# Simple MOS current mirror bias circuit

Consider the basic MOS current mirror bias circuitry

I am really trying to understand on how Vbias when M3 is 'on' can be any other value other than Vt because M3 is diode connected (Vds= Vgs) and this means that the Vds of M3 will always be equal to Vt.

So all that I can vary is only the W of M3 to control the current sunk because Vds=Vbias=Vt is always fixed ?

• I don't understand what your question is! Are you saying that you would expect Vbias to be always Vt? And what is the meaning of the last paragraph? Jun 28, 2017 at 12:10
• @nickagian Yes, when the MOSFET is diode connected I expect the Vbias to be equal to Vt always. The last paragraph I have just asked if assuming that Vbias is fixed at Vt then with only W/L I can control the current sunk. I know that my understanding is skewed but kindly correct me. Jun 28, 2017 at 12:13
• Vds=Vbias: yes (obvious) but Vds=Vt no that's not true. However, Vds will be close to the value of Vt when Ibias is very small. As soon as Ibias is "larger" then a bit more voltage (Vgs) is needed than Vt, so Vds will be larger: Vds = Vt + Voverdrive This Voverdrive is needed to open the NMOS far enough to allow the current Ibias to flow. Jun 28, 2017 at 12:14

You have a misconception at this point. That the transistor is diode connected only means that $V_{ds} = V_{gs} = V_{bias}$, but that's all! This by its own doesn't mean that $V_{gs}$ will be equal to $V_{t}$.

$V_{t}$ is just a threshold that should be at least achieved for the transistor to conduct. You shouldn't mix this with the case of a normal diode. For the transistor, we just say that in this configuration it's I-V characteristic "looks like" that of a diode, but it doesn't mean that $V_{gs}$ will remain equal to $V_{t}$ no matter how much current it flows. The "diode configuration" just makes sure that the MOSFET will be in saturation at all case, once it starts conducting since $V_{ds}$ aka $V_{gs}$ will always be higher than $V_{gs}-V_t$

What you actually achieve with the current mirror, is that by forcing the current through the transistor to be $I_{bias}$, you basically define how much $V_{gs} = V_{bias}$ will be. Take a look at the typical I-V characteristics of a MOSFET. You can see it like this: You define how much $I_d$ will be with your bias circuit above the MOSFET. Then draw a horizontal line at this value of $I_d$ and you find how much $V_{gs}$ will be forced to be.

So, all in all, with $I_{bias}$ having a specific and constant value set by your bias circuit, $V_{bias}$ will also have a specific and constant value that will be, normally, higher than $V_{t}$ and that depends on how much $I_{bias}$ is. In this way, your only free parameter to play with is indeed the Width of the transistors.

Your misconception seems to be that MOSFETs are ideal switches. They are not. The FET isn't suddenly magically full on with gate voltage above Vth and full off below Vth.

Vth is the gate threshold voltage. Actually read a datasheet and see how that is defined. It is usually the gate voltage at which some minimum drain current can flow. It is NOT a magic on/off threshold.

The half of a current mirror you show produces the gate voltage that is required to exactly conduct the bias current for that particular FET. The other half of the current mirror you don't show ideally has a identical FET. It's gate is also tied to Vbias, so it is turned on to the level of conducting the same drain current a M3.

Of course no two FETs are identical, especially when they are discrete parts. In a single IC, often FETs with the same geometry will be matched close enough so that a basic current mirror as described above works well enough. For discrete designs, the matching can't be relied on as well, so often source resistors are added to make the current as a function of the gate voltage more consistent. This comes at the expense of a little more voltage drop. There is no free lunch.

• If I have understood you right, you mean to say that even if a MOSFET is diode connected, the Vds which is supposed to be equal to Vt (voltaage accross a diode) will not be so and the Vds can vary quite far apart ? Because I understand the non idealities of a MOSFET and that it would still conduct at sub threshold conditions and so on but since this MOSFET is diode connected, the Vds = Vbias cant vary a lot from Vt. Or what have I understood wrong is my humble question. And sure I understand with the matching. This circuit is to understand current mirrors in ICs. Jun 28, 2017 at 11:21