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How measure hie of a bipolar transistor?

tsopa

New Member
I d like to design an amplifier with a transistor. For that I need to know the hie of the transistor but this parameter was not written in the datasheet. Can anyone tell me how to measure the hie of a transistor?
 

duffy

Well-Known Member
The hie parameter is the change in base-emitter voltage over the change in base current for a constant collector-emitter drop voltage. You sure you don't mean hfe?
 

tsopa

New Member
Thanks for the reply Duffy,
I am looking for hie because the voltage gain of a common-emmiter circuit is given in function of the load resistance, RC, hfe and hie (http://www.electro-tech-online.com/custompdfs/2012/06/transistor_amplifier.pdf) and hfe is already written the datasheet but not hie (I wonder why??).
So, if I understand you, hie can be measured by calculating the the change in base-emitter voltage over the change in base current [LATEX] dV_{BE}/dI_{B} [/LATEX] with [LATEX] V_{CE} [/LATEX] constant. However, if I change [LATEX] I_{B} [/LATEX] then the collector current [LATEX] I_{C} = h_{fe} I_{B} [/LATEX] would change and I've read in an article that [LATEX] h_{ie} [/LATEX] changes in function of [LATEX] I_C {/LATEX}.
 
Hi tsopa,
perhaps it is helpful to use words instead of symbols only.
Therefore: hie (some call it h11) is the so called "short circuit input resistance". Here, short circuit means "signal short" (Vce=const).
This parameter hie is in direct relation to the dc bias base current Ib and, thus, to the dc collector current Ic.
Because of the known relation between Ic and Ib and because of Shockleys equation (resulting in the transconductance g=Ic/Vt, Vt=thermal voltage)
you can measure Ic, calculate the transconductance g and than use the equation hie=hfe/g.
 
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duffy

Well-Known Member
tsopa: That .pdf you posted seems to have been written by someone who never actually worked with a transistor before in his or her life. The hie of a transistor is never anywhere near the 3k, 1k or 1.4k values given in the examples.

These values are ABSURD.

In reality, it is a fraction of an ohm. It is negligible compared to the impedance shown through the base by the product of the emitter resistor multiplied by hfe, and negligible in comparison to the value of the smallest base limiting resistor you will probably ever use. That's why it's not in the spec sheet - it is so trivially low it can safely be ignored.
 
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tsopa: That .pdf you posted seems to have been written by someone who never actually worked with a transistor before in his or her life.
The hie of a transistor is never anywhere near the 3k, 1k or 1.4k values given in the examples.
These values are ABSURD.
In reality, it is a fraction of an ohm, so low it is negligible. It is a miniscule fraction of the impedance shown by the emitter resistor multiplied through the hfe of the transistor, or even in comparison to the value of the base limiting resistor. That's why it's not in the spec sheet - it is so trivially low it can safely be ignored.
Duffy, are you really sure about that?
Please, follow my recommendation and realize the physical meaning of hie. As indicated in my posting#4 it is the input resistance at the base node.
And in this case, values in the kohm range are normal!
 

duffy

Well-Known Member
The base-emitter drop is never much more than .7V. If it was anywhere NEAR those values it would be much higher.
 
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duffy

Well-Known Member
Guys, seriously -

1. Go get an NPN transistor.
2. Connect a 1k base resistor.
3. Connect a power supply to the base resistor and emitter.
4. Check the base-emitter drop while you dial the voltage up and down.

If it was anything like 1k, then the equation shown in Eric's ref as Zb = hie + (1 + hfe) RE will give you something higher than the .7V drop. You won't see that. You will see about .7V.
 
Duffy, do you know what you are talking about? We speak about the signal input resistance between base and emitter.
Each textbook will tell you (and explain) why the value - depending on the chosen dc collector current - normally is in the lower kohm range.

When I am not absolutely sure I would be very cautios with statements like yours:

That .pdf you posted seems to have been written by someone who never actually worked with a transistor before in his or her life. The hie of a transistor is never anywhere near the 3k, 1k or 1.4k values given in the examples. These values are ABSURD.

Additional remarks:
For your information:
The formula Zb = hie + (1 + hfe) RE gives you the increased input resistance at base node for a BJT with negative feedback due to an emitter resistor RE.
 
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duffy

Well-Known Member
The formula Zb = hie + (1 + hfe) RE gives you the increased input resistance at the base node for a BJT with negative feedback due to an emitter resistor RE.
Section 3.3 of tsopa's pdf file reference shows an hie values lower than the emitter resistor itself. Trying to include RE (not an intrinsic part of the transistor's h parameters) won't even excuse this.

Go hook up a transistor like I said. See if you can measure anything like 1k in the B-E junction.
 
Section 3.3 of tsopa's pdf file reference shows an hie values lower than the emitter resistor itself. Trying to include RE (not an intrinsic part of the transistor's h parameters) won't even excuse this.
Go hook up a transistor like I said. See if you can measure anything like 1k in the B-E junction.
I cannot open this file - however, I am not interested to read the contents.
Tsopa has placed a question and I have tried to give an answer (thereby correcting your false and misleading reply). That's all.
When you believe, all textbooks are in error - I cannot help you.
Perhaps one additional hint: The parameter hie is identical to the inverse slope of the input voltage-to-current characteristics of the BJT. It is like a diode. Does this help?
 
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ericgibbs

Well-Known Member
Most Helpful Member
hi duffy,

The Zb = hie + (1 + hfe) RE applies to a 'connected working' transistor with emitter negative feedback, not the inherent 'hie' ~1K

E.
 

unclejed613

Well-Known Member
Most Helpful Member
with a known current, and by measuring the voltage drop you can figure it out. so build a variable current source and apply it to the junction. then measure the voltage drop and do the math. because the junction has a nonlinear characteristic, the Hie will change with the current. if you measure it at your expected bias current you will be pretty close to what it will average out to with signal applied.

i was reading a book on audio power amplifier design a few years back, and something that stuck with me is that a well designed amplifier makes use of semiconductor physics for it's operation, and minimizes the effects of individual device characteristics. that works ok for an op amp or power amp, but with a single transistor gain stage, you are stuck with calculating the individual device characteristics...
 

The Electrician

Active Member
tsopa: That .pdf you posted seems to have been written by someone who never actually worked with a transistor before in his or her life. The hie of a transistor is never anywhere near the 3k, 1k or 1.4k values given in the examples.

These values are ABSURD.

In reality, it is a fraction of an ohm. It is negligible compared to the impedance shown through the base by the product of the emitter resistor multiplied by hfe, and negligible in comparison to the value of the smallest base limiting resistor you will probably ever use. That's why it's not in the spec sheet - it is so trivially low it can safely be ignored.
It is in many data sheets--for instance (see figure 13):

http://www.electro-tech-online.com/custompdfs/2012/06/2N3903-DPDF.pdf

hie is the small-signal resistance; it's the AC voltage divided by the AC current when the AC voltage applied to the base is small--a few millivolts for example.
 
i was reading a book on audio power amplifier design a few years back, and something that stuck with me is that a well designed amplifier makes use of semiconductor physics for it's operation, and minimizes the effects of individual device characteristics. that works ok for an op amp or power amp, but with a single transistor gain stage, you are stuck with calculating the individual device characteristics...
Hi unclejed613,
just a short remark regarding "individual device characteristics" (tolerances of BJT parameters):
I think, negative feedback ALWAYS reduces the influence of individual amplifier parameters on the gain value. This holds for opamp and also for single BJT stages. For example, this is the main task of the emitter resistor RE that exist in practically all BJT amplifiers. Example: When RE is not bypassed by a capacitor the gain of a common emitter stage is

G=-RC/(1/g+RE)

which can be approximated by the resistor ratio -RC/RE if RE>>1/g (g is the transconductance g=Ic/Vt) .
 
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audioguru

Well-Known Member
Most Helpful Member
hie is not used to calculate the voltage gain of a transistor. It is the input impedance. It is shown on a graph for Fairchild's 2N3904 little transistor and is about 3.8k ohms when the collector current is 1mA and is about 500 ohms when the collector current is 10mA. hie is affected by the amount of hfe and amount of base current of a transistor. Adding an unbypassed emitter resistor increases the input impedance.

The voltage gain is the collector resistance (in parallel with the load resistance) divided by the internal emitter resistance (plus emitter resistor value).
 

rumpfy

Active Member
Hi tsopa,
The hie of a transistor is the common emitter input impedance.
It is a small signal parameter and was of particular interest when the early germanium transistors were developed. Mathematically it is defined as ΔVbe/ΔIb. The parameter varies depending on the operating point. For an old germanium small signal audio frequency transistor type AC125 operating at a collector current of 2 mA and a Vce of 5 volt, hie is between 1 to 2.5 kohm. For a modern silicon type, hie is around 10 Kohm. hie is affected by the current gain of the device; as hFE (DC current gain) gets larger, the required base current for the same collector current becomes less, so the Δib part of the calculation becomes smaller and hence hie gets bigger.
hie is an AC or a signal value, not a DC value. To measure it, set up a transistor at some operating point. If you use an emitter resistor, then bypass the resistor with a large capacitor. Then supply the base circuit with a small 1000 Hz test signal through a series resistor (Rs). Measure the test signal voltage and the voltage across the base/emitter and use the two values to calculate hie.
ie, Vbe = V signal supply/ (Rs +hie). Note that your bias resistors will also affect the measurement so allow for this. The bias resistors will be in parallel with hie.
hope this helps.
 
The voltage gain is the collector resistance (in parallel with the load resistance) divided by the internal emitter resistance (plus emitter resistor value).
Hi Audioguru, don't you think the above formulation (...."internal emitter resistance") is a bit unclear?
I think, a formula can avoid confusions. Thus: "internal emitter resistance"=1/g (see the formula in post#17 ?
 

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