I want to know why both transistors are in active region and not in saturation region with only dc input.

Thread Starter

Ram Chockalingam

Joined Jul 9, 2023
7
I am so dumb and sorry about that but I understood why both BJT's are in active region the moment I rewatched the circuit after posting. So now how I calculate the voltage between collector and emitter in both BJT's ?
 

MrChips

Joined Oct 2, 2009
29,867
Each transistor is an inverter.
What you have is a circuit with positive feedback.

1690639738554.png

What you have as a bistable circuit, i.e. it has two stable states, also known as a flip-flop.
1690639789019.png
 

BobTPH

Joined Jun 5, 2013
8,149
That state may be stable in the simulator, but it is not in the real world. It will flip to one transistor on (saturated) and the other off. It is called a bistable multivibrator because there are two stable states, neither of which has the transistors in the active region.
 

Thread Starter

Ram Chockalingam

Joined Jul 9, 2023
7
That state may be stable in the simulator, but it is not in the real world. It will flip to one transistor on (saturated) and the other off. It is called a bistable multivibrator because there are two stable states, neither of which has the transistors in the active region.
So are you saying like in the real world one of the transistors will go to saturation mode first and thus the other one will go to cutoff mode and that neither of them go to active mode?
 

BobTPH

Joined Jun 5, 2013
8,149
So are you saying like in the real world one of the transistors will go to saturation mode first and thus the other one will go to cutoff mode and that neither of them go to active mode?
Yes, but more accurately, they will not stay in active mode.

If you change one if the collector resistors to 110, you will probably see it go to one state. They will be in active region for a short time.
 

MisterBill2

Joined Jan 23, 2018
16,637
Both transistors are in the active range because the resistor values are incorrect for the transistor types. In the real world it is difficult to read voltages with 8 digit resolution. And even more difficult to have two resistors with the same values out to 8 digits.
 

crutschow

Joined Mar 14, 2008
33,369
The quasi-stable state occurs becomes the simulation is ideal with all elements exactly equal to their given values.
Below is the simulation where I used a 1pF cap to slightly unbalance the circuit during startup which then initially sets it to one of the two (normal) stable states (here Q1 on and Q2 off, as would be expected from the capacitor location).

1690693320856.png
 
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BobTPH

Joined Jun 5, 2013
8,149
When I simulate with LTSPICE, the original circuit flips to the stable state after 600 usec. If I change one of the collector resistors to 110 as I suggested before, it flips to the stable state in less than 1 usec. With 2N2222 it flips in 300 usec with equal collector resistors. I cannot make it stay in a non stable state. I don't know why it suddenly flips.
 

crutschow

Joined Mar 14, 2008
33,369
When I simulate with LTSPICE, the original circuit flips to the stable state after 600 usec. I
My sim doesn't do that.
What exactly, is your original circuit?

Interestingly, I found that even a capacitor as small as 0.1fF will cause the circuit to go to a normal stable state.
 
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crutschow

Joined Mar 14, 2008
33,369
Both transistors are in the active range because the resistor values are incorrect for the transistor types.
How so?
The base current is approximately 1/10 of the collector current which is the usual for saturated switching.

They are both in the active region because of a quirk in how Spice determines the initial circuit state, which gives a quasi-stable condition here.
 
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BobTPH

Joined Jun 5, 2013
8,149
Maybe I run my LTSPICE in a noisier environment and it is picking up interference.


I am using Bordodynov’s library.

Using version XVII, updating now.
 
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