The circuit shows a differential amp used for thermocouples. It is a standard diff amp arrangement with a T network added onto the front of the feedback path.
The test says "It is just the standard differencing amp with the T connection in the feedback path to get high voltage gain (200 in this case) while keeping........:"
Why 200? Without the T network the gain would be 10. How do you get a
20X increase in gain by adding the T network?
What would the formula be for calculating the gain of such an amp? I looked in the chapter where difference amps are covered, but there is no mention of this circuit.
Formula? Smormula! Do the loop and nodal analysis!
Why does everyone think there's always an equation to plug into?
Fortunately it's good for business ;-)
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Well you then should leave electronics to someone else ;-)
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |
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| 1962 | I love to cook with wine. Sometimes I even put it in the food.
Maybe. Then again I lied a bit as I do know what it is. I do not know why I would need to use it in order to calculate the gain of a DC circuit (thermocouple amp).
Your comment is a bit like saying you should not design kitchen furniture unless you are capable of carrying out a detailed stress analysis of the propogation of shock waves in chipboard.
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
No. Customers expect the standard thickness. Experience shows this is a good compromise.
Back to the AoE problem. Winnie put filter caps and integrating caps on the design that are more than adequate for our thermocouple measurement. Now, why do I need to get into the phase domain in order to calculate the gain?
Why can I not get an equation I can use to caluculate the resistors I need for a given gain?
Or perhaps I should say why does my caluculated gain (110) not match the AoE figure of 200?
That's probably not entirely fair to Roger, Jim, as AOE specifically eschews even mentioning loop or nodal analysis. There's probably a good chance that Winfield or Paul used it to derive a few results without even thinking about it, given that it's usually taught in first semester EE classes!
Hmm, and perhaps I am getting mixed up in terminology here. In my first
**term** (we don't have semesters in the UK:) we did circuit analysis with kirchoffs laws and equivalent circuits, I thought Jim was referring to ac analysis with gain and phase plots etc.
Back to my problem, if I consider the case were the non-inverting side happens to be at 0V, the problem appears to be childs play, except that I get a gain of around 110 instead of the stated gain of 200. Where am I going wrong?
In the "Tee" network, Let R1 = the resistor connected to the inverting input. R2 = the resistor connected to the op-amp output. R3 = the shunt resistor. Rtee = the equivalent feedback resistance. Then Rtee = R1 + R2 + R1R2/R3. This also works for an input resistor, provided it's working into a virtual ground. Tee networks are used when you can't find a single resistor with a high enough value. Caution: The output error due to input offset voltage is larger with a tee network than it would be for a single resistor with the same equivalent value. When computing output error due to input bias current, the resistance "seen" by the inverting input is R1 + R2||R3, which is much lower than the equivalent feedback resistance. When computing output error due to offset current, use the equivalent feedback resistance. Regards, Jon
As others have pointed out, you need to do a proper nodal analysis of the circuit to derive the formula for the gain. In its general form, AoE's circuit looks like this (view entire message with a fixed-width font):
R4 is the bridging resistor. For this circuit, the formula for the gain is given below (I hope I haven't made a mistake...).
. Vo R2 + R3 R2 R3 . Av = --------- = --------- + 2 ------- . Vb - Va R1 R1 R4
You can get this by solving the resistor network for the op-amp's (+) and (-) inputs, with the usual condition that they should settle at the same voltage (classical feedback control theory). See Win's post for a quick'n'smart way of computing the gain.
Note the added term 2 * (R2 * R3) / (R1 * R4). Setting R4 to infinity (i.e., removing it from the circuit) yields the usual diff. amplifier formula Av = (R2 + R3) / R1.
Substituting R1 = 25k, R2 = 250k, R3 = 10k, and R4 = 1k we get
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