Hi everyone...
I'm trying to analyse a simple circuit with an LDR but replacing the LDR by 2 resistance values to simulate the bright light and the absence of light!
We have the following:
Vcc = 12V
Rb1 = 10KΩ
Rb2 = 400Ω or 1MΩ (from the datasheet of an NORP-12 LDR)
BJT -> BS170, hFE = 200
VLed = 1.8V
Rc = ??? for Ic max = 20mA
I've calculated Rc for Ic max = 20mA and I got Rc = 510Ω by means of Vce = 0.
So:
12 = 20*Rc + 1.8 + Vce
Rc = 510Ω
Then I've simplified the circuit by means of replacing Rb1 and Rb2 for their equivalent as:
For Rb2 = 1MΩ
Rb = Rb1*Rb2/(Rb1 + Rb2) = 9.901kΩ
and
Vbb = 1MΩ*12/1,01MΩ = 11.88V
From here I can check if the transistor is on the active region or not by checking if Ib = Ic/hFE, which is not because if I assume that the transistor is on the active region, Vbe ≈ 0.66V and Ib = (Vbb - Vbe)/Rb = 1.13mA. But Ib = Ic/hFE = 20/200 = 100μA.
So I think I can assume that the transistor is on the ohmic/triode region and use the net equations to find Vbe... Am I correct???
If so, I could try to do the following:
-Vbb + Rb*Ib + Vbe = 0
Vbe = Vbb-Rb*Ib
How am I going to find Ib and Vbe if there is no relationship between Ib and Ic and Vbe may not be ≈0.66V????
LTSpice confirms my Ib = 1.13mA and Ic = 20mA...
I'm trying to analyse a simple circuit with an LDR but replacing the LDR by 2 resistance values to simulate the bright light and the absence of light!
We have the following:
Vcc = 12V
Rb1 = 10KΩ
Rb2 = 400Ω or 1MΩ (from the datasheet of an NORP-12 LDR)
BJT -> BS170, hFE = 200
VLed = 1.8V
Rc = ??? for Ic max = 20mA
I've calculated Rc for Ic max = 20mA and I got Rc = 510Ω by means of Vce = 0.
So:
12 = 20*Rc + 1.8 + Vce
Rc = 510Ω
Then I've simplified the circuit by means of replacing Rb1 and Rb2 for their equivalent as:
For Rb2 = 1MΩ
Rb = Rb1*Rb2/(Rb1 + Rb2) = 9.901kΩ
and
Vbb = 1MΩ*12/1,01MΩ = 11.88V
From here I can check if the transistor is on the active region or not by checking if Ib = Ic/hFE, which is not because if I assume that the transistor is on the active region, Vbe ≈ 0.66V and Ib = (Vbb - Vbe)/Rb = 1.13mA. But Ib = Ic/hFE = 20/200 = 100μA.
So I think I can assume that the transistor is on the ohmic/triode region and use the net equations to find Vbe... Am I correct???
If so, I could try to do the following:
-Vbb + Rb*Ib + Vbe = 0
Vbe = Vbb-Rb*Ib
How am I going to find Ib and Vbe if there is no relationship between Ib and Ic and Vbe may not be ≈0.66V????
LTSpice confirms my Ib = 1.13mA and Ic = 20mA...
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