Saran said:
So, let me restate what I understood. The current mirror's output
current is used to charge the gate of the transistor. So, depending on
the magnitude of the current, the gate voltage increaser faster or
slower (sorry about my layman language, I just want to be very very
clear).
If I understood this correctly, then:
1) If a current mirror is charging the gate of a transistor, or in
other words, changing its Vgs, the maximum Vgs value is determined by
the saturation voltage for the mirror.
2) We are controlling the rate at which gate is charged in a circuit.
Overall, I'd say you didn't understand it, and it wasn't explained real
well to you, but maybe I just don't understand what you and others have
said.
First off, MOSFETs are generally considered to be voltage controlled
current sources, at least when operated in the saturation regime (Vds >
Vgs - Vt). BJTs on the other hand are current controlled current sources.
Each of these transistors have respective advantages.
In analog circuitry you often deal with both voltage and current, and
since MOSFETs are voltage controlled current sources its important to
consider all aspects. MOSFETs are complicated devices, entire books have
been written simply discussing aspects of how MOSFETs work. To try to
boil them down into one or two sentences will surely leave you with an
incorrect understanding. Even the most cursory treatment of them in an
engineering text book would be at least several pages long.
The purpose of a current mirror is to duplicate (or mirror) a current with
some sort of multiplicative factor (M/N). If you are following good
layout practices, M and N should both be integers. Let's say we have a
simple N-channel mirror, where M and N are both 1. So we have two
identical transistors (M1 and M2). M1 and M2's gates are both tied
together and tied to the drain of M1. M1 and M2's source is tied to
ground. A current, I1 is pushed into the drain of M1. M1 is said to be
"diode connected" because its drain and gate are shorted together.
Now, let's ignore M2 for the moment. If we take M1 to be an ideal
transistor, no current will enter the gate of M1, so all of I1 must enter
the drain. The gate however is capacitive, as such its voltage can be
increased by pumping current in, or decreased by pumping current out.
The gate has access to a current source (I1) to increase or decrease the
gate voltage so as to allow all the current to flow through the drain.
It should be obvious that as its adjusting the gate voltage the current
through the drain will not perfectly match I1, but it should get closer
and closer over time. Eventually the gate voltage will be forced to
settle to whatever voltage will produce drain current equal to I1. Assume
for the moment that M1 is biased into saturation:
Id = K * (W/L) * (Vgs - Vth)^2 * (1 + Lambda*Vds)
Lambda is the channel length modulation and is often quite small. It can
intentionally be made small by increasing L (gate length). K and Vth are
constants that are fixed by the processing. W and L are the gate width
and length respectively and are chosen by the designer.
Now, let's add M2 back into the circuit. M2's gate is shorted to M1's.
M2's source is shorted to M1's. If we look at the Id equation, assuming
identical W/L, the current through the drain of M2 should closely match
that of M1. In real designs the designer may use different W's for M1
and M2 in order to achieve the multiplicative effect I previously
mentioned.
I don't think all that much thought is given to the amount of current
required to charge the gate of the MOSFET, unless speed is a concern in
which case there are much more advanced and accurate topologies that can
be used. Typically current mirrors are used for providing a static bias
current to multiple points within a circuit based on a single reference.
In this case, the reference rarely moves, and speed simply isn't a
concern, and hence no one thinks about the charge rate. There are other
instances when an AC signal is being mirrored in which case it may be more
of a concern however.
So the current mirror's *INPUT* current is used to charge the gates of the
mirror transistors. The output current could be used for a variety of
things depending on why a mirror was even employed to begin with.
Regarding your assertion about the maximum voltage... The only criteria
on a current mirror and voltage is that the transistors (both of them)
must remain in saturation (i.e. Vds > Vgs - Vth). If the transistors
slip into triode region, then the transistors cease to act as very good
current sources, and start acting more like resistors, as such the mirror
becomes far less effective if the loads on the mirror branches differ
substantially.
dan