I'm studying this circuit:
The author of my book says that the purpose of this circuit, called "Self Biased Current Reference", is to generate a current almost independently of the supply voltage. He also says that, if M3 and M4 are perfectly equal, than I1=Iref. Then it follows that:
from which I can find VGS1 and I1.
Question 1: When the author says that the current mirror is perfectly matched and then I1=Iref, isn't it too much simplified? He neglects channel lenght modulation, but in reality the source-to-drain voltages of M3 and M4 will not be equal (because we have just M1 under M3, instead we have M2 and a resistor under M4)
Question 2: Why is this current almost independent of VDD? The author says:
"The value of the generated current weakly depends on the supply voltage
because there is a high impedance element per branch capable of absorbing
possible supply variations. These two elements are the transistors M2 and M3.
The drain to source voltage of M1 and M4 can not change freely: the former is
two VGS above ground the latter is one VGS below VDD. Any supply fluctuation
is then “absorbed” by the high resistance that we have between drain and
source of M2 and M3"
I don't understand the strong part of the text: why is any fluctuation “absorbed” by the high resistance that we have between drain and source of M2 and M3?
Thank you
Best Answer
1) Yes, channel length modulation must be ignored to get the \$I_1 = I_{Ref}\$ relationship. Otherwise, as you correctly point out, the currents will depend on the voltage dropped across the remainder of the two branches of the circuit.
Channel length modulation can only be ignored if the transistors stay in saturation. If not, the drain current will depend on the drain-source voltage. But as long as \$V_{dd}\$ is high enough and \$I_{Ref}\$ is low enough, the drain-source voltage and consequently the gate source voltage (due to the gate-drain tie) will be high enough to keep the transistors in saturation and the mirror function intact.
2) While the mirror function is due to M3 and M4, the constant current source function is due to M1 and M2. As the equations indicate, the current is dictated by the relatively constant \$V_{Th}\$ and \$R\$. So how is the current kept constant, even as \$V_{DD}\$ varies?
As \$V_{DD}\$ rises, something must "take up" the excess voltage. The author uses "absorb" instead of "take up", which is fair because as well as providing the additional voltage drop, the extra power must also be absorbed, but it's the increase in voltage drop that is important here. M1 and M2 cannot perform this function, because as stated their drain to source voltage is relatively fixed, so it is up to M2 and M3.
Since the current mirror is enforcing \$I_1 = I_{Ref}\$, and the current source is enforcing constant \$I_{Ref}\$, additional voltage due to \$V_{DD}\$ is dropped over the drain to source of M2 and M3. This will remain intact while ever M2 and M3 can continue to produce large enough drain to source voltages. MOSFETs are capable of producing high drain to source "resistance", so the circuit can be made stable.