Re: Op-Amp Design - Stage 2 Current to Voltage
From: Charles Schuler (charleschuler_at_comcast.net)
Date: 12/28/04
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Date: Mon, 27 Dec 2004 22:40:15 -0500
"Monty Hall" <chickenkungpao@hotmail.com> wrote in message
news:fXUzd.7010$by5.1639@newssvr19.news.prodigy.com...
> I'm reading Tom Frederickson's "Intuitive IC Op Amps" - in particular pg
> 15
> concerning the op-amp schematic, and am not sure how the second stage of
> the
> op amp works. This stage that takes the single ended current and converts
> it a voltage before it's off to the output stage.
>
> The current input goes to the base of a transistor whose emitter is
> connected directly to ground and the freq compensating capacitor is placed
> across the collector and base. How does this configuration work -
> especially if the base input current is negative? Current is sourced from
> cap?
>
> Horowitz and Hill's 741 schematic has a 300 ohm resistor on the base
> input.
> Is this where the current to voltage conversion takes place and was
> omitted
> by Frederickson? In either case, still not sure where current comes from
> when current mirror sinks current in stage 1.
>
> If the sinking current mirror pulls from the cap, I would expect the
> mirror,
> when sourcing current, to load the cap by symmetry. But in a sourcing
> configuation, the transistor is now forward biased. If current is being
> sunk
> @ constant rate, wouldn't the magnitude of the voltage across the cap
> ramp wrt time? I expected the output voltage to be proportional to
> input differential voltage and not ramp in an open loop config.
>
> Can somebody explain how current is converted to voltage in stage 2 of an
> op-amp?
Don't have the book you referenced but the most simple op-amp current to
voltage converters feed the current into the - input and the negative
feedback resistor determines the scaling. For example, if the feedback
resistor is 1,000 ohms, the output of the op-amp will swing one volt per mA.
To easily understand this, you just assume the current into the - input of
the op-amp is zero and that the - input is a virtual ground. This is based
on the assumption that the + input is at ground potential and that the
differential voltage is 0 (a valid assumption since the open-loop gain is
typically 100,000 or more). So, the output of the op-amp will respond to
supply (sink) the input current. In this example, 1 V at the output will
cause 1 mA to flow in the feedback resistor (Ohm's Law).
The prime basis of understanding many op-amp circuits is that the
differential input voltage is very small and can be assumed to be zero when
the device is in a linear mode. This is where the virtual ground concept
comes from. If the + input is at ground, then the - input must be very
close to the same (again, assuming a linear condition). When an op-amp is
over driven (to the rails and beyond) this assumption is no longer valid.
Multiple emitter transistors are used in several ways ... one of which is RF
power transistors.
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