Thank you!

 

Hey.

Why is diode-connected MOS/BJT is used in a current mirror?
Even when using a MOS/BJT as a voltage source, its diode-connected.
I dont see why.

Thank you.

 

The diode connected transistor acts as a current sinc, and the current is determined by the series resistor. Since resistors can be made to tight tolerances, the current thru the diode connected transistor is tightly controlled. Also, the other half of the pair is a transistor that who's collector current/base-emitter voltage relationship matches the diode connected one, and since the base-emitter voltages are equal in both transistors, the output current is equal, and determined by the series resistor.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I understand what you said.
You basically described how a current mirror works.

But it could have worked just the same without the diode connected configuration (without connecting D/C to G/B), couldnt it?

 

Um, no becasue without this connection, the current would not be determined by the resistor. If the connection were not made, you would need a whole other bias arrangement to get any current. Now, you could probably find another bias that works, and would by perfectly legitimate. This is just the bias the current mirror uses. It's simple and effective.

PS: I edtied my post, and I know you're online. You might want to re-read before responding.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Oh, i see now.
Thank you very much

 

Everything you need to know about current sources and mirrors:

http://focus.ti.com/lit/an/sboa046/sboa046.pdf

__________________
Natural laws have no pity.' -Robert Heinlein

 

Hey guys,

I got back to the lab and I wanted to see if i indeed built a 18-19mA current source.
So Instead of 150ohm load (R2), I connected a 75ohm load (R2), and the current I got was IDS2 = 26.5mA.

Do you have any idea whats wrong here?
I think that Q2 might not be saturated (I measured VGS = 2.2V), i'll check its VTN and post it up here.

Attached Thumbnails

 

 

According to the datasheet of the BSS670S2L L6327, VTn is 1.2V-2V.
The VGS in the circuit is 2.2V

For R2 = 150ohm (18.7mA), VDS = 1.89V > VGS - VTn.
For R2 = 75ohm (26.5mA), VDS = 2.7V > VGS - VTn.

I dont understand why I got two different currents for two different loads

 

I don't know either. It looks like it should have worked.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Yes its strange.

I conducted the experiment again here are the results:

R1 = 100ohm, IR1 = 26.85mA, ***VCC = 5V.
VGS = 2.3V

Each time I used another load:
R2 = 150ohm, IDS2 = 20.5mA, VDS2 = 1.99V.
R2 = 75ohm, IDS2 = 27.6mA, VDS2 = 2.92V.
R2 = 25ohm, IDS2 = 38.9mA, VDS2 = 4.02V.

I really dont understand whats wrong here.

--
Edit:
I added an important detail that VCC = 5V.

 

R2 = 150 ohm; (4.7 - 1.99)/150 = 18mA
R2 = 75 ohm; (4.7 - 2.92)/75 = 23mA
R2 = 25 ohm; (4.7 - 4.02)/25 = 27mA

The current is tighter than you think. Have you thermally coupled T1 and T2? If not, then do so.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Hey, i'm sorry for not updating you about it.
I changed VCC from 4.7V to 5V.
Again, sorry, I thought only to give you these details.

Anyways, this is the only difference (the VCC).

About thermally coupling Q1 and Q2, I only soldered them as close as i could get.
The gates are touching each other, but the drains are 1cm away from each other.
The sources are 0.5cm away from each other.

 

It is critical in current mirrors that the transistors be in thermal contact. Otherwise, the output won't track the diode connected transistor's current.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Oh I see.

Do you think that its that crucial? (an error of 38.9mA/26.8mA * 100 = 145%).

How do you thermally couple the transistors please?

Thank you for your help and guiding.