DC shift to the fully differential amplifier circuit

Hello,

My differential amplifier work on the condition of the input common mode voltage is equal to the output common mode voltage, since I am using 3.3 V it means my amplifier work very fine when I set VCM_in = VCM_out = VDD/2 = 1.65 V

However, I am facing a problem to interface a signal with different input common mode voltage, say like 1 V or whatever,
is there a circuit that can shift the DC component of the input voltage to the desired level?
I remember such circuits in basic amplifier circuit where RC circuits are used to block the DC component of the input signal and fixing the amplifier biasing voltage to the desired level.

I do appreciate your help
Thank you
Best Regards

Hi,

Doesn´t any diff_in diff_out amplifer do what you want?
No DC blocking needed.

Please show your circuit and the exact type of the diff amplifier.

Klaus

Most fully differential amplifiers will have some form of output common mode control. One with an externally accessible feedback reference might give you what you need, within reason. Op amps lose gain and pick up offset as they approach either rail.

Capacitor blocking can level-shift AC signals. Resistor dividers can translate DC up or down. Both have a modest cost to signal amplitude / phase / frequency response.

The next stage with the lower input common mode range might be put on a diode pedestal perhaps, depending on whether it's a piece part with independent supply pins, or integrated so it's "stuck", and what -its- successor wants to see....

You can try pmos source follower to go from 1v to about once Vgs higher. This will be a fully DC coupled solution.

AC coupling is possible. The lower corner had to be lower than the smallest frequency. So AC coupling is usually avoided in base band circuits.

You can even try common source stage if the swings work out.

Dear friends,

Thank you very much for your reply

here is the circuit I am using from the paper "
A CMOS Fully Balanced Differential Difference Amplifier and Its Applications"

Below is the schematic circuit



and here How I am connecting it as proposed by the paper, where the first connection proved me buffer gain non inverting fully differential amplifier and the latter one provide me the non inverting non inverting amplifier



By comparing the schematic to the circuit connection, you will find that always one of the differential pair input (the circuit has two differential pairs) is connected to the output side. Since the output is referenced to the VCM which is VDD/2 in my case, it means the pair will onl ybe palanced if the common mode voltage is equal to VDD/2, otherwise if I select it different it will be obvious that the part of higher vcm will be fully saturated and the other one is off and the complete circuit fail.

According to this, the circuit work perfectly to readout sensor interface from balanced whetstone bridge with VCM = VDD/2, but can not work for example to read the voltage current for current sensing and power management application.

Hi,

Good pictures, that exactly show what I mean:

The big schematic of post#5 shows the input "Vcm" (most right)
But the other schematics don´t have this input. So your pictures don´t match.

--> I recommend to use a diff amplifier as shown in the big picture and apply the desired Vcm voltage.
Then you neither need input capacitors nor output capacitors.

Klaus

KlausST said:

Hi,

Good pictures, that exactly show what I mean:

The big schematic of post#5 shows the input "Vcm" (most right)
But the other schematics don´t have this input. So your pictures don´t match.

--> I recommend to use a diff amplifier as shown in the big picture and apply the desired Vcm voltage.
Then you neither need input capacitors nor output capacitors.

Klaus

Click to expand...

Dear Klaus,

Thank you very much for your help
Before I go into your solution let me clear one thing to you as you already presented,
The VCM on the big schematic post5 is the common mode voltage that we apply externally to the diff amplifier and it is not shown in the second circuit plot. This VCM is usually fixed to the VDD/2 as ADC requires and we have control over it.

However, we don't have a control over the common mode input voltage which is basically (vin+ + Vin-)/2, this component comes from the sensor reading like bridge for example.

The presented circuit only work when the VCM in = VCM out = VDD/2

Therefore I didn't understand what you mean by the "apply the desired Vcm voltage"

Thank you
Best Regards

Hi

Junus2012 said:

Therefore I didn't understand what you mean by the "apply the desired Vcm voltage"

Click to expand...

If you want Vcm to be VCC/2 then just do it. Connect Vcm to VCC/2

****
But I´m confused. Which schematic relates now to your problem?

Klaus

KlausST said:

Hello klaus

Click to expand...

back to you again again

let us discuss about this one below, assume that VCM of the CMFB circuit is internally connected to VDD/2 and not shown on the schematic this terminal

Now please refer to the inputs, if those are coming from sensor with different common mode voltage than VDD/2, it will unbalance the internal differential pair and the DC operating point will be lost.
may be my understanding about the common mode voltage is wrong

thank you

KlausST said:

If you want Vcm to be VCC/2 then just do it. Connect Vcm to VCC/2

****
But I´m confused. Which schematic relates now to your problem?

Klaus

Click to expand...

Hello

Hi,

I think I don´t understand what´s your exact concern.
If Vcm (usually the ouput_Vcm) is connected to VDD/2 then I expect the output_Vcm to be VDD/2.
It does what it should do.

And output_Vcm does not depend one input_Vcm at all ... on a fully diff_in diff_out amplifier. That´s the benefit of it.

Input_signal_DC is not not affected and will be transferred as output_signal_DC (both differentially).

Indeed the common_mode voltage (input as well as output) is not important and does not need to be accurate nor precise ... it just needs to be in a range that ensures proper device operation.

***
Some example with values and the schematic of post#9:
Let´s say: VDD = 5.0V, Gain = 2.
if Vip = 1.1V and Vin=1.0V then I expect Vop = 2.6V and Von = 2.4V
if Vip = 2.1V and Vin=2.0V then I expect Vop = 2.6V and Von = 2.4V
if Vip = 3.1V and Vin=3.0V then I expect Vop = 2.6V and Von = 2.4V
all have a DC input voltage of 0.1V and output voltage of 0.2V. All output Vcm is VDD/2 = 5V/2 = 2.5V

Klaus

Hello Klaus,

Thank you very much for your explanation,

I have tried your numirecal values and I will tell you exactly the problem

Since I am working with 3.3 V, I have set the VCM to the CMFB =1.65V, hence the output is referenced to 1.65 V

Now take the value if Vip = 1.1V and Vin=1.0V, I have tried this and my circuit as I am facing failed to give Vop = 1.75 V and Von = 1.55V as you assumed.

Back to the big schematic of post5 and to the circuit connection of post 9, you will see that one input of the differential pair for both pairs are connected to the output voltage that is referenced to 1.65 V. The other terminal is connected to the input

The common mode voltage between 1.1 V and 1 V is 1.05 V which is far below 1.65 V, leading eventually to pairs to saturate due to the difference between the input common mode voltages , not due to the differential.

in order to confirm my assumption, I have changed the VCM to 1.05 then the result came as per your calculation.

I hope I made it now more clear to have your further discussion

Thank you a lot
Regards

KlausST said:

Hi,

I think I don´t understand what´s your exact concern.
If Vcm (usually the ouput_Vcm) is connected to VDD/2 then I expect the output_Vcm to be VDD/2.
It does what it should do.

And output_Vcm does not depend one input_Vcm at all ... on a fully diff_in diff_out amplifier. That´s the benefit of it.

Input_signal_DC is not not affected and will be transferred as output_signal_DC (both differentially).

Indeed the common_mode voltage (input as well as output) is not important and does not need to be accurate nor precise ... it just needs to be in a range that ensures proper device operation.

***
Some example with values and the schematic of post#9:
Let´s say: VDD = 5.0V, Gain = 2.
if Vip = 1.1V and Vin=1.0V then I expect Vop = 2.6V and Von = 2.4V
if Vip = 2.1V and Vin=2.0V then I expect Vop = 2.6V and Von = 2.4V
if Vip = 3.1V and Vin=3.0V then I expect Vop = 2.6V and Von = 2.4V
all have a DC input voltage of 0.1V and output voltage of 0.2V. All output Vcm is VDD/2 = 5V/2 = 2.5V

Klaus

Click to expand...