In Neamen's microelectronic book, there's a problem using this current mirror

MOSFET current mirror

There is a 5V DC voltage source above the current source. The reference current is 250 microAmp. The threshold voltage is 1V. The transistors are matching (W/L is 3 for both). The Kn' is 80 microAmp/volts squared.

Also, Instead of ground, there's a -5V DC voltage.

What is the VGS of the two transistors?

from the figure, it seems that the Vgs should be 10V. Since assuming there's no voltage drop acroos the current source, V+ - V- = Vgs. However, getting the Vgs from the drain current equation

enter image description here

yields a 2.44V.

Which one is corrent and why?

EDIT: And no this is not a homework question.


To figure out \$V_{GS}\$, all you need is the MOSFET current equation. You have all the parameters you need for that, and you did the calculation correctly:

$$I_D = \frac12 k' \frac WL (V_{GS} - V_t)^2$$

$$250 \mathrm{\mu A} = \frac12 \left(80 \mathrm{\frac{\mu A}{V^2}}\right)(3)(V_{GS} - 1 \mathrm V)^2$$

$$V_{GS} \approx 2.44 \mathrm V$$

Now, if you want, you can use KVL to figure out the other voltages in the left branch of the circuit:


simulate this circuit – Schematic created using CircuitLab

$$5\mathrm{V} - V_{C} - V_{GS} - -5\mathrm{V} = 0$$

where \$V_C\$ is the voltage across the current source. You can plug in your \$V_{GS}\$ to get:

$$5\mathrm{V} - V_C - 2.44\mathrm{V} - -5\mathrm{V} = 0$$

$$V_C \approx 7.56\mathrm{V}$$

Normally you don't need to figure out the voltage across the current source for a homework problem. (In real life, you need to make sure you have enough voltage, especially for more complex circuits like a Wilson current mirror.) You might have to find the voltage across the output transistor, since that will let you take channel length modulation into account and determine the error of the mirror.

  • \$\begingroup\$ Thanks so much Adam. One tiny question. You said "In real life, you need to make sure you have enough voltage". For what though?. I think it's to keep the transistor in saturation, correct? \$\endgroup\$ – Mustafa May 4 '15 at 18:17
  • 1
    \$\begingroup\$ On homework-like questions we generally prefer to just give the OP enough to figure it out on their own rather than do all the work for them. \$\endgroup\$ – The Photon May 4 '15 at 18:18
  • \$\begingroup\$ @Mustafa One example would be if you're using a resistor to generate the reference current. You need enough voltage across the resistor to satisfy Ohm's Law. \$\endgroup\$ – Adam Haun May 4 '15 at 18:52
  • \$\begingroup\$ @ThePhoton He (?) said it wasn't a homework question, and he already got the correct answer on his own. \$\endgroup\$ – Adam Haun May 4 '15 at 18:53
  • \$\begingroup\$ Yes but it is a homework-like question, even if OP wasn't assigned it as homework. \$\endgroup\$ – The Photon May 4 '15 at 19:17

Since assuming there's no voltage drop acroos the current source

This is a wrong assumption. An ideal current source can have any voltage drop at all across it, and will produce whatever voltage drop is needed to achieve its output current. Its I-V curve is a straight horizontal line, with equal current for any voltage drop.

  • \$\begingroup\$ So how does this explain the differences in voltage?. This question is solved assuming there's no voltage drop across the current source. \$\endgroup\$ – Mustafa May 4 '15 at 16:32
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    \$\begingroup\$ Who says the "question is solved assuming there's no voltage drop across the current source"? If you throw away that assumtion you should get the right answer. \$\endgroup\$ – The Photon May 4 '15 at 16:41
  • \$\begingroup\$ The proffesor says that. Fine, throwing away that assumption, how do I get the voltage drop across the current source? \$\endgroup\$ – Mustafa May 4 '15 at 17:04
  • \$\begingroup\$ Vdd - Vgs = 10 - 2.44 = 7.56V. How it is not a homework question boggles my mind. \$\endgroup\$ – ilkhd May 4 '15 at 17:28
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    \$\begingroup\$ @Mustafa, you said you had a conflict between calculating one way (assuming voltage across current source is 0) and another way (using the FET equation). I told you the assumption about the voltage across the current source being 0 is wrong. That leaves you with one correct answer and no conflict. \$\endgroup\$ – The Photon May 4 '15 at 17:54

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