Floating voltage points

[teknora]

I am attaching 3 figures from a well known book (Electronic Devices by Thomas L. Floyd). How does a multimeter measure µV and mV on Figures 4-39 a and f? How does this part of the transistor behave as a floating voltage point? And also how does it measure 3V on the base side and 2.5V or more on the ground side of the transistor on Fig. 4-44?

Floating voltage points.pdf

[studiot]

With reference to my attachment showing comments.

 

A modern voltmeter for this purpose is likely to have an input around 10 megohms.

This is in series with the meter.

I have shown this in my fig1.

 

My fig2 shows how this voltmeter is connected in your circuit 4-39a and identifies the floating voltage pont B at the base of the transistor.

 

If you look at my version you can see that the base is not connected to any voltage but is earthed (connected to zero volts) through the 10M of the voltmeter.

In theory the voltage at the base is indeterminate since it is not connected to anything. (In theory it is totally isolated from the base and emitter by the internal structure of the transistor.

In practice the leads and internal circuitry of the voltmeter form a crude antenna which picks up stray (mostly radio) voltages across its terminals.

You can see this effect on the voltmeter by simply turning it onto a low micro or millivolt range and watching the random readings.

As soon as the voltmeter is connected into a complete circuit this effect will disappear or be swamped by the usual action of the live circuit.

However in the connection shown it is not connected to a complete circuit since the base is not connected to anything, except the transistor base.

One of the voltmeter's terminals is tied to earth, so the random pickup voltage will appear at the other one ie the one connected to the transistor base, thus varying the apparent voltage at the base.

Note this voltage is only likely to be present when the voltmeter is connected since the base leads and circuit board pads are short.

 

My fig3 shows the voltmeter connected to read the open emitter voltage as in your fig 4-39f

 

Here we have a different situation since with the voltmeter connected the transistor has full connectivity, albeit through an excessively large emitter resistor.

This 10M emitter resistor restricts the collector current to slightly under 1 microamp, bu that is enough to almost turn on the transistor as we see that the base is at 3 volts and the emitter at 2.5 volts + , just slight under the normal 0.6 to 0.7 base emitter voltage for a correctly biased 2N3904 in the active region.

 

I have noted that the book explanation says this rather more briefly.

 

Edited December 7, 2015 by studiot

[Danijel Gorupec]

Note this voltage is only likely to be present when the voltmeter is connected since the base leads and circuit board pads are short.

 

We could debate this. It might actually be higher than with the connected voltmeter (but certainly below some 0.5V). In any case it is quite floating and so who cares...

 

...

 

Regarding the picture 4-44 in the OP, I think the book might be partially in error. Depending on the voltage supplied at the transistor base, the voltage at the collector lead might be higher than few mV. It is actually similar to the 4-39f case.

[studiot]

 

We could debate this. It might actually be higher than with the connected voltmeter (but certainly below some 0.5V). In any case it is quite floating and so who cares...

 

You are welcome to explain your thoughts and propose a mechanism.

[Danijel Gorupec]

I suppose it depends on the doping levels in the transistor... The collector-base junction is reverse biased, but it still leeks some tiny current (in a diode case that would be called 'reverse saturation current'). This current then goes through the base-emiter junction toward GND... The base-emiter junction voltage rises up slightly so that the system reaches equilibrium (reverse saturation current of the collector-base junction equals to the forward current of the base-emiter junction). The voltage needed is small, but I suspect that in some cases (transistor doping levels) it might be above few millivolts.

[studiot]

Indeed there will be a small leakage, which is why I pointed out that the 2N3904 is a modern transistor. By modern I mean it was introduced after we learned to make near ideal small signal transistors.

My tables give the leakage at less than 15 nanoamps. Now the intrinsic emitter resistance is less than 100 ohms so the voltage developed by this mechanism, iw 15nA passing through 100 ohms is 15 x10^-9 x 10² = 1.5 microvolts.

[teknora]

 

I am attaching 3 figures from a well known book (Electronic Devices by Thomas L. Floyd). How does a multimeter measure µV and mV on Figures 4-39 a and 4-44? How does this part of the transistor behave as a floating voltage point? And also how does it measure 3V on the base side and 2.5V or more on the ground side of the transistor on Fig. 4-39f?

 

I am mistaken. I have corrected the original post.

Edited December 7, 2015 by teknora

[studiot]

Well if my explanation wasn't enough, perhaps you would be good enough to say why?

[Danijel Gorupec]

But I don't think that the base resistance is the main component that generates the base voltage... In addition to base resistance, the 15nA current must also pass through p-n junction (base-emitter junction). The U/I curve of this p-n junction is exponential, and some "considerable" voltage difference is needed for the p-n junction to pass 15nA.

 

Now, I didn't find any base-emiter voltage/current diagram for any real transistor. But for a real diode a very rough calculation says that it will take about 25mV of forward voltage to pass such currents. Okay, I admit that this can still be called 'a few millivolts'.

 

(Edit: this was my answer to Studiot's post #6)

Edited December 7, 2015 by Danijel Gorupec

[studiot]

What you have found is the Shockley Equation for the dynamic (ac) resistance of the junction.

 

The Schockley equation says the dynamic resistance = 25mv/I_E.

Since there is no ac signal we want the static or dc resistance.

 

I will dig out a 2N3904 tomorrow and post some real measurements (not SPICE sims that will depend on the model and is probably outside the range of the formulae)

[teknora]

Thank you very much Studiot for theoretical and practical explanations together with nice illustrations. I also thank Danijel for treating the question from other aspects.

To Studiot:

Why does the author call “floating point” specifically only the cases of Figs 4-49a and 4-44? As far as I understand, the ground is floating ground ,which isn’t connected to Earth. In this case, aren’t all voltages measured floating, Vcc or Vbb?

In your statement “In theory the voltage at the base is indeterminate since it is not connected to anything. (In theory it is totally isolated from the base and emitter by the internal structure of the transistor.)” Shouldn’t it be “… from the collector and emitter…”?

So, based on this explanation of yours, in the non-conducting state of the transistor, I understand that all the leads can approximately be considered as separated parts not touching each other. Am I right?

I would like to ask a general question: If we consider a transistor as being consisted of internal resistors (internal resistances between collector, base and emitter), would the voltages measured be like those in true resistors? For example, would the measured voltages on points between collector and base be between 9V and 3V?

To Danijel: In your statements:

“We could debate this. It might actually be higher than with the connected voltmeter (but certainly below some 0.5V). In any case it is quite floating and so who cares...”, what do you mean by some 0.5V? Also, what do you mean by “In any case it is quite floating and so who cares...”?

“Regarding the picture 4-44 in the OP, I think the book might be partially in error. Depending on the voltage supplied at the transistor base, the voltage at the collector lead might be higher than few mV. It is actually similar to the 4-39f case.” In this case, the collector lead is also grounded like the emitter lead by multimeter ground. Is this is the reason why more than a few mV?

“…The base-emiter junction voltage rises up slightly so that the system reaches equilibrium…” Could you clear up this for me please?

[studiot]

OK, I said I would post some measurements today.

 

Here is the overall setup rig.

Note all the white leads connected to the negative terminal of the 9volt battery.

This is my earth. I will return to this to answer your question about this later.

 

 

To start with let us set up the circuit with no disconnections.

 

 

The measurements show the collector at 1.5 volts and the base at 0.7 volts, as expected.

The transistor has a measured gain of 310 on the meter in transistor test mode.

 

So the current in the base resistor is (3-0.7)/56000 or 41 micro amps

Similarly the current in the collector resistor is (9-1.5)/560 or 13 milliamps.

So the gain is 317 as confirmed by the meter in transistor test mode.

 

 

Here we can see the transistor connected to the 9 volt battery but the base is not yet connected to the 3 volt supply.

The voltmeter is reading the the 3 volts supply (2.77 since the batteries are weak)

 

 

As soon as we connect the base to the 3volt supply through the 56k resistor the transistor turns on and the base emitter voltage drops to 0.72 volts

 

 

If we remove the 3volt supply from the base resistor the transistor now turns off and the base voltage drops to 1.5 millivolts as described in 4-39a of your post.

Note the range switch on the meter is now set to millivolts, not volts.

 

This is important because there is no longer a forward biased junction between the base and emitter, just the intrinsic resistance of the silicon between the two.

 

 

Moving on to the second situation, 4-39f.

The base drive has been reconnected but and the emitter is connected to ground.

The meter is now monitoring the emitter voltage.

It is reading almost zero. But not quite.

This is typical with the 'rats nest' lashup shown.

 

 

The emitter is now disconnected from ground and its voltage rises to 2.43 as described in your textbook.

 

 

To return to the issue of 'floating' look again at the first picture and note that allthroughout the tests I am using the negative terminal of the battery as a reference.

That is what earth (british) or ground (american) is.

A full discussion of what an earth is (most get this wrong) really deserves its own thread.

Using the negative terminal as a reference means that the common of the voltmeter is always conencted to this point, although the other terminal connection is moved about the circuit to suit the required measurement in the pics.

When the base is connected to the 3volt supply that supply (in combination with transistor action) is forcing it to adopt certain voltages as measured.

If the base is not connected to anything it is free to adopt any voltage (or no voltage) it latches onto. This is the floating condition.

 

As regards your second question,

 

Yes you can consider a transistor as a Y connection of three resistors and apply Kirchoff's laws.

But at DC only.

 

As soon as transistor action is need to be modelled you must introduce an active source and either use the Pi or T models.

including a controlled current or controlled voltage source.

The Pi and T models correspond to the Delta-Wye or Delta-Star transformations in circuit theory.

 

 

[Danijel Gorupec]

To Danijel: In your statements:

“We could debate this. It might actually be higher than with the connected voltmeter (but certainly below some 0.5V). In any case it is quite floating and so who cares...”, what do you mean by some 0.5V? Also, what do you mean by “In any case it is quite floating and so who cares...”?

In the NPN transistor case, the base voltage cannot be much higher than the emitter voltage. Higher the voltage difference, larger current will pass through the base-emitter junction. This dependance is exponential, and so once the voltage difference becomes higher than cca 0.7V the current might become so large that it might destroy the transistor. On the other hand, if the voltage difference is lower than cca 0.5V the current will be very small, too small for most practical purposes. That is why we say that transistor base-emitter voltage drop is about 0.6V-0.7V - it is because for all practical base currents, the base-emitter voltage drop is somewhere in this range.

 

As Studiot said, the leakage current from collector-base junction is only about 15nA - this is a seriously small current. That is why I am pretty sure that the base-emitter voltage drop that passes such small current is (much) below 0.5V.

 

I said "who cares" because the question - what is the base voltage when the base is non-connected - has no practical value. Once you connect anything (including a voltmeter) to the base, the voltage level will probably change significantly (that is why i said the voltage is 'quite floating').

 

“Regarding the picture 4-44 in the OP, I think the book might be partially in error. Depending on the voltage supplied at the transistor base, the voltage at the collector lead might be higher than few mV. It is actually similar to the 4-39f case.” In this case, the collector lead is also grounded like the emitter lead by multimeter ground. Is this is the reason why more than a few mV?

Yes. If, for example, the base voltage is at 0.7V, and you put a voltmeter between collector and ground, a small current will start to flow from base to collector to voltmeter and finally to the ground. This current will generate maybe 0.5V (just a guess) voltage drop in the base-collector junction and the voltmeter might show some 0.2V... However, making an experiment would be a better way to get the exact answer (as it always is).

[teknora]

With reference to my attachment showing comments.

 

A modern voltmeter for this purpose is likely to have an input around 10 megohms.

This is in series with the meter.

I have shown this in my fig1.

 

My fig2 shows how this voltmeter is connected in your circuit 4-39a and identifies the floating voltage pont B at the base of the transistor.

 

If you look at my version you can see that the base is not connected to any voltage but is earthed (connected to zero volts) through the 10M of the voltmeter.

In theory the voltage at the base is indeterminate since it is not connected to anything. (In theory it is totally isolated from the base and emitter by the internal structure of the transistor.

In practice the leads and internal circuitry of the voltmeter form a crude antenna which picks up stray (mostly radio) voltages across its terminals.

You can see this effect on the voltmeter by simply turning it onto a low micro or millivolt range and watching the random readings.

As soon as the voltmeter is connected into a complete circuit this effect will disappear or be swamped by the usual action of the live circuit.

However in the connection shown it is not connected to a complete circuit since the base is not connected to anything, except the transistor base.

One of the voltmeter's terminals is tied to earth, so the random pickup voltage will appear at the other one ie the one connected to the transistor base, thus varying the apparent voltage at the base.

Note this voltage is only likely to be present when the voltmeter is connected since the base leads and circuit board pads are short.

 

My fig3 shows the voltmeter connected to read the open emitter voltage as in your fig 4-39f

 

Here we have a different situation since with the voltmeter connected the transistor has full connectivity, albeit through an excessively large emitter resistor.

This 10M emitter resistor restricts the collector current to slightly under 1 microamp, bu that is enough to almost turn on the transistor as we see that the base is at 3 volts and the emitter at 2.5 volts + , just slight under the normal 0.6 to 0.7 base emitter voltage for a correctly biased 2N3904 in the active region.

 

I have noted that the book explanation says this rather more briefly.

 

voltest1.jpg

To Studiot:

 

Again, I have a question related to Figure 4-39a. No current through the transistor because the base is open. The collector side of the transistor is 9V, the emitter side of the transistor is 0V. The both sides of the resistor (Rc) are 9V. Both resistor and transistor are electronic components. But here the transistor behaves like an open (all voltage drops across it but no current). What are the physics in the transistor and resistor here. What makes them act so different? You have said that in DC voltages the transistor can be considered as an assembly of resistors. Also, why is there no voltage on the open base tip almost in the middle of the transistor? Think an open in a simple circuit with only a battery and a resistor. In the open still there is an electrical field. Because of this field, the is a voltage in the open. If we put a charge there, it will move because of that voltage. So, why is there no voltage on the open base almost in the middle of the transistor?

[studiot]

To Studiot:

 

Again, I have a question related to Figure 4-39a. No current through the transistor because the base is open. The collector side of the transistor is 9V, the emitter side of the transistor is 0V. The both sides of the resistor (Rc) are 9V. Both resistor and transistor are electronic components. But here the transistor behaves like an open (all voltage drops across it but no current). What are the physics in the transistor and resistor here. What makes them act so different? You have said that in DC voltages the transistor can be considered as an assembly of resistors. Also, why is there no voltage on the open base tip almost in the middle of the transistor? Think an open in a simple circuit with only a battery and a resistor. In the open still there is an electrical field. Because of this field, the is a voltage in the open. If we put a charge there, it will move because of that voltage. So, why is there no voltage on the open base almost in the middle of the transistor?

 

 

I thought this question was going to be more difficult than it turned out.

 

To understand the voltage distribution use the simplest model for a transistor.

 

This is a pair of 'back to back' diodes as shown in my sketch (correct for an NPN 2n3904 transistor).

 

This means that the pair block in both directions between collector and emitter when nothing is connected to the base.

 

So any current flowing will be the collector - emitter leakage with abse open circuit I_ceo

 

Now look first at the BE diode.

 

It is forward biased, that is it allows current to flow in the direction from the +9 volts to 0 volts.

An ideal diode has zero forward resistance so both terminals will be at the same voltage.

The actual diode will be very close to this so the emitter is set at zero so the base will be slightly more positve as observed.

 

Now look at the CB diode

 

This is reverse biased so blocks current.

The collector is set at +9 volts. This is the cathode of the upper diode

If this were an isolated diode you could tie the anode of this diode to any voltage and it would be blocked.

In this case the anode is also the base of the transistor which is set to near zero by the forward biased lower diode.

 

As regards to the electric field part of the question.

 

The electric field and potential in a resistor change (perhaps linearly) along the resistor from one end to the other.

Components acting like this are said to be acting in resistive mode.

 

The field and potential change abruptly across a reverse biased diode whose action is blocking.

This is like a capacitor and components acting like this are said to act capacitively or in capacitive mode.

 

It may be of interest to know that this ability of a reverse biased diode to act capacitively is exploited in so called varicaps or voltage dependent capacitors.

 

Edited January 28, 2016 by studiot

[teknora]

To Studiot:

 

Thank you very much for taking time and instructive explanations. I think you are the right person to ask. I want to learn very well the voltage distribution in circuits (either simple or complex). Could you give me a good elementary reference for this purpose.

Edited January 29, 2016 by teknora

[studiot]

To Studiot:

 

Thank you very much for taking time and instructive explanations. I think you are the right person to ask. I want to learn very well the voltage distribution in circuits (either simple or complex). Could you give me a good elementary reference for this purpose.

 

This is difficult to answer since I don't know where you are, what your interest in electronic/electric circuits or what you have access to.

 

Most of the books I would recommend are British, simply because most of my library is from British publishers, though I have some good American ones as well.

 

First to distinguish between electric circuits and electronic circuits.

 

The main applications of electricity are power (which includes domestic drills, heaters etc) and information/communications.

 

Electric circuits are more geared towards power engineering and most books with this title may or may not have a chapter or two on electronics.

 

Conversely Electronic circuits tell you about , well electronic circuits, have scanty basic information about power applications.

 

Both are intensely practical subjects so the next question is what do you want this for?

 

If you are following a recognised course towards professional application then you will need much more theoretical detail.

If this is an amateur interest then a more practical approach that offers the right amount of the right theory would suffice better.

 

For an amateur and even professional introduction to electronics with a wide ranging coverage the books by Michel Tooley and his Brother are brillaint.

They offer a great deal of the type of 'why are we doing this' explanation I put into my posts above.

 

'Electronic Circuits' was one of the originals but there have been several new ones and revisions.

 

http://www.amazon.co.uk/Electronic-Circuits-Fundamentals-Applications/dp/0750669233

 

Two other good books in the professional + amateur category are

 

A Practical Introduction to Electronic Circuits

 

Martin Hartley Jones

 

Electronics and Electronics Systems

 

George H Olsen

 

Before you pay for the oft recommended american

 

'The Art of Electronics' by Horowitz

 

Get it from the library and check it out. I cannot recommend it.

 

A brilliant american book is

 

Microprocessors and digital systems

 

Douglas V Hall

 

A good modern electrical circuits first serious book is

 

Electrical Circuit Analysis and Design

 

Noel M Morris

 

Finally try this site, they were a really good bunch before they were taken over by a commercial enterprise, so beware the advertising, but they have the internet rights to the famous e-textbook by Tony Kuphaldt.

 

allaboutcircuits.com

 

Also beware the original Kuphaldt book uses the 1990 fashion of electron flow as the positive direction for current flow which can cause confusion.

This is not a good idea and but was fashionable in the 1990s.

In fact always check the work of any author anywhere before reading as to whether he uses what is known as conventional current (postive in the direction positive to negative) or electron current (positive in the negative to positive direction)

 

Post again if you need more.

Edited January 31, 2016 by studiot

[teknora]

To Studiot:

 

Thank you a lot. Right now I am analysing the references.