Gain mV/Volt

The equation 1000mV = 1V / 20 = 50mV * Shunt mV(.002) = 100µV * 20 = 2mohm works forwards and backwards to satisfy 20x gain = 1V/V for 2mohm shunt.

Sort of the point for reducing V/V into mV/mV to compare shunt mV apples. So gain = 20 rather than 20V/V which makes no sense to shunt shoppers. I have questioned A1 mV/A output seems greater than simply multiplying 20 * shunts(mV).

Table-1 SBOS662B 15mohm shunt 20V/V is closer to 300mV/A, not 150mV/A. For example a 7mohm shunt x13 gain should produce 91mV/A. 

Hi,

Your comment about the way Gain is specified is noted.

This is how it is done according to industry convention. As a user, you can use any equivalent form that makes sense to you as you please.

Regards, Guang

Guang Zhou said:

However this is how it is done according to industry convention

So the 2mohm shunt for 20V/V=50mV/A or 40mV/A? That was the point of the post as it seems other amplifiers only have xN gain value and not V/V.

Is all a bit confusing as Table 1 seems to make the same mistake?

To make matters worse the scope probe V/A appear to require Peak or 100mV/A. Otherwise the readout Amp scale values are ridiculously high. The scope vendor will not comment on this undocumented channel probe setting but it seems to be Peak only, not PTP. So who would know that unless they say somewhere or we know the exact V/V gain for 2mohm shunt.

A volt is a volt (1 volt is implied) and a millivolt is a fractional value of a volt. Volt per volt means, 1 volt = 1 volt relative to gain x20 or 1:1 ratio of 1000 millivolts. Again both sides of the equation have to be equal prior to the math calculation as taught in basic algebra and electronics. So the use of the symbol infers a reduction must take place in order to down convert the volt gain into millivolt gain. 

Unless electronics theory has suddenly changed, Ohms law E=I/R. Using the 20V/V symbols infers amplifier gain 20 was determined at 1 volt, not 1 millivolts. Seemingly the shunts millivolt amplifier gain is 20/1000 = 0.02/1mV gain from applied shunt differential (+/-IN), not simply x20 gain at 1 volt applied.

Why even 20V/V if it has no significance in factoring amplifier gain? Would it not be less confusing to simply state amplifier has x20/mV gain, does it?

The point of 20*1V/1V gain has not been answered.

Hi,

I can only say how Gain of 20 (or any other for that matter) is defined and measured.

The gain is defined as the ratio of the output to the input while the amplifier is kept in its linear operation range. The unit “V/V” certainly doesn’t mean the input is 1V.

Regards, Guang

Guang Zhou said:

The unit “V/V” certainly doesn’t mean the input is 1V

That point is debatable sense the gain results of a 2mohm shunt 20V/V do not seem to produce 40mV/A, rather 50mV/A or more.

Guang Zhou said:

The gain is defined as the ratio of the output to the input while the amplifier is kept in its linear operation range

Seemingly lab engineers calculated the INA gain factors via 1 volt CM sinusoidal applied as +/-IN rated +/-80V CM. And V/V seems to suggest A1 gain of 20 was determined 1 volt peak sinusoidal wave into +/-IN. The symbolic Symantec 20V/V does not directly justify the actual gain behavior of INA +/-IN CM=mV as any other amplifier gain is a simple math calculation of (+INmV x gain). So the shunt formula is directly linked to the laboratory method of gain calculation.

Please present a formula TI suggests to use to match a millivolt shunt to INA V/V gain value. Again datasheet example is far to vague and seems to make a mathematical mistake complicating this issue.

A 10mV INA output difference is a huge shunt error percentage is it not? You did not answer how INA WEB calculator shunt error determined, in millivolts or volts gain? Obviously it makes a difference how the math formula determines error % in both cases. Your ratio linear operation range seems to suggest exactly what I am debating. We need to factor 1V into (1000mV) prior to multiplication x20 gain or there is a 10mV error in precision no matter what. The 10mV/A matters as the ADC full scale is directly factored from the INA gain expressed as V/V and not more proper mV/mV to match shunt E=I/R.

That seems to explain why the scope channel and current clamp both disagree with INA shunt gain math factored as 1:1 mV shunt values into +/-IN. There will always be a 10mV difference from real to imaginary if the ratio of Volts is not first translated to 1000mV for INA V/V from the shunts perspective of gain. If that is indeed true even the FE who wrote up INA datasheet was just a naive and suggested incorrect math information in the process. 

Hi,

The gain of INA240 is specified by the ratio of output voltage over input voltage.

Shunt resistor comes in many forms, and accuracy varies widely. It is impossible for us to take this into account. That’s why the gain is specified the way it is, ie, through voltage-voltage not current-voltage gain.

Because INA240 is a high precision device, the shunt will determine the system accuracy.

Regards, Guang

Guang Zhou said:

Shunt resistor comes in many forms, and accuracy varies widely

Yes they do come in a variety and precision (PPS/sec) yet every single shunt is rated millivolts, not volts. This thread is more about how to determine ADC full scale relative to INA V/V gain rating for any selected shunt resistance.  

Guang Zhou said:

The gain of INA240 is specified by the ratio of output voltage over input voltage.

Right but the V/V symbolism is misplaced by your synopsis and seems to hide certain necessary facts from the observer. If the V/V symbolism has no meaning relative to INA amplifier gain why was the number 20 placed in front of it? Again where are the TI formulas for designers to calculate shunt gain via millivolts relative to the V/V symbolism? Obviously there is more to the symbolism than even you are aware of.

Why not investigate why typical amplifier datasheets do not rate amplifier gain using V/V symbolism?

Guang Zhou said:

If you don’t like “20 V/V”, you can convert to anything that makes you comfortable - 20mV/mV, 20KV/KV, 20000mV/V, 0.02V/mV, 0.00002KV/mV…

Again the reason to convert is to keep both sides of the equation (V/V) the same scale as used to determine amp gain 20V/V, is not the same as millivolts scale. That dynamic changes the output by adding 10mV from the scale reduction of volts into millivolts. I'm looking for reason/equation that explains why simply multiplying shunt millivolts x20 ends up producing >ADC full scale by nearly 10mV/A or 50mV/A, versus 40mV/A using simple math.

Difference of 10mV/A is a huge value is it not?

Guang Zhou said:

Please send links to the “typical amplifier datasheets”

Newer data sheets such as OPA607 express gain in terms of (GBW) or 6-V/V and often include formulas to calculate any offset that occurs for given gain relative to VS. Such as TLV2460-Q1 section 8.3.2 has formula to calculate the output offset (Voo) that occurs for the value of Rg/Rs +/-IN signals.

Perhaps INA output has inherent offset adding 10mV/A no matter how REF1/2 are configured? So the INA 20V/V use for any shunt millivolts does not produce the expected mV/A output using the simple math calculation. When it seems to simple it likely is more complicated, Razors edge.

https://simple.wikipedia.org/wiki/Occam%27s_razor

Guang Zhou said:

If your “system gain” differs, there is often an explanation, especially if the deviation is big.

Yet a sine wave input on 2mV shunt with A1 seems to produce 50mV/A as the first formula I posted. I was thinking number 20 was a factor of gain at 1 volt applied, not 20 volts +/-IN. The part that had tripped me up was the volt per volt and 20 of them.

Sound like your saying 20 volts IN for 1 volt OUT, makes since to reduce 1v to 1000 millivolts. Then divide by 20 and the difference in gain equals exactly 10mV relatively.

That 10mV difference does not happen if we simply multiply shunts mV value x20. Discovered some time ago it reduced output by 10mV/A but didn't know why. Tina model seems to do a direct x20 multiplication as if 20x +IN current gain, 10mV difference had to be explained mathematically, Tina looses.