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#### an_82

##### Member level 2
Hi every body,

I need a simple analog circuit consist of only MOS transistors to do this:

Vo=(V1+V2)/2

Where V1 and V2 are DC voltages less than 1v. By the way the V2 can be negative e.g. the circuit should be work for (V1-V2)/2.

If you have idea by using resistors it works for me. I can replace the resistors by MOS transistors in linear region.

thanks

#### xiahanzh

##### Junior Member level 1
Try using current mirror voltage adder shown below.
This circuit is for Vout = V1 + V2, so you need to change it to have function of divide by 2. It should be simple.
Please notice if you want negative value, it should be dual power supply, Vss should be a negative value instead of gnd.
V_sublinear is just a bias voltage and forget about the confusing name.

an_82

### an_82

Points: 2

#### an_82

##### Member level 2
Dear xiahanzh,

Could you please explain how it works? Is it possible to make it simpler? (less no. of transistors)

#### xiahanzh

##### Junior Member level 1
Two inputs V1 and V2 control currents flowing through M4 and M5, then currents flowing together into M8 and mirror to M9. M7 tries to convert it back to voltage so Vout = V1 + V2. To make it simpler, you can try to remove M2 and M6. Or you can even remove all transistors of right mirror (M6,7,2,3,9) and create Vout directly from M8. But I tried it before, accuracy would be much lower than this circuit.

#### an_82

##### Member level 2
To have better accuracy I will use your original circuit.

two question:

1. What do M2 and M6 do?
2. Why do you put M0,1,2,3? I guess you utilized them to have cascode structure and more accuracy, right? Should we connect all of their gate to GND?

#### xiahanzh

##### Junior Member level 1
1. You can see left part of the circuit is identical to the right part of the circuit. M6 gate is connected to gnd, which means 0. Therefore, V1 + V2 = 0 + Vout so that it can do voltage add.
2. You can give them as any bias as you need. Only thing is they must be biased to the same voltage. Gnd is just for simplicity and for my application, it is good enough.

#### an_82

##### Member level 2
M6 gate is connected to gnd, which means 0. Therefore, V1 + V2 = 0 + Vout so that it can do voltage add.

1. So I can eliminate M2 and M6, and directly have V1 + V2 = Vou. right?

2. Do you use M0,1,2,3 for having more accuracy? ( cascode structure)

#### xiahanzh

##### Junior Member level 1
1. You can try that. I did not do it before.
2. I think they are used to control current. And I also think M4 - M7 should operates in linear region so they can have linear addition instead of square law. I cannot remember it exactly but you can play with it.
BTW, you should remove V_sublinear port, that is for my application and it should not apply to yours.

#### LvW

Why not use the classical diffeential amplifier configuration?
The output voltage at one of the drain nodes is

Vout=0.5*(v1-v2)*gm*Rd with gm=transconductance and Rd=drain resistance.

1.) It should be possible to make gm*Rd=1
2.) For addition: If necessary one of the input voltage can be inverted (additional FET) prior to enter the amplifier.

#### an_82

##### Member level 2
Dear LvW,
I can not use differential pair amplifier, because my desired output voltage should be single ended (one node).
In fact I want to realize the circuit which do what you said: Vout=0.5*(v1-v2) but not double ended.

#### LvW

Dear LvW,
I can not use differential pair amplifier, because my desired output voltage should be single ended (one node).
In fact I want to realize the circuit which do what you said: Vout=0.5*(v1-v2) but not double ended.

But I think, you are not forced to use both outputs, do you?
A differential amplifier provides two output possibilities with opposite signs and you are free to use both or only one.

an_82

Points: 2