Basic circuit problem about BJT and buzzer

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  • #1
ZoeDale
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New user has been reminded to always show their work on schoolwork problems.
Homework Statement
Find the time when the the buzzer first emits in the circuit below.
Relevant Equations
Noton theorem;
1701560257581.png

I don’t understand what does the v_t mean?
 
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  • #2
Vt is the threshold voltage of the MOSFET. When the gate to source voltage (same as the capacitor voltage) is greater than Vt, the transistor will conduct current from the drain to the source. If it's less than Vt the transistor is off.
 
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  • #3
Welcome to PF.

R1 = R2 = 20 kW ; kilowatt ?
Maybe those should be 20 k ohms = 20k
 
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  • #4
For your example, that 20kW should be either 20kΩ or 20,000Ω

Ω is the uppercase Greek letter Omega, which is used to indicates Ohms; lower case Ω is ω.

It looks like whoever did the typesetting for that book got rather confused between Greek, Latin, uppercase, lowercase! :rolleyes:

Oh well, at least you learned a bit of trivia. :oldwink:

Cheers,
Tom
 
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  • #5
Baluncore said:
Welcome to PF.

R1 = R2 = 20 kW ; kilowatt ?
Maybe those should be 20 k ohms = 20k
Yes, I guess there is a type
 
  • #6
DaveE said:
Vt is the threshold voltage of the MOSFET. When the gate to source voltage (same as the capacitor voltage) is greater than Vt, the transistor will conduct current from the drain to the source. If it's less than Vt the transistor is off.
Got it! So until the capacitor is charged to Vt, the transistor is off, which means I just analyze the circuit in which R1 and (R2//C) are in series. Now I guess I know how to solve it. Thanks much!
 
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  • #7
ZoeDale said:
.... which means I just analyze the circuit in which R1 and (R2//C) are in series.
Another view of the problem.
( R1 // R2 ) = Rt, the Thevenin resistance at Vt.
Rt is in series with C1.
 
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  • #8
Baluncore said:
Another view of the problem.
( R1 // R2 ) = Rt, the Thevenin resistance at Vt.
Rt is in series with C1.
Yes, the easy way, LOL.
It also requires that you find the equivalent (Thevenin) voltage source. 5V in this case.

Most simple transient problems can be solved with just knowing the initial state, the final state, and the time constant. But they don't really teach that most places.
 
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  • #9
DaveE said:
Yes, the easy way, LOL.
It also requires that you find the equivalent (Thevenin) voltage source. 5V in this case.

Most simple transient problems can be solved with just knowing the initial state, the final state, and the time constant. But they don't really teach that most places.
Oh!!! So the time constant is decided by the equal circuit of Rt in series with C? So time constant \tao = Rt* C = 10k * 100m = 1000 ???!!! I need to go back to learn how I can derive that "( R1 // R2 ) = Rt, the Thevenin resistance at Vt." Thanks, Baluncore and Dave!
 
  • #10
ZoeDale said:
Yes, I guess there is a type
typo
 
  • #11
DaveE said:
Yes, the easy way, LOL.
It also requires that you find the equivalent (Thevenin) voltage source. 5V in this case.

Most simple transient problems can be solved with just knowing the initial state, the final state, and the time constant. But they don't really teach that most places.
Oh!!! So the time constant is decided by the equal circuit of Rt in series with C? So time constant \tao = Rt* C = 10k * 100m = 1000 ???!!! I need to go back to learn how I can derive that "( R1 // R2 ) = Rt, the Thevenin resistance at Vt." Thanks Baluncore and Dave!
Baluncore said:
Another view of the problem.
( R1 // R2 ) = Rt, the Thevenin resistance at Vt.
Rt is in series with C1.
hi Baluncore, I didn’t get it how to derive the equivalent circuit of (R1//R2) in series with C, can you please tell me more about it? Or if it is convenient, can you show me the equivalent circuit? Thanks much!
 
  • #12
You'll want to search for "Thevenin's Theorem" and "Thevenin and Norton source transformation" to learn more about these analysis techniques. This is the easy way to solve simple networks, IMO.

I think some offline study is your next best step. We can't really teach this with short comments. However, it's not difficult once you are familiar with KVL/KCL analysis.
 
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  • #13
Tangential, I must hope that is a benign 'solid-state' sounder rather than a 'trad' make-break buzzer. Latter would rapidly kill the transistor as-is. Would need sundry spike-quenching components, as if for a small brush-commutated motor. eg protective diodes, 0.1 uf capacitors, back-to-back zeners etc etc...
 

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