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schematic

simulate this circuit – Schematic created using CircuitLab

I've tried to Google it but my keywords don't really yeild anything.

I'd like to read more about this type circuit ; what applications its useful in ? How to select C1 ? From what I recall, it provides a DC gain of 1, but provides an AC gain set by the feedback resistors.

Does it have a name ?

Added I should have emphasized that the focus for this question is C1 and its location in the circuit.

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Yes - it has a name. In control theory this circuit is known as a PD-T1 unit. It has a proportional-derivative behaviour with a certain delay term T1. In filter terms, it works like a first-order high-pass with a superimposed constant gain.

The transfer function is \$H(s)= 1 + sR1 \cdot \dfrac{C}{1+sR2C}\$

This device is used to enhance the phase (for stabilizing purposes) in a certain frequency range. Please note that application as a PD-T1 element requires \$R1>R2\$.

More than that, the shown circuit is used as a simple non-inverting amplifier (gain: \$1+R1/R2\$) for single-supply operation. For this purpose, the non-inv. input is dc biased with 50% of the supply voltage - with the consequence that the input signal must be coupled via an input capacitor. Because the dc gain remains unity, the bias voltage is transferred to the output also with the gain of "1".

BODE diagram: The magnitude starts at unity and begins to rise at \$wz=\dfrac{1}{(R1+R2)C}\$, then it stops rising at \$wp=1/R2C\$ at a gain value of \$1+(R1/R2)\$. The rising of the gain is connected with a corresponding phase enhancement.

Because of the mentioned phase enhancement properties the PD-T1 block is also known as a "lead controller".

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  • \$\begingroup\$ It is utterly incredible that anyone would downvote this answer. It either shows malicious intent or a complete lack of engineering understanding (or both). \$\endgroup\$ – Russell McMahon Mar 4 '15 at 0:45
  • \$\begingroup\$ Russell, thanks for editing. Question: Generally, is it preferred to correct/edit the text directly rather than to add something called "EDIT" (as I did) ? \$\endgroup\$ – LvW Mar 4 '15 at 9:01
  • \$\begingroup\$ preferences vary. I felt this was a somewhat special case. Usually I would not have made an edit like that. I especially wanted to remove your comment re a correction as mistakes happen and once you had the correct formula in place the old text was history - AND some [poor benighted soul] has downvoted your answer - which is ridiculous - so I wanted to minimise any opportunity for specious justifications. Usually I stick to the technical stuff :-) \$\endgroup\$ – Russell McMahon Mar 4 '15 at 12:36
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I'd just call that a non inverting amplifier.

Calculating the transfer function is quite easy if we can consider the op amp ideal.

In DC C1 is open, so you don't have current in R1 nor R2, so the op amp is in the buffer configuration and the gain is 1.

When f gets very big C1 is closed, the gain of the circuit is the usual 1+R1/R2 leading to a 2 gain for your values.

You then expect a finite pole and a finite zero, the zero comes first than the pole kicks in. You can calculate the pole with the "seen resistance" method: C1 sees R1+R2 so \$\omega_p=\frac{1}{(R_1+R_2)C_1}\$. You can now calculate the zero pulsation as \$\omega_z=\omega_p\frac{A_0}{A_\infty}=\omega_p\frac{1}{2}\$

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  • \$\begingroup\$ This is the most direct and correct answer +1. \$\endgroup\$ – Olin Lathrop Mar 2 '15 at 13:54
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    \$\begingroup\$ I want the "you made Olin happy" badge :P \$\endgroup\$ – Vladimir Cravero Mar 3 '15 at 7:18
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    \$\begingroup\$ I can't imagine why Olin commented as he did, given that LvW and Vladimir's answer are both of good quality. LvW's gives the actual transfer function, advises the name assigned to the configuration in some environments, and comments on general response with frequency. Both answers are good and should be left to stand n their own merit. If answers are manifestly inferior but being upvoted it may be worth commenting but in cases like this Olin has already had his +1, some moron (I assume it was not Olin) has downvoted LvW's answer and patronage amounts to attempting to sway the system. \$\endgroup\$ – Russell McMahon Mar 4 '15 at 0:43
  • \$\begingroup\$ People just hate, there's little we can do about it... \$\endgroup\$ – Vladimir Cravero Mar 4 '15 at 6:06
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There is no generic name for it I believe. It has unity gain at DC and, at some point in the spectrum the gain will have risen to 2. This is dictated by: -

High frequency gain = 1 + R1/R2

The frequency where the gain is nearly 2 (the 3dB point) is determined by R2 and C1. When the reactance of C1 equals R2 this is the 3dB point and the reactance of the capacitor is: -

Xc = \$\dfrac{1}{2\pi f C}\$ = R1

therefore f = \$\dfrac{1}{2\pi R_1 C}\$

For your values this is 1591 Hz.

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