Author Topic: Voltage divider biasing NPN transistor solution.  (Read 279 times)

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Offline leftek

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Voltage divider biasing NPN transistor solution.
« on: October 16, 2018, 02:41:46 am »
Hello,
I would like to solve one exercise on the schematic in attached file.
The request is what is the value of Vcollector if the resistor R2 is OPEN.
I have make this in simulator and the values are as showing. But i can't find HOW i can calculate these if the R2 resistor in open.
Can anybody help me? 
« Last Edit: October 16, 2018, 02:57:54 am by leftek »
 

Online Hero999

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Re: Voltage divider biasing NPN transistor solution.
« Reply #1 on: October 16, 2018, 03:35:38 am »
The impedance seen at the base is a diode in series with R3*Hfe, which will be very high, compared to R1, so can be ignored.

The transistor will just turn on as much as it can, giving a saturation voltage of a couple of hundred mV. The voltages across R3 and R4 can be calculated using Ohm's law.

I = (V1-VSAT)/R3+R4)
V = I*R

Where VSAT is the saturation voltage of the transistor, which will be around 200mV. If you're unsure, presume it to be zero, calculate I, then look up the value on the data sheet and recalculate. This will give as gooder result as you're going to get, as the saturation voltage isn't 100% predictable and can vary from device to device.
 

Offline Kevin.D

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Re: Voltage divider biasing NPN transistor solution.
« Reply #2 on: October 16, 2018, 05:16:29 am »
The impedance seen at the base is a diode in series with R3*Hfe, which will be very high, compared to R1, so can be ignored.


I think he wanted to take into account the extra base current . Thus when operating in the  saturation region  there is insufficient Ic to cutoff the base emitter diode then the b-e diode now just looks like a low impedance diode and more current will flow into the base until the b-e diode reaches cutoff (or the base drive can supply no more current).
In saturation that circuit can be roughly approximated (ignoring Vce(sat) and re') by this, then just apply superposition or your favorite method to  find the currents and voltages.

« Last Edit: October 16, 2018, 05:18:20 am by Kevin.D »
 

Online Hero999

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Re: Voltage divider biasing NPN transistor solution.
« Reply #3 on: October 16, 2018, 06:28:26 am »
The impedance seen at the base is a diode in series with R3*Hfe, which will be very high, compared to R1, so can be ignored.


I think he wanted to take into account the extra base current . Thus when operating in the  saturation region  there is insufficient Ic to cutoff the base emitter diode then the b-e diode now just looks like a low impedance diode and more current will flow into the base until the b-e diode reaches cutoff (or the base drive can supply no more current).
In saturation that circuit can be roughly approximated (ignoring Vce(sat) and re') by this, then just apply superposition or your favorite method to  find the currents and voltages.


Yes you're right. I missed that. Not only do you have the potential divider consisting of R3 and R4, but there's also R3 in series with a diode, in parallel with R3.
 

Offline leftek

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Re: Voltage divider biasing NPN transistor solution.
« Reply #4 on: October 16, 2018, 03:46:25 pm »
Thank you very much. You helped me to become understood.
 


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