# Transistor Resistance

#### SamR

Joined Mar 19, 2019
4,122
Came across this interesting topic concerning transistor usage.

Using a transistor as a voltage divider! What a novel idea! So I built the circuit using my chinesium Decade Resistor Box. I got what I thought was a flyer or two "bad" readings so I hooked the box up to my Fluke bench meter and was amazed at how accurate it was. Less than 1% for the dozen or so test measurements I did with it. Here are the circuit results. Which give ~330mW @ 0ΩRc so slightly above the 1/4? W rating and adding 100Ω is also slightly over so maybe some heating effect going on. But as I increase by 100Ω steps the V/A does not increase by 100Ω? I know temperature affects transistors but what else is going on here? In particular the only 4.0Ω delta from 200-300Ω? Is this all temp effect or what? I just noticed I did not record the V I used for the V/A calculation but it is correct.

The first measurement @ 0Ω for the Rc is ~300Ω.

#### nsaspook

Joined Aug 27, 2009
9,444

#### SamR

Joined Mar 19, 2019
4,122
the resistance between two terminals, respectively collector and transmitter, is modified by changing the base voltage

But I didn't change the Base Voltage! All I did was change the value of Rc... I looked at the Base current and it was surprisingly much lower than its max value.

Edit: The measured hFE using the chinesium tester is 425, not the 100 as in the text.

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#### nsaspook

Joined Aug 27, 2009
9,444
the resistance between two terminals, respectively collector and transmitter, is modified by changing the base voltage

But I didn't change the Base Voltage! All I did was change the value of Rc... I looked at the Base current and it was surprisingly much lower than its max value.
Then the question is how, where and under what conditions is the bias voltage set.

#### BobaMosfet

Joined Jul 1, 2009
1,944
the resistance between two terminals, respectively collector and transmitter, is modified by changing the base voltage

But I didn't change the Base Voltage! All I did was change the value of Rc... I looked at the Base current and it was surprisingly much lower than its max value.
You can't take things in isolation- every component is working in unison, each affected by the other. The statement in the text about R2 while correct is misleading. The purpose of R2 is theoretically to act both as pull-up when the transistor is OFF, and as protection against a short to ground when it is ON. This is the entire concept behind 'open-collector'. The output from the collector is not driven by the BJT, but by the pull-up.

Consider this- any two components, irregardless of what they are form a voltage divider if you take a signal from between them.

#### SamR

Joined Mar 19, 2019
4,122
Not that I expected it to be linear but in particular the "dip" measuring deltaΩ of Vce/Ice between 200-300Ω Rc is strange. That and the "jump" in Ice around the 400-600Ω Rc.

#### Audioguru again

Joined Oct 21, 2019
4,524
Why do you ask on a forum instead of looking in the datasheet of a very old BC108B?
Its current gain is not 100, it is from 220 to 480 when its current is 10mA and its collector voltage is 5V.
It is guaranteed to saturate like a switch when its base current is 1/20th its collector current, current gain is not used for saturation because its gain drops as its collector voltage drops.

#### SamR

Joined Mar 19, 2019
4,122
Yes, I substituted a recommended transistor that I had in stock and I did look at its PDF. I saw the 2mA @ 5V x the hFE of 200-450 (measured 425) and with 10V and 100kΩ expected to see 42.5 mA and I measured 33.18mA. Ok, in the ballpark.

It is guaranteed to saturate like a switch when its base current is 1/20th its collector current,
K but this part I don't understand and no curves were shown on the PDF. Not sure how I would extrapolate this to cover what I was using.

I didn't measure the Base I or V but the 10V/100kΩ should be 0.1uA @ 10V ignoring any voltage division by the be junction and the transistor was apparently ON since I was getting Vce and Ice.
The datasheet gives a small signal Impedance @ 1kHz but no DC resistance. And, as I said, I didn't expect it to be linear but do not understand what appear to be anomalies in the results and cannot explain them. Needless to say that I am very low on the learning curve here as about all I've used transistors for in the past has been as switches.

#### BobaMosfet

Joined Jul 1, 2009
1,944
@SamR

May I recommend :

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

It has good coverage of BJTs in many configurations, as well as tests & answers, and might help you

And then of course:

The Art of Electronics 3rd Ed.
Author(s) Horowitz & Hill
ISBN-10: 9780521809269

They have a good discussion about BJTs and their transconductance covering exactly what you are dealing with.

#### michael8

Joined Jan 11, 2015
275
> The collector of a transistor is a constant current source ...

This is true under some conditions. Under normal bias conditions, not cutoff nor saturated, the collector current of a transistor depends on Vbe (voltage beteen it's base and emitter) and to a lesser sense on the specific transistor as well as it's temperature and Vce (voltage between the collector and emitter).

The circut in the original posting doesn't keep Vbe constant, it keeps Ibe (base current) constant which isn't the same.

Also the current gain Ic/Ib isn't a constant and isn't usually very well isn't very well specified -- the transistor data sheet will usually give a range or just a minimum.

It's Vbe which controls the transistor collector current.

Different transistors if not matched will have different Vbe for the same Ic. Two transistors which are identical with the same Vbe will have the same collector current.

Try this circuit, if you vary R4 (the collector resistor) you will find that the current through it remains about the same.

#### LvW

Joined Jun 13, 2013
1,428
Using a transistor as a voltage divider! What a novel idea!
Not so novel.....however, the voltage at the common node between both resistive parts (Rc and the C-E path, respectively) will practicaly not vary with a variation of the supply voltage Vcc (10V).

#### LvW

Joined Jun 13, 2013
1,428
Try this circuit, if you vary R4 (the collector resistor) you will find that the current through it remains about the same.
Yes - thats correct, of course.
Because the SamR did mention the use of such a circuir as "voltage divider":
Vary R3 - and you will observe a drastic change in the currents. So - R3 and R4 behave differently.

#### SamR

Joined Mar 19, 2019
4,122
Try this circuit,
Yes, I am familiar with that as a "standard" voltage divider (R1&R2 to set Vb) common emitter circuit. What I found novel was varying Rc to set the output voltage which hasn't been mention in the several BJT texts that I've used previously. Which while I didn't expect it to be linear was surprised at how it moved up and down as a function of V/I.

#### nsaspook

Joined Aug 27, 2009
9,444
Yes, I am familiar with that as a "standard" voltage divider (R1&R2 to set Vb) common emitter circuit. What I found novel was varying Rc to set the output voltage which hasn't been mention in the several BJT texts that I've used previously. Which while I didn't expect it to be linear was surprised at how it moved up and down as a function of V/I.
That effect (the field effects inside the transistor are linked) of causing variations in the base bias is why bias stabilization emitter feedback and voltage divider Vb circuits are used.

#### LvW

Joined Jun 13, 2013
1,428
What I found novel was varying Rc to set the output voltage which hasn't been mention in the several BJT texts that I've used previously.
I rather think that every time we are designing a common emitter stage we make - of course - use of this property (voltage division between Rc and the coleector-emitter path), dont we?
We select a suitable collector resistance which - together with the desired collector current Ic - allows a DC voltage at the collector node (operational point) that allows a sufficient signal voltage swing around this point.

#### michael8

Joined Jan 11, 2015
275
OK, try this, no emitter resistor. The current through R2 is still constant since Vbe of Q1 is constant. Ignoring the base current of Q1, Q2's current will be about 1mA (10V/10K) and it's Vbe will adjust to make it's collector current enough to hold it's Vbe
at the voltage needed for 1 mA of collector current. The same Vbe on Q1 will make it's collector current the same. This is called
a current mirror...

#### k1ng 1337

Joined Sep 11, 2020
420
Consider this- any two components, irregardless of what they are form a voltage divider if you take a signal from between them.

Is this always the case? If all atoms are electromagnetic and in motion it would seem your statement satisfies KVL and the amount of calculation involved to determine the atomic interactions in my house is mind boggling

#### Danko

Joined Nov 22, 2017
1,269
What I found novel was varying Rc to set the output voltage which hasn't been mention in the several BJT texts that I've used previously. Which while I didn't expect it to be linear was surprised at how it moved up and down as a function of V/I.
See, how your experiment looks in graphical form:

Resistance of R1→

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#### LvW

Joined Jun 13, 2013
1,428
Yes, I am familiar with that as a "standard" voltage divider (R1&R2 to set Vb) common emitter circuit. What I found novel was varying Rc to set the output voltage which hasn't been mention in the several BJT texts that I've used previously. Which while I didn't expect it to be linear was surprised at how it moved up and down as a function of V/I.
I think, up to now the following difference betwen the "classical" voltage divider" and the "transistor-based divider" (see the diagram in the 1st post) was not mentioned:
* There is a considerable difference between both as far as the divider function is concerned because in one case the driving voltage Vo is kept constant and in the other case we should assume a constant current (Ic, transistor case).

* Case 1: Two equal resistors R1=R and R2=R
Voltage across R1 is V1=Vo/2 (constant voltage Vo).
Variation of R1 using a tuning factor k does not influence the value of R2.
Voltage change across R1: V1k=Vo*k/(1+k)
Ratio V1k/V1=2k/(1+k).

* Case 2:
Again two equal resistive parts R1=R and R2=Rce=Vce/Ic=R
Voltage across R1 is V1=Ic*R1 (constant current).
Variation of R1 (tuning factor k) now changes also R2=Rce (Vce changes because of Vo=V1+V2=const)
Voltage change across R1: V1k=Ic*k*R1.
Ratio: V1k/V1=k.

* Evaluation: Variation of one resistor gives a divider function which is non-linear for two ohmic resistors and linear when one resistive part is realized as a transistor (current source).