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Technique Using Artificial Capacitance to Frequency Compensate an OP AMP Internally

IP.com Disclosure Number: IPCOM000036691D
Original Publication Date: 1989-Oct-01
Included in the Prior Art Database: 2005-Jan-29
Document File: 2 page(s) / 41K

Publishing Venue

IBM

Related People

Braden, JJ: AUTHOR

Abstract

The circuit in Fig. 1 uses only the devices and nodes numbered. These are Q1 through Q7, R1 and R2. The remainder of the devices make up a standard high gain OP AMP connected in a non-inverting mode with a gain of approximately two. The OP AMP is included in this circuit as an aid to its understanding.

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Technique Using Artificial Capacitance to Frequency Compensate an OP AMP Internally

The circuit in Fig. 1 uses only the devices and nodes numbered. These are Q1 through Q7, R1 and R2. The remainder of the devices make up a standard high gain OP AMP connected in a non-inverting mode with a gain of approximately two. The OP AMP is included in this circuit as an aid to its understanding.

Q1, Q2 and Q3 form a current mirror, and Q4, Q5, Q6 and Q7 also form a current mirror but of an opposite polarity. These current mirrors are configured to be temperature-compensated and track each other with power supply changes.

Node 3 has equal and opposite current sources connected to it

(Image Omitted)

from a DC aspect. R2 is specifically a high resistance (40 K ohms) to isolate the AC signal at the base of Q6 from the other components. In other words, R2 supplies the required DC current to the base of Q6 but is sufficiently large to isolate the AC signals.

Node 3 appears in the simplified circuit in Fig. 2. Node 3 is the frequency- compensated node; it appears as a current source connected to a capacitor to ground. When the OP AMP attempts to cause a transient on node-3, a current flows through Cc-b of Q6 and Cc-b of Q7. Almost all of the current through Cc-b of Q6 goes into the base of Q6 because R2's large resistance blocks it. The current into the base of Q6 is multiplied by beta squared and is drawn from node 3 into the collectors of Q6 and Q7 because they are connected as...