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Old 2nd June 2004, 01:49 PM   #1
hjelm is offline hjelm  Sweden
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Default Yadis: Yet another discrete IV stage

Hi all,
Inspired by Jocko's simple iv stage and Bricolo's suggestion , which he scrapped when the input impedance was too high. I thought a bit on how to fix the input voltage to a fixed value.

I am not sure this circuit works in real life but in simulation it performs quite good. ~-70db THD at 10k
If i replace the transistor models with sa970 and sc2240 instead and optimize the resistor values and look at 1k it says ~-90dB THD, third order dominating.
Now i do not beleive distorsion measurements in simulations but they are the only thing i have.

Can anyone comment on the concept at least, does it work at all. A friend of mine said the startup could be problematic and might require extra circuitry i.e if there is no current there is no way to get a current.

The idea to lower the input impedance lied in the fact that the Vbe is correlated to the Ic so i needed to copy the Ic of the input transistor to the one supplying the reference point to the input transistor base this meant i couldnt take this current to drive a output so i mirrored the current both ways and got fairly good results.

The circuit works down to +-5v if you limit the output swing but +-15 seems to give better distortion figures. The emitter resistors of both the input and the "mirror" are not equal to acheive optimum performance on both distortion and inout impedance.
R4_10 is slightly bigger than R4_9 and R4_11 to compensate for the base currents in the mirror. The R1_12 is a bit bigger than R0_13 to compensate for different currents in the devices.

The cascode on the output isn't necessary but the distortion went down a lot when i included it, optimum seems to be to place it some 4V from the rail (less if the rail is 5V of course).

Looking at transient step responses it looks very controlled and settles nicely.

If anyone wants to i'll post the simulation files, spice cir files.

And to intercept all those who will say just go build the damned thing and measure it, i would if i had the time, too many kids and dogs requireing attention! All i have time to do is come up with ideas and try them out in a simulator, i do not even have the measurement equipment to do the measurements.

Anyway here it is and if this is nothing new then just ignore my mad ramblings.
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Old 2nd June 2004, 03:33 PM   #2
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Default Why the 100 ohm resistor?

You want the inout Z as low as possible.

Simualtions are one thing, but in real life, the limit on mine was -80 dB at full output.

Jocko
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Old 2nd June 2004, 03:59 PM   #3
Bricolo is offline Bricolo  France
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If you want to go with current mirors, Mr Hawksford has a good article about IVs, and on page 20 you'll find a very well designed one based on mirors.

It uses a 1:1 miror, and a buffer after the I/V resistor. But there's certainly a possibility to modify it into a 1:3 or more curent miror

http://www.essex.ac.uk/ese/research/...0amplifier.pdf
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Old 2nd June 2004, 05:10 PM   #4
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Question Iv

Hi hjelm,
I agree with Jocko that the inputimpedance should be as low as possible. The common base amplifier (also called grounded base) has the lowest input impedance of all three possible configurations. I believe that using a Sziklai pair as input has even lower input impedance, hope so.
I would connect Vref to the base of the input transitor. In your circuit a lot of current goes into the Vref pin.
The folded cascode is a nice way of level shifting and having a extra common-base amplifier.
I think Jocko's circuit is much more elegant as that of the Hawksford guy.
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Old 2nd June 2004, 05:11 PM   #5
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A bit more respect for Hawksford, Elso
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Old 2nd June 2004, 09:03 PM   #6
Pedja is offline Pedja  Serbia
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Quote:
Originally posted by hjelm
If anyone wants to i'll post the simulation files, spice cir files.
Sure Hjelm, just drop them in, please.

Quote:
Originally posted by Jocko Homo
Simualtions are one thing, but in real life, the limit on mine was -80 dB at full output.
Jocko, you kinda should try harder. It can be done better. Trust me.

Quote:
Originally posted by Elso Kwak
I believe that using a Sziklai pair as input has even lower input impedance, hope so.
Elso, in the circuit you see above (or in the Hawksford's paper) the current mirrored back to the Q1 (or the first transistor of the "super pair" in the Hawksford's paper) yields practical zero input impedance of the circuit (100R-12 resistor must be removed, of course).

Pedja
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Old 2nd June 2004, 10:23 PM   #7
Bricolo is offline Bricolo  France
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Elso,

I didn't test Hawksford's IV yet, so I can't relly tell if it's good or not.
But I've read the article, and in it, he explains every choice he made. And he does this very well, the explanations are easy to understand, and seem very wise.

With all his explanations, he convinced me that the choices were good. That's a good point IMHO
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Old 3rd June 2004, 07:35 AM   #8
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Lightbulb Common base IV

Quote:
Originally posted by Pedja

Sure Hjelm, just drop them in, please.


Jocko, you kinda should try harder. It can be done better. Trust me.


Elso, in the circuit you see above (or in the Hawksford's paper) the current mirrored back to the Q1 (or the first transistor of the "super pair" in the Hawksford's paper) yields practical zero input impedance of the circuit (100R-12 resistor must be removed, of course).

Pedja
Hi Pedja,
The input impedance is not zero as it should be as the the DAC current source has limited compliance. (Horowitz page 184).
Input impedance of a common base goes down when the collector-emittor current goes up. For example the CB-CB pair in Low Noise Electronic Design, Motchenbacher and Fitchen, has a Ri of 260 Ohm with 100µA of current. But the CB-CC pair with 1mA of current in the CB transistor has an input iimpedance of 29 Ohm.
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Old 3rd June 2004, 07:40 AM   #9
hjelm is offline hjelm  Sweden
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The thing is that i get a lower zin with the 100ohm resistor since i have the same emitter resistor in the transistor that gives me the reference voltage for the base of the common base transistor.
And it gives me a lower distortion as well.

Consider the case when the current through both the q1 and q2 are the same i get the same voltage drop over the resistors and provided that the Vbe is the same i have 0 volt on the input!
In simulation i trim it down to 110uv PP for 2.4mA PP in.

I admit that the circuit is not very suitable to a Vref other than ground since i feed back the current through it.

Admittedly when the current mirror cannot keep up i will have worse performance on input impedance than without but i choose to do that to get better distortion figures.

Didnt actually read the hawsford paper until now but the concept on the input is similar and yes the Zout is the 1k resistor so it needs a buffer.

I tried this configuration with wilson current mirrors as well but got worse figures, actually they were close but i found out i had made a mistake and when i did the mirrors as they should the distortion increased .

And Jocko yes i wish i had the time to build it .

As for the attached files the newivsa970.cir is the one with best performance on input impedance and distortion. The newiv.cir is the version with the BC850C BC860C models unoptimized.

I think there is a need to edit the include lines to adapt the path to the models (the relevant ones are included). The zenermodel include can be deleted.

Thanks for all the answers.
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File Type: zip ivversions.zip (4.2 KB, 141 views)
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Old 3rd June 2004, 07:44 AM   #10
hjelm is offline hjelm  Sweden
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Elso you cannot think of it as a CB only. Since you are moving the vref on the base (due to the mirroring of the current) you are lowering the input impedance.
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