage Vbe. If so, this is a conceptual error, on your part, that contributed to your lower resistor value. Note, that the collector voltage can go below the base voltage when the transistor is in saturation. I often use 0.2V as a rule of thumb, but the true value depends on many things, and I agree (in hindsight) with Bill to use 0 V in this case for a worst case calculation. Still, you have the right approach. To me the correct resistor is 800 Ohms. For 5% tolerance you would round up to 820 Ohms, and for 1% i think it is 806 Ohms. My personal opinion is that 820 Ohms should not be the correct answer unless the problem (or the teacher) instructs you to use 5% tolerance. But, that's debatable I suppose. Steve
Not too complex, most of us are home hobbiests, and we use local parts outlets. I happen to have an excellent one that is almost local, so I was able to build a complete resistor kit in 5% values. In my experience only industry keeps the shear numbers of resistors you would have to keep to have the full range of 1% parts. Even then, when I was working for Collins Radio, we didn't do it when we should have. The fundimental issue is quantity and price, I paid $0.02 / resistor on my kit. Check this out.
Bill, Ah, I understand. Now that I think about it; I, and the companies I've worked for, use 1% resistors all the time, but don't stock all values. If the part needs to be precise, we order a reel of that value, but if the tolerance is not critical, we just use the closest value in stock. Also, for my hobby work, I'm spoiled since I collected an ungodly amount of electronics components during the technology crash in 2001. Thanks, Steve
Ice = β Ibe is correct. Steve was also correct. I had to work on it to figure it out. For the problem 5c you have to add in the transistor saturation drop, which is .2V Collector to Emitter. Use this to calculate the problem.