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.
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J
Jim Thompson
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.
R
Roger
Maybe if I knew what the loop and nodal analysis thing was I would not be asking for a formula ;-)
J
Jim Thompson
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.
R
Roger
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.
J
Jim Thompson
Is that not so ?:-)
...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.
R
Roger
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?
J
Joel Kolstad
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!
R
Roger
But not to the point. My problem is that I do not get the same answer. Is it a mistake or am I doing something wrong!
R
Roger
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?
F
Fred Bartoli
'coz electronics is black magic... ...and there are always formulas involved in magic.
Thanks,
Fred.
J
Jon
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
C
Costas Vlachos
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