Solving Transistor KFY34 Switching Load Homework

AI Thread Summary
The discussion revolves around solving a homework problem involving the KFY34 transistor, which requires drawing a schematic, calculating base current (Ib), and determining circuit voltages and currents under different conditions. Participants clarify that the maximum voltage specifications initially mentioned were incorrect, and they discuss the need to calculate Ib based on the worst-case beta (h21E) values found in the datasheet, which range from 35 to 125. The importance of ensuring the transistor operates in saturation is emphasized, with recommendations to design for safety margins in Ib calculations. Additionally, there is confusion regarding the time-dependent nature of the required plots, which participants clarify should reflect the transistor's on/off states rather than actual time variations. The conversation concludes with suggestions for further reading and practical applications of the concepts discussed.
-SJ-
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Hello,
I've spend a lot of time at following example of my homework but still unable to find a solution. In fact I have no idea how to solve that, any help would be very appreciate.

Homework Statement


Transistor KFY34 has to switch a load of nominal supply voltage 24V and resistance Rs=60Ω.
a)Draw a schematic of electrical circuits.
b)Determinate Ib while the load is switched on.
c)Determinate currents and voltages in the circuit I B = f(t), IC= f(t), UBE = f(t) a U CE = f(t) with the load switched on and off and draw them to a graph.
d)Decide if additional cooling is necessary.
KFY34: NPN, h21E = 35 až 125, UCB0 = 70V, UCER = 50V, ICmax = 500 mA, UCES ≈ 1,5V, PC = 800mW


Homework Equations


Ib=Ic/β
Rb=(Uin-Ube)/Ib


The Attempt at a Solution


obr_zek.jpg
 
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The voltages and currents you have drawn are the maximum allowed specifications, not the values that the transistor experiences in this circuit. How could you get 50V or 70V across the transistor, with only a 24V power supply? :biggrin;

Calculate the voltages in the circuit assuming two conditions: 1) the transistor is cut off (Off state), 2) the transistor is in saturation (On state).
 
Thank you for your answer, 70V with a 24 power supply is a great mistake of mine, of course :)
1)if the transistor is cut off, there would be a full voltage at it, wouldn't it?
2)in saturation mode there's should be voltage drop of 0,6-0,7 due to diode-juction.
 
-SJ- said:
Thank you for your answer, 70V with a 24 power supply is a great mistake of mine, of course :)
1)if the transistor is cut off, there would be a full voltage at it, wouldn't it?

Correct. Vce would be 24V in cutoff. What would Ic be?

-SJ- said:
2)in saturation mode there's should be voltage drop of 0,6-0,7 due to diode-juction.

Somewhere around there, perhaps Vce would be a bit lower than 0.6V. So what does that mean for Ic? And then how can you work back to finding the Ib needed to support that Ic?
 
If 24V be cutoff, there will be none Ic.
And there would be relative hight Ic while transistor switched on determinated by Rd. But not sure how to calculate it because I don't know beta neither Rd.
 
-SJ- said:
If 24V be cutoff, there will be none Ic.
And there would be relative hight Ic while transistor switched on determinated by Rd. But not sure how to calculate it because I don't know beta neither Rd.
Compared to 60 Ohm, you can hopefully neglect the internal resistance of the transistor. You can probably assume it conducts perfectly (just with the internal voltage drop).
 
So, Ic=400mA. Hm, but what about Ib if assumed we don't know beta?
 
-SJ- said:
So, Ic=400mA. Hm, but what about Ib if assumed we don't know beta?
In practice, you'll never know β. You design for the worst case, plus allow an extra margin for safety. If you don't observe this cautious approach, then you risk losing the transistor.

So, what worst-case β will you design for?
 
Ic can't get over 500mA, so beta should be appropriate to that. But we don't know Ib, do we?
 
  • #10
-SJ- said:
Ic can't get over 500mA, so beta should be appropriate to that. But we don't know Ib, do we?
You are trying to determine what value you will set IB ... such that the transistor is guaranteed to operate in saturation, regardless of who builds it. So you base your calculations on the worst-case you can anticipate. You will need to refer to the device's specs for this.
 
  • #11
I've read some pages about transistors but nowhere found how to calculate it. In fact, I have no idea how to set Ib and any guidance would be really appreciate.
 
  • #12
You cited the relevant equation in the OP.

Edit: Well ok you need to rearrange it a bit. Ic=Ib*β

However β isn't a constant. It varies because the manufacturing process isn't perfect. If you need to achieve a certain Ic you had better assume β is at the lower end of the specified range when you calculate how much Ib you need to feed it. Otherwise not all transistors of that type would work.
 
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  • #13
But which value of beta I should assume to determinate Ib? I've tried to look up it's range in a data sheet but it is not stated there. How can I find it out?
 
  • #14
-SJ- said:
But which value of beta I should assume to determinate Ib? I've tried to look up it's range in a data sheet but it is not stated there. How can I find it out?

You are not finding Beta information in the datasheet? Can you post a link to the datasheet?
 
  • #15
Google suggests this transistor dates from the 1970's and found what I think is a Czech data sheet.

http://www.datasheetarchive.com/dl/Scans-048/DSAGER000349.pdf

Looks like the static Current Gain is "Proudove zesileni" and is the parameter h21E.

Note that the gain is specified at various values of IE and temperature.
 
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  • #16
-SJ- said:
But which value of beta I should assume to determinate Ib? I've tried to look up it's range in a data sheet but it is not stated there. How can I find it out?
Refer to part (d) in your first post.
 
  • #17
Hello,
I am sorry for my very late response due to my work responsibilities. Thanks for help with the current gain, I've find some data sheet but was looking for the greek symbol beta and didn't realized that h21e may be the same.
It looks like the current gain is 35-125 for this particular transistor manufactured by Tesla (indeed wide range seems to me), so the base current would be Ib=4mA while h21e=125 and Icmax=500mA. But is there a guarantee the transistor will be in saturated region with so low current? If the internal resistance is negligible are we able to determine the voltages across the transistor how we are asked?
 
  • #18
4mA would be the best case. You have to plan for the worst case where you need more base current.
 
  • #19
-SJ- said:
It looks like the current gain is 35-125 for this particular transistor manufactured by Tesla (indeed wide range seems to me), so
... the base current would be Ib= ... mA while hFE=35 [/color]and Icmax=500mA.

To ensure the transistor will be saturated for all β in the range, you have to take the conservative approach.
 
  • #20
OK, for h21e=35 is the base current Ib=14mA. But what happens if we set this value and the h21e will be 125?
 
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  • #21
-SJ- said:
OK, for h21e=35 is the base current Ib=14mA. But what if we set this value and the h21e will be 125?
What do you predict will happen? Base this on your understanding of transistors, including their characteristic graphs (which you should be able to sketch from memory).
 
  • #22
I would say it will damage the transistor, but so it is not more safety to set the Ib at lower current?
 
  • #23
Why would it damage the transistor? Suppose β was 1,000,000 do you think the collector current would be 14,000A? Unlikely. What else might limit the current?
 
  • #24
-SJ- said:
I would say it will damage the transistor, but so it is not more safety to set the Ib at lower current?
Which transistor rating would be exceeded by making IB to be 14mA? Only by exceeding some rating can you expect to damage the device.
 
  • #25
The current will be also limited by the collector load resistance. Ic=24/60=400mA, so the base current would be Ib=400/35=cca 12mA to ensure the transistor will be open. Is that correct?
 
  • #26
-SJ- said:
The current will be also limited by the collector load resistance. Ic=24/60=400mA, so the base current would be Ib=400/35=cca 12mA to ensure the transistor will be open. Is that correct?
That's the idea. But recognize that in practice that exact choice would be sailing too close to the wind. To ensure there is a margin of safety, you'd probably round out to 16 - 20 mA.

Your use of "open" is not correct, though I know what you mean. A better word is "closed", because when a switch is in the conducting state we say it is closed. An open switch is an open circuit = no current. Though I'd prefer "on" because the transistor is being used as a switch here, so it is either ON or OFF. (Well, it will be either ON or OFF if it's been designed properly.) :smile:
 
  • #27
OK, thank you. I'll be using either ON or OFF :-) But can be calculate the voltage between base and emitter and between collector and emitter? I was looking for it on the Internet but never find it (again:-).
 
  • #28
-SJ- said:
OK, thank you. I'll be using either ON or OFF :-) But can be calculate the voltage between base and emitter and between collector and emitter? I was looking for it on the Internet but never find it (again:-).

Look at the transistor datasheet. There will be specs and curves for Vbe and Vcesat as functions of temperature, current, etc. Ballpark numbers are a diode drop (0.6V-0.7V) for Vbe, and a few hundred milivolts (0.3-0.4V) for Vcesat.
 
  • #29
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?
 
  • #30
Functions of time?? Where do I see that?

They shouldn't be.
 
  • #31
In our homework is written "c)Determinate currents and voltages in the circuit I B = f(t), IC= f(t), UBE = f(t) a U CE = f(t) with the load switched on and off and draw them to a graph." Or perhaps, it mean more likely temperature?
 
  • #32
-SJ- said:
In our homework is written "c)Determinate currents and voltages in the circuit I B = f(t), IC= f(t), UBE = f(t) a U CE = f(t) with the load switched on and off and draw them to a graph." Or perhaps, it mean more likely temperature?
I'd say what is intended is that you make base current a function of time, ON & OFF, then the others follow along. Are you doing the base switching manually, or using a square-wave?

You don't have a way to measure the silicon temperature, anyway, I'm sure.
 
  • #33
We haven't had any special labs for transistor switching. Is there any difference between base switching manually an using square-wave?
If the transistor is ON, there will be appropriate voltage changeless in time, won't be?
 
  • #34
I agree with NascentOxygen. Fix some times where you switch the transistor on/off and then plot the currents for the on/off states. There is no capacitor or inductance involved, so you can assume the switching goes instantly. Ignore the details of the switching, focus on the states before and after it.
 
  • #35
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?
 
  • #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.
 
  • #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).
 

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