# COMMON EMITTER WITH CURRENT MIRROR

#### Ghina Bayyat

Joined Mar 11, 2018
129

there is no problem with the solution i already did it but what i don't understand is why the transistor in the saturation ? since Rc = 2260 Ω and VRc ≈ 11.3 v and Ve ≈0.7v so vce ≈ 50mv
the simulation did work well too as if there is no problem at all but i don't understand how could a transistor work as an amplifier in the saturation region ??

#### Zeeus

Joined Apr 17, 2019
597
Noob here :
operating in the FAR? please what's FAR
Maybe 50mv Vce does not saturate it?

And collector of Q2 is 0.7v? not necessary I think

what was peak to peak of input signal?

//Making the circuit

Wilson mirror might do

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#### Ylli

Joined Nov 13, 2015
965
Why did you set Rc = 2260 ohms? Don't you want to set the collector of Q1 at about 1/2 Vcc?

So you would want to drop about 6 volts across Rc, with a current (as determined by the mirror) of 5 mA. R = E/I = 6/5 = 1.2K.

#### Wolframore

Joined Jan 21, 2019
1,748
Lots of missing information. I was able to kind of simulate it but it’s clipping. I couldn’t get it to linear amplification of a small input signal. But it is dropping out on the low. So sort of an signal gain. If you play with it you can make upside down Batman looking things with it.

#### Zeeus

Joined Apr 17, 2019
597
Why did you set Rc = 2260 ohms? Don't you want to set the collector of Q1 at about 1/2 Vcc?

So you would want to drop about 6 volts across Rc, with a current (as determined by the mirror) of 5 mA. R = E/I = 6/5 = 1.2K.
Maybe he is an experimenter just testing...And saw today in diff amp that not always to put it at 1/2 Vcc

#### Ylli

Joined Nov 13, 2015
965
Maybe he is an experimenter just testing...And saw today in diff amp that not always to put it at 1/2 Vcc
When configured as a full diff amp, the output is usually the current, and the 'Rc' is actually a real resistor and the B-E junction of the next transistor.

#### Zeeus

Joined Apr 17, 2019
597
Hmmm..An expert is needed.... Thanks

used CA3096
Rref = 2.2k Rc = 2760
There is clipping! not working like you said

Dc voltage across Cac3 was not constant!! (0.86 - 9v)
Added 100k resistor to base of q1..Voltage across cap still not constant same as base of Q1 : not constant

Removed the Rbias from vcc and added to collector : voltage across cap still not constant

Changed Rc to 1k (voltage at collector of Q1 (4v - 9v) ) : across Cac3 still not constant

No idea why not constant : I doubled check wiring and multimeter is not bad..

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#### Ylli

Joined Nov 13, 2015
965
Not an expert, but:

#### Ylli

Joined Nov 13, 2015
965
Emitter of Q1 can be higher, but the output will not be able to go as low.

#### Zeeus

Joined Apr 17, 2019
597
Emitter of Q1 can be higher, but the output will not be able to go as low.
Yeah your R1 is 1.2k..Using 2.2k..it can go higher than 1v?

And thanks...Using 100k and 20k to bias the base of Q1 in original diagram keeps voltage across Cap constant

#### Ylli

Joined Nov 13, 2015
965
Fixing the base voltage with a divider keeps things a bit more stable. You need the divider to set the base voltage at > 2 diode drops (one for Q1, and one for the CC transistor). A higher base voltage will work OK, keeping in mind the output will not be able to drop much below the base voltage. Assuming you are looking for a voltage output, Rc needs to be set to maximize output swing.

#### crutschow

Joined Mar 14, 2008
25,098
Why are you trying to solve this obvious homework problem for the TS?

#### Zeeus

Joined Apr 17, 2019
597
Fixing the base voltage with a divider keeps things a bit more stable. You need the divider to set the base voltage at > 2 diode drops (one for Q1, and one for the CC transistor). A higher base voltage will work OK, keeping in mind the output will not be able to drop much below the base voltage. Assuming you are looking for a voltage output, Rc needs to be set to maximize output swing.
Used 100k and 100k as bias but it was not stable compared to 100k and 20k
Good night! Chemically unstable : not thinking right

#### Ylli

Joined Nov 13, 2015
965
Why are you trying to solve this obvious homework problem for the TS?
Whoops. Just got interested in it and forgot.

#### Zeeus

Joined Apr 17, 2019
597
Why are you trying to solve this obvious homework problem for the TS?
Because not everyone is an expert that can attempt every question

#### Wolframore

Joined Jan 21, 2019
1,748
it was a fun challenge. Good night.

#### crutschow

Joined Mar 14, 2008
25,098
Because not everyone is an expert that can attempt every question
That may be true, but beside the point.
For homework, we help the TS determine the answer, not do the problem for him.

#### WBahn

Joined Mar 31, 2012
25,899

there is no problem with the solution i already did it but what i don't understand is why the transistor in the saturation ? since Rc = 2260 Ω and VRc ≈ 11.3 v and Ve ≈0.7v so vce ≈ 50mv
the simulation did work well too as if there is no problem at all but i don't understand how could a transistor work as an amplifier in the saturation region ??
Let's take a step back.

What value did you come up with for Rref?

How did you determine that value?

What is the "recommended voltage drop" that you are shooting for across Rc? Is it really 11.3 V?

How did you come up with the value of Rc that you did?

What value of beta are you assuming for your transistors?

#### Ghina Bayyat

Joined Mar 11, 2018
129
Why did you set Rc = 2260 ohms? Don't you want to set the collector of Q1 at about 1/2 Vcc?

So you would want to drop about 6 volts across Rc, with a current (as determined by the mirror) of 5 mA. R = E/I = 6/5 = 1.2K.
i did it this way :
Rref = 12-0.7 / 5m = 2260 ohms and since Iref = Ic so i chose Rc = Rref but now i knew my fault if it is the same value then there will be the same voltage drop across it
i sould have chose a different voltage drop across Rc
i did a stupid mistake
thank u for the help

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