INA240-Q1: About the output impedance of REF1 and REF2

Noritomo Harukawa

Intellectual 990 points

Part Number: INA240-Q1

Q1.

INA240-Q1 is used in the following configuration.

Resistance values of R1 and R2 are 2.2kΩ.
In this case, how much does it affect CMRR and gain error?

Q2.

I found the following E2E.

From this answer I understood the internal resistance value of INA240-Q1 as follows.
In this case, are the output impedance of REF1 and REF2 correct with 50kΩ + 125kΩ?

-Harukawa

over 7 years ago

0 Peter Iliya over 7 years ago

TI__Mastermind 32395 points

Noritomo,

Our expert and I are going over all this now and trying to clarify all of the information. We will respond shortly.

Best,

Peter Iliya
Current Sense Applications

0 Guang Zhou over 7 years ago in reply to Peter Iliya

TI__Mastermind 33130 points

Hello Harukawa,

Q2 – The 125K Ohm resistance highlighted in the diagram really should be 500K. Let’s call it Rf.

Q1 –The divider appears as parasitic resistor to the INA, and it breaks the balance of the original resistor network, as a result both CMRR and Gain Error will be affected, the magnitudes depend on the resistor values of the divider. It should be as small as possible.

The sampling configuration above cancels the error contribution from Vs, otherwise introduced by the divider. In other words, if single ended sampling were used, the common mode error would be bigger.

The picture is a very simplified block diagram. The real circuit has a multistage architecture. The output referred common mode error is found to be R1/(2*Rf)*2.5V. And the output referred differential error term, neglecting other 2^nd order terms, is ~R1/(4Rf)*Vdiff*20V, where Vdiff is the differential input voltage.

With the resistor divider you selected, the numbers work out to be

Common mode error = 2.2/(2*500)*2.5=5.5mV;

Differential error = 2.2/(4*500)*Vdiff*20=0.022Vdiff.

The above error terms are identical for all 4 gain versions, therefore when referred to input, A1 would be the most impacted as it has the lowest gain.

500K is nominal value; I would further substitute it with 400K to be conservative.

Regard, Guang

Apps Support-Current Sensing

0 Noritomo Harukawa over 7 years ago in reply to Guang Zhou

Intellectual 990 points

Thank you for your reply.

I understood that CMRR and gain error can be calculated by the following formula.

CMRR: R1/(2 * Rf) * 2.5V
Gain Error： R1/(4 * Rf) * Vdiff * 20V

Q1.
> The 125K Ohm resistance highlighted in the diagram really should be 500K.

Regarding 500 kΩ, which of figure a or b is correct?

*Figure a

*Figure b

Q2.
How much is the internal resistance of INA240-Q1 respectively?

-Harukawa

0 Guang Zhou over 7 years ago in reply to Noritomo Harukawa

TI__Mastermind 33130 points

Hello Harukawa,

Q1 – Figure “a” is correct.

Q2 – a)=500K, 200K, 100K and 50K for A1/2/3/4 respectively. b)=475K; c)=50K

Regard, Guang

Apps Support-Current Sensing

0 Noritomo Harukawa over 7 years ago in reply to Guang Zhou

Intellectual 990 points

Thank you for providing information on the internal resistance value.

I am so sorry that I have so many inquiries.
I have two questions below.

(1)
I got an answer about the Common Mode Error formula before.

　(R1/(2*Rf))*2.5 (V)　　　　　note:Rf=500

Q1.
Could you tell us about the calculation formula of Common mode rejection error?

(2)
From the previous answer I understand that the output impedance of REF1/2 is 500 kΩ.

Q2.
Is the output impedance of "OUT" correct at 500kΩ?

-Harukawa

0 Guang Zhou over 7 years ago in reply to Noritomo Harukawa

TI__Mastermind 33130 points

Hello Harukawa,

Not a problem at all.

Q1 – the derivation is rather lengthy and not suitable for a post, but is straightforward otherwise given you have all the resistors value. The only thing to note is that the common mode voltage of the output stage is 2.5V assuming a 5V supply. This is where that 2.5V in the equation comes from.

Q2 - No, the output impedance is different than the resistor values. What we’ve been discussing are “resistor values”, not output impedance. I believe you had a previous post specifically addressing this topic. But the feedback resistor connected to the OUT pin is indeed 500K nominal, if this is what you're asking.

Regard, Guang

Apps Support-Current Sensing

0 Noritomo Harukawa over 7 years ago in reply to Guang Zhou

Intellectual 990 points

I'm sorry to bother you over and over.
The product that the customer uses this time is INA240A2-Q1.

Q1.
I got an answer of internal resistance = 500 kΩ.
But from the customer this value is the opinion that it is the value of INA240A4-Q1.
(The customer seems to have referred to the following E2E contents)
e2e.ti.com/.../559241

The internal resistance of INA240A2-Q1 is 500 kΩ.
Is this correct?

-Harukawa

0 Guang Zhou over 7 years ago in reply to Noritomo Harukawa

TI__Mastermind 33130 points

Hello Harukawa,

You're correct, only A4 feedback resistor is 500K, and the corresponding value for A1/2/3 are 50K, 125K and 250K respectively.

Regard, Guang

Apps Support-Current Sensing

0 Noritomo Harukawa over 7 years ago in reply to Guang Zhou

Intellectual 990 points

Thank you for your reply.

I understood that the internal resistance of INA240A2-Q1 is 125 kΩ.
(I understood it with the contents of the attached file.)

Q1.
If this is correct,
Rf of the following formula is 125.
Is this correct?

CMRR: R1/(2 * Rf) * 2.5V
Gain Error： R1/(4 * Rf) * Vdiff * 20V

-Harukawa

INA240A2-Q1_Internal resister.pdf

0 Guang Zhou over 7 years ago in reply to Noritomo Harukawa

TI__Mastermind 33130 points

Hello Harukawa,

Yes confirmed, Rf=125K for INA240A2 and this value should be used in the above equations.

Regard, Guang

Apps Support-Current Sensing

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INA240-Q1: About the output impedance of REF1 and REF2