hfe of a npn bjt transistor is given by collector current / base current so to calculate hfe you need to know the collector current and base current (which can be set using resistors) . I have to ask , what is the point of doing this ? You can simply calculate the emitter current using Kirchhoff current rule . This is obviously pointless since the transistor acts like a resistor between collector and emitter and unless you know the resistance of the resistor you can't find the current .
I see in transistor datasheets a maximum and minimum value of hfe . This leads me to believe that hfe is a constant and I do not undersand the point of a varying constant ; I think for a single transistor only a single value of hfe exists which must be measured using a multimeter and the collector current is dependent of the base current so the collector current can be measured .
And also , if the collector current is entirely dependent on the base current (Ic = Ib*hfe) then what is the point of the adding a resistor to the collector end of the transistor ? Surely there must be some change in current if a greater resistance is added to the collector end . I checked this with a multimeter with base resistance of 10k ohms and collector resistance of 330 ohms and I found no change in current when I connect the multimeter in common emitter topology (I think , I connect the emitter to ground and tested the current flowing through the collector )
I think there is a big hole in my understand or a big misunderstanding in my not understanding . Please help
Thank you
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\$\begingroup\$ Each transistor has a different beta/hfe. It is not specific to the model, but also to each specific piece, as well as to the conditions of the transistor! The constant is very volatile, so we try NOT to depend on it. \$\endgroup\$– OFRBGAug 3, 2015 at 1:49
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\$\begingroup\$ A.Ganesh, yes, there is a "hole" in your understanding. Perhaps it helps to follow the discussion in electronics.stackexchange.com/questions/182192/… \$\endgroup\$– LvWAug 3, 2015 at 7:23
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\$\begingroup\$ @OFRBG Then what's the point ? isn't a factor of 100 - 300 important and circuit changing , won't an increase by that factor fry your components ? \$\endgroup\$– VriskAug 3, 2015 at 12:29
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\$\begingroup\$ @Anirudh - A good engineer will design the circuit to accommodate those beta variations and yet keep the circuit working to specifications. There are many circuit configurations that will work but in general you NEVER just bias a BJT with a constant base current. The base current is just an annoying side-effect of transistor operation and is not intrinsic to the functionality. \$\endgroup\$– Kevin WhiteAug 3, 2015 at 16:33
2 Answers
"... since the transistor acts like a resistor between collector and emitter ... "
No, not really. The collector of a bipolar transistor acts like a current source (or sink) whose value is determined by the base current and the hFE of the device. However, the external circuit can limit the current to something less than this value, in which case the effective hFE is lower.
"I see in transistor datasheets a maximum and minimum value of hfe."
Yes. The actual value varies considerably from device to device, even from the same manufacturing batch, and it also varies somewhat with the operating parameters (voltage, temperature, etc.) of the device. You really can't depend on having a particular (or even a constant) value, so you design your circuits so that they work over a range of values.
"... then what is the point of the adding a resistor to the collector end of the transistor?"
This is part of the circuit design. When you're creating a voltage amplifier, you use the collector current of the transistor to develop the desired voltage across the external resistor. This resistor is called the "load resistor", and it gives you a definite value of output impedance — the transistor by itself has a very high effective output impedance.
Example:
collector emitter voltage is 9 v ,Hfe = 100 , base emitter voltage is 9 , a resistor at the collector has resistance of 330 ohms and one at the base has resistance 10k ohms, tell me the current at the collector with steps.
OK, assuming you mean that 9V is applied to the base through a 10K resistor, 9V is applied to the collector through a 330Ω resistor, and that the emitter is grounded, the steps are as follows:
The base current is \$I_B = \frac{V_{BB} - V_{BE}}{R_B} = \frac{9.00 V - 0.65 V}{10k \Omega} = 0.835 mA\$
Assuming the transistor is not saturated, the collector current \$I_C = h_{FE} \cdot I_B = 100 \cdot 0.835 mA = 83.5 mA\$
The voltage across the collecor resistor should be \$I_C \cdot R_C = 83.5 mA \cdot 330 \Omega = 27.5 V\$
Since that value is higher than our supply voltage, the assumption made in the second step must be false — the transistor is saturated. Therefore, the collector current is determined entirely by the collector resistor and the collector supply voltage: \$I_C = \frac{V_{CC} - V_{CE(SAT)}}{R_C} = \frac{9.00 V - 0.3 V}{330 \Omega} = 26.4 mA\$
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\$\begingroup\$ "...whose value is determined by the base current"...Dave Tweed, did you follow the discussion in the following link? electronics.stackexchange.com/questions/182192/… \$\endgroup\$– LvWAug 3, 2015 at 7:19
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\$\begingroup\$ @LvW: Yes, but this OP is not looking for that level of detail. He's asking specifically about \$h_{FE}\$, AKA "current transfer ratio", as given in a transistor datasheet. \$\endgroup\$ Aug 3, 2015 at 10:45
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\$\begingroup\$ Didn't the scientists at Bell lab call first call the transistor the transformative variable resistor and shorten it to transistor ? Besides , if the collector current varies depending on the base current , can't we think of the transistor as a resistor that changes value depending on the base current ? What's the difference ? I think an example should dispel the confusion , collector emitter voltage is 9 v ,Hfe = 100 , base emitter voltage is 9 , a resistor at the collector has resistance of 330 ohms and one at the base has resistance 10k ohms, tell me the current at the collector with steps \$\endgroup\$– VriskAug 3, 2015 at 12:26
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\$\begingroup\$ "can't we think of the transistor as a resistor that changes value depending on the base current?" No. A resistor is a simple two-terminal device whose current depends only on the voltage across its terminals. There are no conditions under which the collector-emitter pair of a BJT works in this way. \$\endgroup\$ Aug 3, 2015 at 13:48
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\$\begingroup\$ "... base emitter voltage is 9, a resistor ... at the base has resistance 10k ohms ..." I assume you mean that the 9V is applied to the end of the 10K resistor, not directly to the base? \$\endgroup\$ Aug 3, 2015 at 13:49
This leads me to believe that hfe is a constant and I do not understand the point of a varying constant ; I think for a single transistor only a single value of hfe exists
Hfe varies between individual transistors because they cannot all be made identical. Often the manufacturer will make a batch of transistors and then sort them according to gain, giving each group a different suffix or even a completely different code. A single transistor's Hfe will be 'constant' somewhere in the specified range, but also only under certain conditions. It reduces at very low or high current and when collector voltage is low, and may vary quite strongly with temperature.
To get a true picture you should examine the curve traces for Hfe vs Ic and Ic vs Vce. In the example below you can see that Hfe is never truly constant, but for a BC107 is relatively flat in the 1-10mA range. The groups 'A', 'B', and 'C' are for the 3 different gain suffixes. Any individual transistor will have a similar curve, closer to its group than the others.
In the right-hand graph you can see that collector current is almost independent of voltage, except at very low voltage when the transistor is saturated.
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\$\begingroup\$ +1. Bonus points for posting BC107's datasheet excerpts. Brings tears to the eye! :-) 20+ years ago, together with BC109 and BC177, it was the workhorse of the electronics class I attended at uni (and it was still on the way of the dodo back then). That's vintage! :-) \$\endgroup\$ Aug 3, 2015 at 5:36
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\$\begingroup\$ I see . I notice in the first graph that the collector current is the dependent variable . So Hfe depends on the collector current and not the base current ; ie the base current depends of collector current ? I thought that the transistor acts like a variable resistor (I'm quite sure I read somewhere that the scientists of Bell labs first called it Transformative variable resistor which was then shorted to transistor ) between collector and emitter but here it seems like it's a variable resistor between the base and the emitter . <br> I have to ask , What is the point of Hfe , practically? \$\endgroup\$– VriskAug 3, 2015 at 12:02
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\$\begingroup\$ Regarding the term "transistor". It is an abbreviation for "transfer resistor" which means: INPUT voltage determines OUTPUT current. (Normal resistor: Voltage across R determines current through R). Insofar it is NOT something like a classical resistor. As a consequence, the input-output relationship is described by the "mutual transconductance" (inverse transresistance) dIc/dVbe=gm. \$\endgroup\$– LvWAug 3, 2015 at 14:42
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\$\begingroup\$ A bipolar transistor is essentially a current amplifier, and Hfe (Ic/Ib) is the gain (Ic = Ib * Hfe). The point of the graphs is to show that 1. this relationship is not perfectly linear, and 2. it is insensitive to Collector voltage (or Base voltage, which is not shown because it has no effect). This behavior is nothing like a resistor, so don't think of a bipolar transistor as being like a variable resistor - it isn't! \$\endgroup\$ Aug 3, 2015 at 18:47