Could you please calculate the Vce maximum , or show me how to calculate it ,I really don't know how to find it.

 

The worst-case collector voltage is when the AC signal peak is high enough to completely turn off the transistor.
Assume the emitter voltage doesn't change significantly from this due to the capacitor bypass across the emitter resistor.
I'll let you determine what the Vce is at that point.

 

I guess

Vce = Vcc peak

 

I guess

Vce = Vcc peak

Guessing is not a viable approach in solving technical problems.
You need to think about what I said in Post #22.
What is Vc when the transistor shuts off?
What is the emitter voltage (which you calculated)?

 

Vc = 20.23 mA × 100 = 2.023 v

This is what I understand

Vcc = Vc + Vce + Ve
24 = (20.23 mA × 560 ) + ( 10.64 ) + ( 20.23 mA x 100 )
The transistor resistance = 10.64 ÷ 20.23=526 ohm

When on peak
Vcc peak = 24 × 1.41 = 33.94 V
If the emitter voltage does not change then
This voltage should affect the current flowing into the Rc and the transistor and greater current should flow in.
Or maybe the current through the Rc remain the same but we shall experience a higher voltage across the transistor then higher current will flow into the transistor which may damage it.
33.94 - 24 = 9.94 v
This is the extra voltage due to the peak voltage
We shall add this to the Vce
9.94 + 10.6 = 20.54 v
I don't know how far this is correct.

 

Vcc peak = 24 × 1.41 = 33.94 V

How can the collector voltage become greater than the supply voltage in that circuit?

 

The supply voltage peak is
24 x 1.41 = 33.94

 

The supply voltage peak is
24 x 1.41 = 33.94

I might be wrong, but isn't the +24V a DC voltage?

 

The supply voltage peak is
24 x 1.41 = 33.94

Not according to your schematic.

 

Yes ..you are right
It is a dc voltage.
I really mixed up.....
Could you guys give a clear method which enables me to find the Vceo?

 

Current flowing in the emitter and collector when the base is open is
I = 24 ÷( 100 + 560 ) = 36 mA
Vc = 36 × 560 = 20.36 v
So what should I do forther?

 

You really need to think about what is happening in the circuit, otherwise how are how are you going to solve other similar problems?
You can't have someone else solve them all for you.

You've determined the DC bias voltage across the collector and emitter.
When you apply an AC signal at the input it generates an amplified AC signal at the collector.
The maximum peak AC voltage you can have is when the transistor shuts off and the collector ries to the supply voltage.
The AC voltage doesn't appear at the emitter due to the capacitor bypass so the emitter voltage doesn't change from the bias value.

So then what is the maximum voltage across the transistor?

 

Hi Jony,

I saw your post somewhere else. I'm trying to import 4538 one-shot model to LTspice. I downloaded the "extra"
zip-file you uploaded, but it failed to work. Can you help with this? My email is jxm841@case.edu.

Thanks

 

Hi Jony,

I saw your post somewhere else. I'm trying to import 4538 one-shot model to LTspice. I downloaded the "extra"
zip-file you uploaded, but it failed to work. Can you help with this? My email is jxm841@case.edu.

Thanks

Here you have all you need. Open the CD4000 exampel

 

Hi Jony130,

Thanks for your reply. I'm having trouble simulating it. I put the CD4538.asy in the C:\Users\J**** M***\Documents\LTspiceXVII\lib\sym, and the CD4000.lib file under dir lib/sub, but the system shows "unknown subcircuit" called in when simulating. Can you show me more how you make it work in more details?

Thanks very much