Please clarify one point: You stated that the maximum capacitor voltage "is proportional to the resistance (series)" ? Or did you mean "inversely proportional to the series resistance'? If the latter is true, then the capacitor leakage resistance is the reason as others have said. If the series resistance is R1 and the leakage resistance is R2, then the capacitor final voltage V2 is fixed by the divider action between R1 & R2 [ie., V2=V1*R2/(R1+R2)] and will thus never reach the input value V1 ! .Also note that only when R2 is very large, the capacitor voltage V2 is close to the applied voltage V1. Alternatively, R1 has to be very small compared to R2 for V2 to reach V1.

In series RC circuits,

voltage across capacitor = (power supply voltage) - (voltage drop across resistor)

where, voltage drop across resistor = Resistance x current passing through the circuit

That's why you are getting voltage across capacitor less than expected and that is proportional to resistance value.

As Nishant Saxena points out, you have current in the circuit, which should not be the case after the ideal RC circuit has reached steady state. However, we have to work with real capacitors and they always have a minute leakage, which becomes significant when the resistor is large enough.

What kind of leakege current you have in the circuit depends on the capacitor type. For example, electrolytic capacitors have larger leakage compared to ceramic capacitors. A very significant leakage points to faulty capacitor.

I can think of several possibilities that might explain the result you are obtaining. I can categorize them into two as follows:

I) You may have an unintentional parallel resistor (Let's call it Rx) across the capacitor, which is comparable with your R resistor. This Rx may be the loss resistance of the capacitor (is ideally infinite but practically may be no so large) or the input resistance of your voltmeter (ideally infinite, but still very large if it's an electronic voltmeter. Old voltmeters may have small input resistance, especially for low voltage measurement settings). It may be also due a partly damaged capacitor or a problem on your circuit board.

II) When you apply the DC voltage (say, 1V), some time is required for the capacitor to be charged up to 1V. In theory, infinite time is required; nevertheless within several RC times it will reach to 1V (That's why R*C product is called time constant). If RC is large (e.g. 1 second), then you need to wait long (several seconds) for the final value to be established to a good approximation. However, for such a case, you should have noticed the "ongoing" change on the measured value. So, I think this one is less probable.

Please clarify one point: You stated that the maximum capacitor voltage "is proportional to the resistance (series)" ? Or did you mean "inversely proportional to the series resistance'? If the latter is true, then the capacitor leakage resistance is the reason as others have said. If the series resistance is R1 and the leakage resistance is R2, then the capacitor final voltage V2 is fixed by the divider action between R1 & R2 [ie., V2=V1*R2/(R1+R2)] and will thus never reach the input value V1 ! .Also note that only when R2 is very large, the capacitor voltage V2 is close to the applied voltage V1. Alternatively, R1 has to be very small compared to R2 for V2 to reach V1.

A real pleasure for me was to read these wonderful comments!

"In series RC circuit with dc power supply the maximum capacitor voltage should be equal to power supply but in my experiments it's less than expected value and it's proportional to resistor,"

If an experiment does not give the expected results, I think it would be helpful to show and describe the experimental arrangement.

There could be more to this than meets the eye. If the prime source of the DC voltage is already well-smoothed, after some (presumed) rectification process, or alternatively coming from some sort of battery, then all the foregoing comments are valid. But if the case in point involves rectification of an AC source, then a great deal depends on whether that source is, first, a single-phase or three-phase supply, and second, the type of rectification circuitry that is used.

As a simple example, the average value of a full-wave-rectified, single-phase AC voltage having an RMS value of 100V (that is, a peak value of about 141V), then the average value after smoothing will only be about 63.7% of the peak value, that is, 90V. (The quantity 0.6366 is 2/pi).

Farzane, if can provide more detailed information about your system I feel sure that we can offer more precise assistance.

Barrie, while your theory of average voltage is correct, I dont think it is applicable here, since one needs a LC filter circuit in general, to extract the average value (DC) and the inductor current must be in continuous conduction to get the average voltage. Discontinuity will cause the capacitor to charge up towards the peak input voltage rather than the average value ! RC filter circuits can be used also to extract the average (DC) value, but their output voltage will depend upon the output loading. RC filters however are used in AC low-pass filters with less problems. But I do agree that circuit details are needed to give a precise answer.

Dear Farzane

Study chapters 3 and 4 of "Electrical Engineering" by Allan R. Hambley.

I see in this forum bright names on the research gate

I agree with Ali and Sujet.

This question concerns the interpretation of the deviation between the theoretical and corresponding experimental results. It is a typical question showing how the insertion of the measuring instruments like the dc voltmeters can affect the measurement accuracy. The capacitors have normally a very small DC leakage current  and consequently very high DC leakage   resistance .When connecting the DC voltmeter across the capacitor  with its normally much smaller resistance  value , Then the total DC resistance will be mostly of the  voltmeter. Consequently the voltage division between the capacitor and the series Resistance will be determined by the voltmeter resistance and the series  resistance. Ideally, in order to measure the DC voltage across the capacitor, you have to use an infinte resistance voltmeter. In practice there are Dc voltmeters that have very high input resistance, they are called electrometers. There are companies which produce such instruments such as Keithly. It is then the matter of using the suitable measuring instrument

This i really a good question and a common observation.

Dear Abdelhalim,

I was about to mention the same thing.

I felt greatly honoured to see Barrie Gilbert's response on the same issue that I responded. I just couldn't believe it !

We analog integrated circuits designers learnt a lot from his works. He has always been a kind of legend for me.

(I am sorry for moving out of the scope of the question, but I couldn't help myself)

I fully agree with Lutz, Barrie and Sujit that it is good to see the measuring arrangement. But this will immediately give an answer to the mystery... the tension will drop... and the so interesting discussion will cease:) So, it is much more interesting and useful for our human imagination to guess what the problem is before to see the arrangement. Then let's continue to exercise our thinking...

The Farzane's observation that "it's less than expected value and it's proportional to resistor" does not talk about any "movement" in time (charging), but rather about a relatively constant voltage that depends on R1. This brings me to two assumptions - either the resistances R1 and R2 are quite low (the equivalent time constant R1||R2*C is small), or simply the capacitor is disconnected (for example, from the ground)... and there is only a humble voltage divider...

Ali, my full respect to Gibert. He is the inventor of the Gibert cell in analog circuits. But i intended also Cyril Meckhov and Lutz von Wngenheim, We shared the answer of not small number of question on research gate and sometime hot discussions led by Cyril.

Dear Cyril, 

good evening

You mean that the circuit connection itself is defective. The ambiguity arose from the statement that the measured voltage is proportional to the series resistance, which can not be fully explained by the limited voltmeter resistance. Increasing the series resistance and keeping the voltmeter resistance same  would reduce the measured voltage as sujit and I argued.

After this discussion, it is good that Frazane speaks. It will be very useful, otherwise we will be like the Greek philosophers.

wish you the best. 

Whats the real behavior of rc circuits?

The capaciors store energy as charge. The resistors use energy, heat ;-)  

Best wishes of a Merry Christmas and a Happy New Year to all of you  ;-)

As all the friends state that it is simply transient and steady state responses. These responses can be obtained using different modeling techniques. I also used TLM technique to obtained these responses for both linear and nonlinear. All the best, Okan

Continuing the philosophical comments.

Concerning modelling our basic assumption is quasi-stationary circuits so lumped elements could be used. Transient and steady state analysis is the same. Steady state oscillators are just circuits which can't find a DC bias point as the result of the step response of the battery. Chaotic steady state behavior is more common than limit cycle behavior   ;-).

Hi Eric - I think your last sentence deserves a special discussion. Where and how?

Lutz vW

Hi Lutz, Thank you for your response. Please take a look around in the universe. You may observe chaotic oscillating systems everywhere. To me noise in electrical manmade systems is just chaotic oscillations due to parasitic components. 

Erik, would you translate in a simple human language the written above? Because I have the feeling that I am in the office of a doctor who explains to me in Latin how to cure of a common cold:)  I think, in this way, we will surely frighten Farzane and she will never reveal the secret of her measurement arrangement:)

Cyril - apart from your request to Erik - I am afraid that Farzane is not interested anymore in an answer because she did not inform us about her measureent set.

in reality we never reached the max value of the supply voltage. In addition, all Depond of electric circuit elements (passive or dynamic)

hello every body;

with respect to all great researchers and analog circuit designers who answered  the question, I really don't understand the idea behind of such complicated answers when there is a complete and correct answer which Sujit mentioned. I think his answer completely describes the reason. Of course there is a transient time which theoretically never is reached, but surely Farzaneh knows that and she means the final voltage which -due to leakage conductance of capacitor- is always less than the supply voltage.

best regards

RC time constant shows exactly what kind of behavior the time-domain response of a R+C series circuit. This charge and discharge characteristic  gives you an idea of what the period for on-off phenomenon which translates into frequency generation. For example, the period is 2 x 5 tau (RC time constant) = 10 tau, the frequency you may have is 1/10 tau... 

For signaling/wave shaping, transient response ckt,  impulse, frequency band /bandwidth , break frequency assessment, ac coupling of ampl. analog filter, oscillator , re generator/multi-vibrator , etc

Dear Farzane, the DC-Analysis of your circuit contains only supply and resistors. You can leave the ideal capacitors. Real capacitors have a parallel resistor. This parallel resistor stand for the ohmic losses of dielectric (not dielectric losses). Please measure the ohmic losses of dielectric of all capacitors!  

An RC Series Circuit (with a Lossless Pure Capacitor),when connected to a D.C. Source acts initially at t = 0+ as a Short Circuit and finally at the Settling time  t = 5 * R * C as an Open Circuit,with the Full Supply Voltage appearing across the Pure Capacitor. The Charging of the Capacitor takes Place exponentially with Time Constant = R * C.  Finally the Voltage across the Pure Capacitor being equal to the Supply Voltage, remains the same,without discharging after  t= 5 times the the Time Constant.  This is the Principle we use in Wave Shaping  Circuits.

On the other hand if the Capacitor is a lossy one with Resistance = Rc, then the Charging takes place with Time Constant = (R + R c) * C and after charging fully to Voltage across the Capacitor = V * R c / (R + R c) which incidentally happens to be the  Charging Voltage for the Lossy Capacitance.and then remains the same without discharging. A Lossy Capacitance is represented by its Equivalent Circuit (Rc || C).

P.S.

Another suggestion: the capacitor maybe fractional. In this case the analysis is somehow different. See the paper

Fractional model of an electrochemical capacitor

http://www.sciencedirect.com/science/article/pii/S0165168414000917

It also likely that you are neglecting  the Thevenin's resistance of the supply, which will be applicable if your R value is comparable with it. The voltage available to charge the capacitor will be determined by  V_in x R / (R+R_th), where R_th is the Thevenins resistance of the supply and V_in is the open circuit voltage

Referring to Serkan Gunel's contention V_in x R / (R+R_th),  I may state that R_th is R  in my statement  and R is Rc in my statement with C acting as Open Circuit after Full Charging of the Capacitor..

P.S.

Please send me your data which u obtained by experiments.

M.F

Dear Pisupati, I will only add to your nice explanation that the time constant is R||Rc * C.

Dear Abdelhalim,

You advise Farzane to use a perfect voltmeter (electrometer). In addition to this relatively conventional solution, I would add another more sophisticated and a little extravagant solution - to improve the imperfect voltmeter (or capacitor... or the both) by eliminating the equivalent resistance Rc||Rv. She can do such a magic by connecting a negative resistor (INIC) with negative resistance -Rc||Rv in parallel to the capacitor. The beauty of this solution is that it eliminates en masse all the problems (Rc, Rv, etc.) Here is the idea:

https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#Compensating_resistive_losses_by_N-shaped_negative_resistors

There is only one problem here - how to determine the equivalent leakage  Rc||Rv... perhaps experimentally?

In theory, the series RC circuit will charge the voltage of the capacitor up to 98% of the DC power supply. However, in reality, it is not the case due to many factors, such as kind of resistor used, wire loss, connector loss, alligator clip contact, the capacitor internal resistance, and many factor involved...

Dear Prof. Cyril,

It is a Wonderful idea to Connect a Negative Resistance -Rc||Rv in parallel to the capacitor. -Rc||Rv in parallel to the capacitor. 

This will be Very Useful in Precision Measurements.

Regards.

P.S.

Three factors leads your measurement wrong

1) Leakage current is more result into unexpected drop across R (Ideally it should be zero)

2) Input impedance of your measuring instrument is comparable to the capacitive impedance (Both will be in parallel, hence current gets shared will lead to wrong indication in spite of having correct value).

3) The instrument itself is faulty 

Amipara, the current does not get "shared" between an ideal capacitor and a paralleled "impedance" under steady state in a DC circuit ! Under steady state in DC, ideal capacitors dont carry current and the word 'impedance' has no meaning (in the absence of frequency): only resistance is effective.

Eureka:) A new idea came to me when I realized that we can go even further in this neutralization of positive to negative resistance. Let's see how...

Actually, we have three resistances in parallel - R, Rc and Rv, and the equivalent resistance is Re = R||Rc||Rv. Imagine, we connect a variable (N-shaped) negative resistor in parallel to Re and begin adjusting its resistance RN. Depending on its value, it will "eat" some part of Re and it will dissapear (become infinite):

1. RN = Rv (Rv is neutralized) or RN = Rc (Rc is neutralized)

2. RN = Rc||Rv (both Rc and Rv are neutralized)

3. RN = Rc||Rv||R (all the positive resistances are neutralized)

In case 2, we obtain a perfect exponential shape... and this solves the Farzane's problem. But, as we are curious and bold enough... and decide to go further beyond this question... we see in case 3 a completely linear shape? But this is a kind of "improved Deboo integrator" where an INIC compensates not only the internal source resistance (R) but, in addition, all the leakages! Am I right? If so, this would be a good confirmation of what I wrote yesterday in...

https://www.researchgate.net/post/How_to_make_1st_year_electrical_engineering_student_understand_voltage_current_resistor/3

... sometimes, not the very question is interesting, but its impact on the following discussion...

And, if we are even more curious and persistent, we could ask ourselves, "But what happens if RN

The measurement it self tells that It is NOT AN IDEAL situation. I mentioned shared current are flows of leakage current and flow through meter. Unless there is a current flow, cannot be voltage drop & cannot be voltage loss.

Hi Cyril, I am sorry but I can't translate my "philosophical" statement into "simple human language" ;-).  

Erik, still we teachers should be able to do this...

What do you think about my super-neutralization idea that occurred to me while bathing in the tub as Archimedes:)? Maybe I should not have to share it here with simple words... but to keep it for a highly scientific article written in your sort of language:)?

Regards... and Merry Christmas, Cyril.

Cyril, since "neutralization" describes a certain kind of singularity, how can it benefit from a superlative? What might seem to be an exception - the term "super-regenerative" - is actually not so, because it describes a step beyond the "mild", Q-enhancing feedback of an RF signal to a point beyond, at which the system becomes a relaxation oscillator, having certain beneficial properties. But then, the prefix "super" is of universal utility. 

Now look how far we've drifted from the dear lady's innocent question! Please, Farzane, let us know if we have helped - at all - and whether we might help you further.

Dear Barrie Gilbert,

Thanks for your support in maintaining this discussion. As you probably guessed, my idea is, while we are waiting for Farzane to respond, to fill the time with interesting discussions (relevant to and further developing the initial problem). Of course, I could ask a separate question... but the result will not be better... even worse...

Since it became clear the problem was in all sorts of leakages (represented by a total positive resistance Re in parallel to the capacitor), an idea (speculation) to compensate them with an equivalent negative resistance RN arose. As a result of this exact neutralization (RN = -Re), all the leakages will disappear (as though they are replaced by an infinite resistance).

The next my idea was that if we continue decreasing this "destroying" negative resistance beyond this point of exact leakage neutralization, it will begin "eating" a part of the positive resistance of the "useful" resistor R... and finally, it will destroy all the resistance R. This means that the resistor R as though already has an infinite resistance... and behaves as an ideal current source... Actually, this is the idea of the Howland current source and its special case here - Deboo integrator. But while in the classic Deboo integrator the negative resistor (INIC) neutralizes only the positive resistance R, here it neutralizes all the resistances in parallel (the useful R and harmful leakages).

And my final speculation was about what would happen if we continue decreasing the negative resistance even beyond this point of full resistance neutralization. Then the effective resistance (the result of the neutralization) would be fully negative. And here, I suppose, the voltage across the capacitor will begin self-increasing in an avalanche like manner...

This morning, while I was walking in the park and thinking on this, I suddenly made a connection between this excessive (over-, super-, ultra-) neutralization and the refreshing system of DRAM. I realized that the sense amplifier (an RS latch) actually acts as a negative resistor (INIC) when reading the capacitor memory cell. Depending on the contents ("1" or "0"), this negative "resistor" begins accelerating towards the positive (VCC) or negative (ground) rail and finally reach the rail... and thus restores (refreshes) the charge of the cell. So, my assertion is the sense amplifier is a negative resistor.

Finally, a bit of humor - it may seem incredible, but I really have come to the idea of excessive compensation... in the bathtub:) To my displeasure, I noticed the water in the bathtub dropt slightly... the reason was a "leakage" in the siphon. Of course, I opened the tap and began to regulate the water flow. And then I realized that the tap plays the role of a negative resistor which, depending on the degree of the opening, compensated in the corresponding degree (under, exact and over) the siphon leakage...

Best regards, Cyril

I do some of my best thinking - and most of my reading - in the bath-tub. 

In practice, DC capacitors have considerable parasitic resistances (series and parallel) inside. In fact, your circuit is not a simple RC due to parasitic elements.

If you analyse your circuit by consideration of a series and a parallel parasitic resistances of the capacitor again, you will find the answer easily.

good luck

Did you try to use voltmeter of another model? It seems that the internal resistance of the used voltmeter is only one or two orders of magnitude higher than the resistor in the series RC circuit

Hello everybody. I should apologize for my late response due to some involvements. but as Cyril had anticipated, in a moment I faced a collection of ideas and it was great. I read them carefully. Thanks a lot for your attention.

I didn’t use a specific circuit arrangement, But I would explain about my instrument and data. Rectification process was done with a power supply with single phase AC source and as Barrie said, I think the kind of this process is important.

For measuring capacitor voltage, I used both analog and digital scopes and deviation from ideal values (after 5 time constant) was almost the same. So I think using electrometer, in spite of it’s benefits, is not the final solution in my case.

Then It seems that I should pay more attention to capacitor leakage resistance as Sujit and the others said. But what’s the range of this leakage in various capacitors and how can I get rid of this loss? With negative resistance theory of Cyril? perhaps ;)

They are my experimental data in two cases:

1) capacitor and resistor are both variable and are series with a power supply that I explained.

C=100 pF

R=100 Kohm

V=10 V

Measured voltage of capacitor by scope (after 5 tao) is 6 V (and this measured voltage was a function of resistance just) So I should conclude that capacitor leakage resistance is 0.15 Mohm If we use this equality: Vc=V*Rc/(R+Rc)

2) after changing the capacitor I used a ceramic ones and a constant resistor:

C=0.01 MicroF

R=0.24 Mohm

V=5V

Measured voltage of capacitor was 3.4 V. So capacitor leakage is 0.51 Mohm.

Are they reasonable values and do you think that just resistance leakage is response of this deviation?

Thanks

Regards

Thanks Farzane for breaking the silence and ending all our speculations!

First thing that comes to my mind is that you should measure the value of equivalent C & parallel R in a good quality LCR meter in 'parallel mode' of equivalent circuit, at the lowest frequency option for measurement. That will give you the value of the equivalent  parallel leakage resistance.  Then we can compare the data and conclude. The value of leakage resistance depends on the type of capacitor and its capacitance value, as well as on its voltage rating.

Farzane, in practice you have replaced all the elements but the error still remains. I do not know what the purpose of your experiments is, but if you really can vary the values of resistance and capacity (at the same time constant), then select the possibly least resistance and highest capacity...

If you put up with the existence of the leakages, and  decide to compensate (instead to examine and remove) them by a negative resistor with an equivalent negative resistance, connect an INIC to the capacitor (represented by Vs in the attached figure). You can determine the circuit resistances by the equation:

Rc = R3*R1/R2

Practiucally, you can initially choose R3 = Rc and R1 = R2, and then vary some of the resistances so that Vc to reach the power supply (the top of the input square pulses) "after 5 time constant". 

Just one practical consideration - the INIC power supply has to have at least to times higher voltage than the input voltage. And also, another tip - you can connect the osciloscope probe to the op-amp output instead to the capacitor; this will decrease the error (with the probe resistance), and Vc will be amplified.

You can think of this exotic technique as an excessive (over) bootstrapping where the voltage follower is developed into a non-inverting amplifer; as a result, the infinite input resistance is converted into a negative one:

https://en.wikipedia.org/wiki/Miller_theorem#Applications_based_on_subtracting_V2_from_V1

One thing that doesn't seem to have be raised in this thread is the voltage drop in the rectification elements used. I assume that the "lower-than-expected" result has used all the necessary care in making the comparison between the measured AC input and the DC output. Even though the forward-voltage drop becomes very small (a few millivolts) under no-load conditions, it may be as much as 500mV, or even more, for silicon diodes, and of course, many volts for thermionic diode rectifiers.

Another thing to keep in mind is the matter of signal waveform. Many voltmeters (and modern DVMs can, for most applications, be regarded as 'electrometer-grade' voltmeters) do not display the true - that is, the RMS - value of a complex signal, but a sort of surrogate, which is the MEAN value which is then adjusted in the display to read AS IF it were the RMS value. This is true of DC+ [small] AC signals as well as full bipolarity signals, symmetrical or not. 

It's best to think all this through - and test it - for oneself: there is no substitute for self-learning to attain a deep appreciation for a topic.

I doubt leakage in the capacitor is to blame (especially if its value is as small as stated by Farzane in her most recent note)  - although some dielectrics are worse offenders than others, Avoid ceramic capacitors, because, while certain types are not terribly troublesome, most are leaky. Another aspect of poor dielectrics is that they exhibit a slow memory: that is, even after a brief short-circuit discharge, you will find their voltage into an open-circuit - which an instrument-grade DVM will usually emulate - begins to rise again. This is called (and is due to) 'dielectric absorption'.

Two things I don't think have been made clear by Farzane:

(1) Is this 'error' just puzzling, or is truly problematical in some intended application? Of course, a real load will draw current, and thus cause a completely non-mysterious voltage drop across the proposed series-R, whose function, I presume is to limit the surge current into the capacitor.

(2) What level of accuracy is desired to be achieved in all these measurements? When this falls much below 0.1% you can suspect many, many diverse sources of error. In this connection, keep in mind that the error in component values is rarely better than 1%, often as high as 10%, and the value of electrolytic capacitors is commonly specified as -20%/+80% - in other words, they are pretty much guaranteed to exhibit their stated value! One last note about this type of capacitor: they need to be "formed" - that is, they initially have a rather low value of capacitance, which only slowly (many seconds or even minutes) attains the stated value.

May the coming New Year be free of needless mysteries!

Barrie

Quote Barrie: "May the coming New Year be free of needless mysteries!"

Barrie, I am sure you will be disappointed. At the time being I have a controversely discussion in one of the electronic forums about the control mechanism of the BJT. As you certainly know, some people still believe in current-control only. Even books and lecture notes offer the explantion that Ib would control Ic (just a claim without any justification).

And now the "mystery": All persons defending strictly the current-control approach make use from the voltage-control principle during design of a simple gain stage (because that`s the only way !) - without realizing the contradiction!

And, of course, they are not able to verify at which design step they have made use of current-control. But they stick to their claims.

Merry Christmas to all of you.

Lutz

Dear Farzane:

Thank you for explaining what you are REALLY trying to do, namely, determine the value of the effective capacitance of a liquid droplet. (Private communication). Now that I understand your experiment I can offer a few simple hypotheses for why you see an error in your voltage reading. It generally reads too low, correct? 

H1: Your resistance had to be high (about 10 megohms) to produce a reasonably long time-constant in conjunction with this relatively small load capacitance (about 10 pF). Thus, even a small amount of loading on the output will reduce the reading. For example, if your voltmeter has an input resistance of 1 gigohm (109 ohms), you will see an error of -1% in the reading.

H2: A lower-quality DVM may have an input ("bias") current. Thus, even 1 nA into your 10 megohm source resistance will alter the reading by 106 x 10-9 = 1 mV. (In THIS scenario, the reading may be either too high or too low, of coursed. depending on the polarity of the DVM's input current).

H3: Of course, if you are using an oscilloscope without an attenuating probe, its input resistance (to ground) will cause a huge error. For example, using a DC input of 1 V to your  RC circuit, and with R = 107 ohms, you would form a divider network having an output of 106/(107+106) volts, which is only 90.9 millivolts.

H4: Even using a X10 attenuator probe on the oscilloscope (presenting an input resistance of 10 megohms) you will halve the output voltage.

H5: But the most likely scenario is the your liquid droplet whose capacitance you are trying to determine - the 'C' in your 'RC' circuit - is very slightly conductive due to ionic contamination from some source (either the liquid itself or the material of the lower plate on which the droplet sits).

H6: One last possibility is that the droplet is behaving like a weak quasi-battery. Of course, that is easily checked by measuring the voltage across it in total isolation.

You can still determine the capacitance of the liquid droplet using your RC method, but with at least two values of 'R'. If you give it some thought, you will see that you can create a set of simultaneous equations from which the actual value of 'C' can be extracted for any reasonable value of unknown load resistance. But I would suggest using at least three values for 'R' to anticipate both a resistance of the measurement device and the possibility of that instrument also having an input bias current. Using a large set of 'R' values from, say, 5 megohms up to 50 megohns in 5 megohm steps, you could also use a graphical solution, from a correspondingly large set of time-constant measurements.

All this underscore the need for the engineer to be resourceful in the creation of experiments - and indeed, every avenue of exploration. The most important question - which should always be on your mind - is "WHAT IF?". A close ally is the question "HOW ABOUT?". And of course, there is always the basic question "WHY?". These questions should be on your mind a hundred times a day.

Finally, I can suggest a quite different method for rapidly measuring the capacitance of even a stream of droplets. If you wish, contact me directly at barrie,[email protected] and I will explain this method.

And since all this deviates far from the original thread, I suggest the comments have now outrun their useful life.

We have a bold new year ahead of us. Let's move on.

Barrie

Cyril,

Please define your super-neutralization !

Is it just more than 99pct neutralizastion ?

100pct neutralization is a compensation (balance, feed-back, ...)

which of-course may be very sensitive.

I agree with Berrie !

Concerning the modeling of the BJT we may use the Maxwell equations and find distributed parameter models which are in better agreement with our measurements. Current control and voltage control are just the same. The mystery first show up when you try to explain your mathematical model by means of analog physical behavior.

Best regards and Happy New Year to all of you, ERIK

"Current control and voltage control are just the same."

Erik - please, can you explain a litle? I strongly disagree!

Oh, you folks - arguing about how best to model a BJT! That's a VERY important issue and it can easily fill a book - and it has: an early one (1976) was called "Modeling the Bipolar Transistor", written by George Wilson, a co-worker of mine while at Tektronix.My own basic and essential starting point is shown in the attachment.

Modern BJTs are now extremely challenging in developing appropriate and accurate models and then characterizing them by measurements on representative devices.

But surely this thread has deviated FAR from the question posed by Farzane, and I think this ongoing cross-talk should be closed. I will try to encourage such a closure by posing a question of my own. Look for it in the coming moments. 

Barrie

Dear Lutz,

"Current control and voltage control are just the same."

I should have said that they were dual  ;-)

I have asked a separate question about my "100% neutralization idea"...

https://www.researchgate.net/post/Is_it_possible_to_neutralize_all_the_positive_resistances_in_a_circuit_by_an_equivalent_negative_resistance

.... with two purposes - first, at the request of Barrie Gilbert to terminate the irrelevant discussions here... and second, to answer the above question of Erik Lindberg.

Current gain hFE in a bipolar transistor varies a lot from one unit to another.

Tranconductance is much constant, and about 25xE-3/Ic at room temperature. To design a transistor amplifier and calculate its gain you should relay on the transconductance concept. 

Yes - of course, I agree to the former "recommendation". However, for me it is more that a recommendation because "transconductance" not only is a "concept" but the physical truth behind the BJT because:

Ttransconductance gm=slope of the curve Ic=f(Vbe) .

I would like to add to my earlier Answer by way of Correction, that, in the case of D.C. Series R.C. Circuit with a Lossy Capacitor represented by its equivalent Pure Capacitance C I I  Rc  when D.C. Voltage Source Vs  is Switched on, initially at t = 0+  the Pure capacitance C acts as Short Circuit shorting Rc and starts charging Exponentially and the Voltage across it acts as a Voltage Source opposed to the Supply Voltage. Hence the Rc across C can be treated as a Redundant Element as the Voltage across it is known. So, the Capacitor Charges to its Full Voltage of  Vc  =   Vs * Rc / ( R + Rc ) with Time Constant R * C only and there will be no Further Charging.

When the Switch is opened the Capacitor can Discharge only through Rc and the Discharging Time Constant being Rc * C. only.

In the case of Pure Capacitance with no Loss Rc is infinity and the Voltage across the Capacitor Remains at its Final Value, without being able to Discharge.

Hence We can say that the Charging Time Constant is R * C only ,whether the Capacitor is a Lossy or Lossless one ,R being the Total Series Resistance only,.

P.S.

please define: 1) capacitor type  2)capacitor value 3)resistor value

Is it important to see the idea?

This thread was closed some time back. After a private communication 

with Farzane, I determined the original postulation did not convey what

she was really trying to accomplish, and the required help has be given.

Barrie