Solving Transistor KFY34 Switching Load Homework

In summary: So 4mA would not be enough to guarantee saturation. So add a safety margin. Then you're good. But it may be more than 4mA.
  • #36
-SJ- said:
I've read following values fro the graphs: Uce=f(Ic)=3,5V; Ubesat=f(Ic)=1,2V; Ucesat=f(Ic)=3V. Could it be right? But what confuses me are these values as a function of time as listed in the example. They are time independent, don't they?

-SJ- said:
I am affraid I completely don't understand it :/ Do I need to measure it or some calculation to determinate Ube and Uce would be sufficient?

Can you post a link to the datasheet? As has already been mentioned, those numbers make no sense.

Also, have a look at the BJT Switch example on this page: http://www.zen22142.zen.co.uk/Design/bjtsw.htm

That should help you out some.
 
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  • #37
-SJ- said:
I am affraid I completely don't understand it :/ Do I need to measure it or some calculation to determinate Ube and Uce would be sufficient?

The Base-Emitter junction looks like a diode so when the input to the circuit is (for example) 24V then Ube will be around 0.7V.

Ib = (24V-0.7V)/RB

Note that 24V >> 0.7V so Ib depends mainly on the input voltage (24V) and less on the exact value of UBE. No need to measure it.

If the transistor is turned ON UCEsat will typically be around 0.3V for many transistors. So..

Ic = (24V-0.3V)/RS

Note that 24V >> 0.3V so IC depends mainly on the supply voltage (24V) and less on the exact value of UCEsat. Again no need to measure it.

I would pretend that the input voltage is a 0-24V square wave with a frequency of something like 1Hz. Plot Vin on a graph of Vin vs time.

On other graphs with the same horizontal axis plot Ib, Vce, Ic.
 
  • #38
Thank you for the provided link, it's very helpfully website to me. I've read more interesting articles about transistor and I am looking forward for wiring some circuits and do measurements on BJT.
Unfortunelly, I am apparently too dull to recognize what I am asked for. I have no idea how to find out how the time-depending plot should looks like. I've never seen it in my school, in data sheet it's not listed and no online circuit simulator has KFY34 in it's database.
 
  • #39
There is no autonomous time variation of any parameter. You change the base drive, and other voltages & currents simply change along with it. The easiest way to display these on an oscilloscope is to set a square wave to repetitively change the base drive instead of you doing it manually.
 
  • #40
Forget the time-dependence for a moment.
If the transistor is switched on, can you determine the relevant quantities?
If the transistor is switched off, can you determine the relevant quantities?

If you have both, make a plot where from time t=0 to t=1s you have the first set of values (constant in time) and from t=1s to t=2s you have the second set of values (again constant in time).
 
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  • #41
NascentOxygen said:
There is no autonomous time variation of any parameter. You change the base drive, and other voltages & currents simply change along with it. The easiest way to display these on an oscilloscope is to set a square wave to repetitively change the base drive instead of you doing it manually.

And that's exactly what I am confused by. It's a homework and I am not at school anymore. We are not allowed to use labs for our HWs and I haven't heard anyone ever was. Oscilloscope is an expansive device and far not everyone own it. It must be another way how to plot the graphs...
 
  • #42
Perhaps, don't be the square signal deformed by capacity barrier of the transistor? In that case I would be able to sketch it from memory, but without exact time-depending values.
 
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  • #43
-SJ- said:
Oscilloscope is an expansive device and far not everyone own it. It must be another way how to plot the graphs...

http://en.wikipedia.org/wiki/Paper
http://en.wikipedia.org/wiki/Pencil
http://en.wikipedia.org/wiki/Ruler
and if you make a mistake
http://en.wikipedia.org/wiki/Eraser

PS: You are making this problem harder than it is.

CWatters said:
I would pretend that the input voltage is a 0-24V square wave with a frequency of something like 1Hz. Plot Vin on a graph of Vin vs time.

On other graphs with the same horizontal axis plot Ib, Vce, Ic.

mfb said:
Forget the time-dependence for a moment.
If the transistor is switched on, can you determine the relevant quantities?
If the transistor is switched off, can you determine the relevant quantities?

If you have both, make a plot where from time t=0 to t=1s you have the first set of values (constant in time) and from t=1s to t=2s you have the second set of values (again constant in time).
 
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  • #44
You can always try to plot graphs on a computer...
 
  • #45
-SJ- said:
Perhaps, don't be the square signal deformed by capacity barrier of the transistor? In that case I would be able to sketch it from memory, but without exact time-depending values.

At least two of us have suggested drawing graphs with an input of 1Hz.

What "time depending values" do you think might show up at 1Hz?
 
  • #46
-SJ- said:
Perhaps, don't be the square signal deformed by capacity barrier of the transistor?
No, not here. There is no evidence of that because you choose a comparatively low clock rate where such distortions of the squarewave just do now show with the sweep speed involved.
 
  • #50
Looks ok to me. Just need to answer d).
 
  • #51
Actually I've just noticed that the saturation voltage specified for the transistor is UCES ≈ 1,5V rather than the 0.6V assumed. Perhaps worth amending the last graph of UCE.

It also changes the ON collector current slightly but only from 400mA to 375mA. Edit: Perhaps leave it at 400mA because UCES is a maximum figure. It could be lower for the actual transistor used.
 
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  • #52
Best not to assume VCEsat is 0.6 (unless you are sure it will be). Sounds too much like one diode drop, and your examiner may start wondering did you really think it was a VD? To be on the safe side, I suggest something like 1.1V if higher current, or 0.2V if low current, or 0.1V if no current, as reasonable general values for any non-specific circuit.

Otherwise, graphs are what is expected.
 
  • #53
I've amended the last graph to 1,5V but is with this correction Ube remained or changed to 1,5V also?
d)It might look like something like this: P=Ic*Uce_sat=400*1,5=600mW? Pmax=800mW, so no additional cooling is necessary.
 
  • #54
Base current is less than 20mA? So base voltage should be just under 1 volt. Probably close to that I'd say. It's just a p-n junction. Data sheets will show this as a graph between 0.8 and 0.9.
 
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  • #55
I've look into the data sheet and although I haven't found there graph Ube depending on Ib, I've saw there Ube_sat=1,1V and Ube=0,75V for Ic=400mA. Which of these values should be chosen? Why is the voltage over the junction current depending?
 
  • #56
-SJ- said:
Ube_sat=1,1V

That looks to be a typo. You mean Vce_sat (or Ube_sat), right? And even so, that is pretty high for Vce_sat.

Can you post a link to the datasheet you are reading?
 
  • #57
Sure, I was reading from this one posted here: http://www.datasheetarchive.com/dl/Scans-048/DSAGER000349.pdf
Page 21 and 22.
 
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  • #58
-SJ- said:
I've look into the data sheet and although I haven't found there graph Ube depending on Ib, I've saw there Ube_sat=1,1V and Ube=0,75V for Ic=400mA. Which of these values should be chosen? Why is the voltage over the junction current depending?

-SJ- said:
Sure, I was reading from this one posted here: http://www.datasheetarchive.com/dl/Scans-048/DSAGER000349.pdf
Page 21 and 22.

So those curves look quite normal. They show Ic as a function of Vce and Ib, and the last one shows Vbe as a function of Ic and temperature. The graph at the bottom of page 21 is the best one to use to estimate Vce_sat for your Ic...
 
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  • #59
So, for Uce_sat=1,1V should be the P=Ic*Uce=400*1,1=440mW, right?
 
  • #60
-SJ- said:
So, for Uce_sat=1,1V should be the P=Ic*Uce=400*1,1=440mW, right?
Is that 1.1V maximum or typical?

Don't forget the power into the base, it's not always negligible.

A device's power rating is under specific heat-sinking & ambient conditions. You need to consider how close to that your conditions will be. Exceeding about 250mW for a 440mW device would be unwise, at the best of times. If there exists an engineering mantra, it is to ALWAYS err on the side of caution. :approve: :approve:
 
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  • #61
That Uce should be typical. The KFY34 transistor has Pmax=800mW, so 440mW should be fine and no additional cooling is necessary.
OK, that's it. I'll send it to my examiner to check it out and be with hopes he will accept it despite the terrible time spent at solution.

I would like to say THANK YOU REALLY VERY VERY MUCH for patient guidance in this long-suffering thread. Now, I am hopefully a bit wiser about calculations with transistor and looking forward to learn more.
Thank You.
 

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