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The Designer's Guide Community Forum › Design › Analog Design › Common mode and diff mode exitation of current

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Common mode and diff mode exitation of current (Read 5877 times)

raja.cedt

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Common mode and diff mode exitation of current
Oct 24^th, 2015, 8:32am

Dear All,
I have to excite a fully differential trans impedance amplifier (diff current to diff voltage converter), with 3mA fully diff current. Since I have learned diff and common components concept from voltage as input I tend to apply the same way (+1.5mA and -1.5mA on both sides) as shown in fig-2. But at the same time differential excitation means applying input between two input (for the time being please ignore how do we configure photo diodes like this) like fig-1. But circuit gains differ by 2 times.

With the simple differential pair concepts, I would guess fig-2 correct.

I would like to know whether people prefer fig-1 or fig-2, it's more what's the standard practice in the industry when they have to design diff TIA.

Thanks in advance.
Raj.

Untitled_1.png

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sheldon Community Fellow Offline Posts: 751	Re: Common mode and diff mode exitation of current Reply #1 - Oct 25^th, 2015, 4:25pm Personall, I use a balun, in ADE the analogLib ideal_balun. The inputs are the differential and common-mode signals. The outputs are non-inverting and inverting. See the "The Designer's Guide to SPICE & Spectre", page 115.
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aaron_do

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Re: Common mode and diff mode exitation of current
Reply #2 - Oct 25^th, 2015, 5:44pm

Hi,

In fig. 1 you have 3 mA flowing out of the top node and 3 mA flowing into the bottom node. In fig. 2 you have 1.5 mA flowing out of the top node and 1.5 mA flowing into the bottom node. i.e. its not a gain difference. Bottom line is you should configure your test bench to most accurately represent the real situation.

Aaron

there is no energy in matter other than that received from the environment - Nikola Tesla

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raja.cedt

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Re: Common mode and diff mode exitation of current
Reply #3 - Oct 26^th, 2015, 2:48am

Hello all--
Thanks for the answers. What I am trying to say here is both circuits are different and hence all the parameters are diff(ex gain=2X), that's clear. Since I am new to diff current world and old to diff voltage, I am trying to use the same analogy for both voltage and current (I could be wrong). if some one ask me to excite an voltage amplifier with 3mV then I can apply
1. 3mV source between inverting and non-inverting terminals
2. +1.5mV at one side and -1.5mA on other side.
3.+3mV on one side and zero voltage (idea gnd) another side.

All of the above excitations will give exactly same voltage gain, so no need to ask the for the particular method, but looks like same method may not be applicable in case of current.

Hope this time I have explained correctly.
Raj.

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aaron_do

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Re: Common mode and diff mode exitation of current
Reply #4 - Oct 26^th, 2015, 5:52pm

Hi,

first of all,

Quote:

What I am trying to say here is both circuits are different and hence all the parameters are diff(ex gain=2X), that's clear.

I would say that this is not clear. The gain is still the same, only the excitation has changed. Anyway that's unimportant and you can continue using whichever definition you are most comfortable with.

See my attachment. IMO, the problem is the way you are splitting up the circuit. The differential amp naturally splits down the center to go from differential to single-ended in the voltage sense. For current, you would have to split it differently.

regards,
Aaron

CurrentVsVoltage.jpg

there is no energy in matter other than that received from the environment - Nikola Tesla

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Common mode and diff mode exitation of current Reply #5 - Oct 27^th, 2015, 1:39am Dear Aron-- Thanks for the fig, I will think about it and will try to get where I am going wrong. Regarding your comment--In my first post Fig-1 will have 6Rf diff o/p voltage.Fig-2 will have 3Rf diff o/p voltage. Best Regards, Raj.
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aaron_do Senior Fellow Offline Posts: 1398	Re: Common mode and diff mode exitation of current Reply #6 - Oct 28^th, 2015, 5:34pm Hi Raj., Quote: In my first post Fig-1 will have 6Rf diff o/p voltage.Fig-2 will have 3Rf diff o/p voltage. o/p voltage and gain are not the same thing. Quote: all the parameters are diff(ex gain=2X), that's clear regards, Aaron
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raja.cedt

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Re: Common mode and diff mode exitation of current
Reply #7 - Nov 4^th, 2015, 4:28am

Hello Aaron--
Yes you are correct. Finally I understood where i am doing wrong.

1. Fig-1 is correct, i mean if i have to excite with 3mA diff current, i should use +3mA and -3mA.

2.For a simple differential pair with Vin differential input, one transistor signal current is +0.5*gm*vin and another transistor carries -0.5*gm*vin, hence actual diff trans conductance (o/p diff current/input diff voltage) is 0.5*Gm.

Thanks for you help Aron,
Raj.

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A Kumar R

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Re: Common mode and diff mode exitation of current
Reply #8 - Nov 6^th, 2015, 5:40am

Hi Raja,

One basic question about your last post.

You said diff. voltage of Vin..that means single-ended voltages would be...+Vin/2 & _Vin/2.

If you give these voltages to a normal diff. pair, it will produce io1 = +Vin/2*gm & io1 = -Vin/2*gm..

The output diff. current is io1-io2 = gm*Vin & the input diff. voltage is Vin.

So, the effective Gm is gm itself.

Where did you get 0.5*gm?

Bye

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raja.cedt

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Re: Common mode and diff mode exitation of current
Reply #9 - Nov 8^th, 2015, 1:16pm

Dear Anil--
I didn't understand The output diff. current is io1-io2 = gm*Vin & the input diff. voltage is Vin.. Basically diff current is absolute value of (Io1,Io2) rather I01-I02. This is a very subtle concept, but it's not difficult to understand on your Won. Basically we have been spent years of analysing Diff voltage, so we all tend to use same convention by mistake.

Thanks,
Raj.

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