Why won't this new current mirror work?

Question

This basic current mirror is well known:

It has its flaws, such as failure to deal with channel length modulation since \$V_{DS_1}\$ is not same as \$V_{DS_2}\$. As well as, external circuity above \$M_2\$ may cause significant swing in \$V_{DS_2}\$

To counter the first flaw, I propose this circuit:

This circuit will ensure that \$V_{DS_1}\$ is equal to \$V_{DS_2}\$ at all times, ensure that channel length modulation would not cause any issue when \$I_1\$ is copied over. (Assuming that the two MOSFETs are identical).

I couldn't find this rather obvious solution in any standard textbook. What is wrong with it, other than the fact that external circuitry may mess with \$M_2\$?

Answer 1

You have misunderstood the components of the current mirror.

The reason why M1 is diode connected is to produce the right gate voltage M1 needs to pass I1.

We let the loop figure out the right bias conditions

Since M2 should be identical to M1, then M2 should behave as a current source with a current of I1 flowing through it. In other words, M2 in series with some load (a common sourced transistor, for example) must ensure I1 is flowing through its branch. How are you going to ensure that with your circuit?

With your circuit, now M2 behaves more as a voltage source.

Why? Because, as you've drawn it in the last picture, M2 is now in parallel with M1. It's as if you have doubled the size of the transistor in IC design.

Answer 2

The arrow in the middle of the symbol indicates the substrate connection and PN junction.

You've made a PN diode on the left-hand side.

Put another way: notice one MOSFET is upside-down the other.

Answer 3

It's not a new current mirror, it is just a short circuit.

The 1A current is pushed through both transistors and whatever voltage there may be or currents flowing in the circuit, it provides bias voltage and current to the drain of M2.

Answer 4

Philosophy of the current mirror

Initially, the concept of replicating a current using a current mirror might seem redundant. After all, why duplicate a current when we already have one at our disposal? This is similar to questioning the need to duplicate a voltage source.

Here are a few reasons why we use them:

• A current mirror is a perfect current-to-current converter whose input maintains a constant voltage and whose output maintains a constant current. Therefore, we can drive it with an imperfect (resistor) current source and connect a varying load to its output.

• A current mirror reverses the direction of the current. According to our understanding, if a current enters a device, it should then exit it. However, with a current mirror, both the input and output currents are either entering or exiting.

• The current mirror has very low voltage losses across its output transistor (very good compliance voltage).

How to make current mirror

A transistor (FET, BJT...) is a voltage-to-current converter. If we connect the drain to its gate, we force it to start acting as the inverse current-to-voltage converter. By connecting a "reverse" and "normal" transistor one after the other (cascading), we obtain a current-to-current converter or current mirror. Interestingly, although the converters individually are nonlinear, the resulting converter is linear. The reason for this is that their transfer characteristics are the exact opposite of each other.

"Old" current mirror

Circuit

It consists of four parts:

Input current source: I have set the input current by a simple resistor-type current source with the odd name Ri=1k. It is built by the supply voltage Vdd and an imperfect ammeter with an internal resistance of 1 kΩ. Combining a resistor and an ammeter into something like a "visualized resistor" simplifies the circuit (see my related question and answer).

"Reverse" transistor: M1 adjusts its Vg (Vd) voltage to allow the entire input current to flow through. In this way, current and voltage swap roles – the drain current becomes an input quantity and the gate voltage becomes an output. Since this configuration keeps a relatively constant voltage (like a diode), it is known as a "diode-connected transistor".

"Normal" transistor: M2 changes its drain (output) current according to its gate (input) voltage in accordance with its "forward" transfer characteristic.

Load: To simplify the schematic, I have used the same "resistor + ammeter" trick by combining a variable resistor and an ammeter into something like a "visualized load".

Operation

Let's examine the circuit with different load resistance values to see how good a current source (sink) it is.

RL = 1 kΩ: We can start with RL = Ri = 1k. We see that the circuit is completely symmetrical - not only the currents but also the voltages are equal.

^{simulate this circuit – Schematic created using CircuitLab}

RL = 500 Ω: Now let's decrease the load resistance ("increase the load"). M2 reacts to this disturbance by increasing its Rds "static" resistance. The sum of the resistances of the two elements, and consequently the current, hardly changes. Only the voltage drops across them change - Vd2 increases and VRL decreases.

^{simulate this circuit}

"New" current mirror

I will consider your invention under the assumption that flipping M1 vertically is a typo. Then the only difference between it and the "old" current mirror is in the connection in red.

RL = 1 kΩ: As you can see, the circuit's two halves are in parallel. So, as in Schematic 1.1 above, both currents and voltages are equal.

^{simulate this circuit}

RL = 500 Ω: The difference between the two configurations manifests itself when we start to change the load resistance.

^{simulate this circuit}

Since it is now connected to a voltage rather than a current source, at very low values of the load resistance, unrealistically high values of current are obtained. Of course, this only happens here in the simulator; in real circuits, currents are limited.

Answer 5

In your circuit, the current may now be equal through the two transistors, but that alone isn't very useful.

Now the current through the load could travel through M1 or the current through the current source could travel through M2. So now your load could draw any amount of current it desires and this current will evenly distribute through the two transistors (in superposition with the current from the current source being evenly distributed between the two transistors). Another possibility is that the current source could push current through the load in the reverse direction without it going through the transistors at all.

So yes the two transistors have the same current, but you're now unable to control the current through the load at all. You want the transistor M2 to act like a current source for other parts of the circuit, not a current "consumer".