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-   -   Computing output impedance (Cordell's book) (http://www.diyaudio.com/forums/solid-state/232477-computing-output-impedance-cordells-book.html)

Dan Moos 21st March 2013 12:24 AM

Computing output impedance (Cordell's book)
 
I am going through bob Cordell's book, and I seem to be missing something.

In chapter 2 when discussing early effect, it is explained that ro=(VA+Vce)/Ic. Later explains that when emitter degeneration is used, Rout=ro*degeneration factor. Using these formulas, when I do the math for his common emitter stage example (figure 2.5b), I get the correct results that match those in the book.

Here is where the trouble starts. Later in the chapter when current sources are discussed, my output impedance results are way off from the book's.

Here is an example:

In figure 2.10c, a 5mA current source is biased with a green LED running at .5mA. The load is 5k, and the emitter resistor is 214 ohms. The positive rail is 50v and the negative rail is at ground. The resulting voltages are 25v at the collector, and 1.08v across the emitter resistor. Bob seems to be giving output impedance results that ignore the load, and my math comes out wrong either way.

Here is how I am doing it.

As stated in the book, I am using a VA of 100 for a 2n5551. Vce is ~24v. Ic is 5mA. (100+24)/.005 = 24,800.

Ro = 24,800

re' = 26/5

re' = 5.2

emitter resistor + re' = 5.2 + 214 which is 219.2

degeneration factor = 219.2/5.2 = 42.1

Rout= ro*degeneration factor = 24,800*42.1

Rout =~1 meg

The book says the Rout for this current source should only be 750k. This is actually closer to my answer than I got with most of the other sources, and it seems that the the higher the stated output resistance of the example, the more off my calculations are. I'm always way too high with my answer.

I suspect my problem has something to do with the finite transconductance of the transistor, but I have no clue how to incorporate that into my figures.

Any help?

counter culture 21st March 2013 03:16 AM

I agree with your calculation.

It says that he obtains the output impedances by spice simulation.

It also says that the output impedance cannot exceed ro*hfe. That would give hfe=30, which is the minimum value given in the datasheet for Ic=50mA, Vce=5V.

In a CE amplifier the output impedance of the device is of little significance, we're primarily concerned with the output impedance of the stage, which is dominated by (generally taken to be equal to) the resistance in the collector circuit.

The output impedance of a current source/sink, however, is its effective figure of merit, so we pay more attention to it. If spice simulation is his chosen route to obtaining the figures, then that would seem a good reason to use the same method. I notice that in practice CCSs are commonly measured to determine their effective output impedance.

I did find a section of a learned article that said that the output impedance of a CE amplifier is to some extent related to the driving source impedance, perhaps the spice model takes this into account. IEEE Xplore - Output resistance of the common-emitter amplifier

You could always ask him the question directly (send him a PM), I'd be interested to know the answer.

Bob Cordell 23rd April 2013 10:49 AM

Quote:

Originally Posted by Dan Moos (Post 3420081)
I am going through bob Cordell's book, and I seem to be missing something.

In chapter 2 when discussing early effect, it is explained that ro=(VA+Vce)/Ic. Later explains that when emitter degeneration is used, Rout=ro*degeneration factor. Using these formulas, when I do the math for his common emitter stage example (figure 2.5b), I get the correct results that match those in the book.

Here is where the trouble starts. Later in the chapter when current sources are discussed, my output impedance results are way off from the book's.

Here is an example:

In figure 2.10c, a 5mA current source is biased with a green LED running at .5mA. The load is 5k, and the emitter resistor is 214 ohms. The positive rail is 50v and the negative rail is at ground. The resulting voltages are 25v at the collector, and 1.08v across the emitter resistor. Bob seems to be giving output impedance results that ignore the load, and my math comes out wrong either way.

Here is how I am doing it.

As stated in the book, I am using a VA of 100 for a 2n5551. Vce is ~24v. Ic is 5mA. (100+24)/.005 = 24,800.

Ro = 24,800

re' = 26/5

re' = 5.2

emitter resistor + re' = 5.2 + 214 which is 219.2

degeneration factor = 219.2/5.2 = 42.1

Rout= ro*degeneration factor = 24,800*42.1

Rout =~1 meg

The book says the Rout for this current source should only be 750k. This is actually closer to my answer than I got with most of the other sources, and it seems that the the higher the stated output resistance of the example, the more off my calculations are. I'm always way too high with my answer.

I suspect my problem has something to do with the finite transconductance of the transistor, but I have no clue how to incorporate that into my figures.

Any help?

Hi Dan,

I'm sorry I'm so slow in getting back to you on this one. Your calculations are basically correct. All of the current source output impedances were arrived at through SPICE simulations, and those impedances do not include the effect of the 5k load resistors typically shown. I should have been more clear, especially in regard to these not being calculated current source output impedances. Its actually quite good that your calculations got so close to the simulated value in 2.10c.

The theory presented earlier in the chapter for these sorts of calculations is more to give insight into the influences of the different circuit arrangements on the output impedance. In fact, output impedance and its relationship to, for example, Early Effect (and the SPICE parameter VA that establishes it) is rather inexact. Many transistor models have fairly low absolute accuracy in modeling Early Effect, and the output impedance is much more a function of transistor operating current and voltage than the simple estimates shown in the book.

I'll try to make this all more clear in the Second Edition.

Cheers,
Bob

Dan Moos 24th April 2013 12:01 AM

Thanks for the info.

So should I use the method that I stepped through in my initial post? Would my results be more trustworthy than the sim results? The one I posted was the one I got closest to the impedance stated in the book. My results were usually significantly higher in the case of some of the more advanced current sources.

Also, I believe it was with the feedback pair current source that you mentioned that the source would work with even less current in the "feedback" transistor, but that it would result in a lower quality current source.

Could you possibly explain why that is?

AndrewT 24th April 2013 10:28 AM

you can breadboard your CCS and then measure the DC impedance over a range of loads.
Sim should then be able to tell you how the AC impedance will vary with respect to that measured DC as a starting value.

Bob Cordell 24th April 2013 11:26 AM

Quote:

Originally Posted by Dan Moos (Post 3466022)
Thanks for the info.

So should I use the method that I stepped through in my initial post? Would my results be more trustworthy than the sim results? The one I posted was the one I got closest to the impedance stated in the book. My results were usually significantly higher in the case of some of the more advanced current sources.

Also, I believe it was with the feedback pair current source that you mentioned that the source would work with even less current in the "feedback" transistor, but that it would result in a lower quality current source.

Could you possibly explain why that is?

Hi Dan,

I would generally trust the SPICE simulation for estimating current source output impedance over a paper-pencil approach, since the latter necessarily involves numerous simplifications.

The 2-transistor feedback current source can work with less current flowing in the feedback transistor, but one generally does not want the base current required by the current source transistor to come significantly into play. Thus, I like to have the current in the feedback transistor be perhaps 10 times the demanded base current of the current source transistor. Also, running the feedback transistor at lower current increases its input-referred voltage noise, resulting in a noisier current source.

Cheers,
Bob

Dan Moos 25th April 2013 01:04 AM

@Bob:

how does the base current "come in to play" exactly? Is it a matter where the base current available is dangerously close to being not enough to create the desired current in the source? I.E., say you have a beta of 100 and want a 10mA source. Are you basicaly making the point that you don't want the base current to be to close to the 1uA needed to get that 10mA, especially since beta isn't assured to even be 100?

@AndrewT:

would the way to test the impedance be to add a resistor to ground and create a voltage divider, measuring the result?

Bob Cordell 25th April 2013 10:34 AM

Quote:

Originally Posted by Dan Moos (Post 3467344)
@Bob:

how does the base current "come in to play" exactly? Is it a matter where the base current available is dangerously close to being not enough to create the desired current in the source? I.E., say you have a beta of 100 and want a 10mA source. Are you basicaly making the point that you don't want the base current to be to close to the 1uA needed to get that 10mA, especially since beta isn't assured to even be 100?

@AndrewT:

would the way to test the impedance be to add a resistor to ground and create a voltage divider, measuring the result?

Hi Dan,

Yes, one certainly does not want the current available to become dangerously small. If you want to be really safe, current available should be 5X to 10X the base current.

Note that a 10mA source with beta of 100 requires 100uA of base current, not 1uA. So 5X the 100uA required nominal base current is a supply current of 0.5mA.

I usually measure output impedance by injecting a small signal current into the node and looking at the signal voltage produced.

Cheers,
Bob

Dan Moos 25th April 2013 01:12 PM

thanks. yeah, I meant 100uA.


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