Single ended voltage controlled current source

[paulmdrdo]

Hi Guys! Can you tell how VL(max) was derived?

I know how to derive the voltage at point B and A.
Va = Vb = Vcc-Vin.
I want to know how the equation for VL(max) was derived,

 

[BradtheRad]

I suspect Vin equals the voltage applied to Q1 bias. It determines the overall voltage A through R. And that happens to be Vcc-Va.

 

[KlausST]

Hi,

All Rs have the same value.
(Sadly you missed to give the resistors individual names .. this makes a public discussion more difficult.)

V_in is the input voltage
Due to Opamp regulation the voltage at the first resistor (bottom) is identical to V_in.

Opamp input current is fairly close to zero.
So if you ignore the base current then the current through the scond resistor (top left) is identical to current through the first resistor.
Then the voltages across both resistors are identical, too (= V_in)

Due to regulation of second Opamp the voltage across the third resistor is identical, too. (Still V_in)

If you say V_CEsat of output BJT to be (close to) zero, then the equation becomes true.

Btw: replace the BJTs with Mosfets and get much better accuracy (due to almost zero gate current and almost zero V_DSsat.)

Klaus

 

[Easy peasy]

Q2 is set up to provide a mirror current of that in the top left "R" ans this current is set by Vin - so even if the load varies - the current in ti will be constant up to the limits of Vcc ...

 

[paulmdrdo]

Hi,

All Rs have the same value.
(Sadly you missed to give the resistors individual names .. this makes a public discussion more difficult.)

V_in is the input voltage
Due to Opamp regulation the voltage at the first resistor (bottom) is identical to V_in.

Opamp input current is fairly close to zero.
So if you ignore the base current then the current through the scond resistor (top left) is identical to current through the first resistor.
Then the voltages across both resistors are identical, too (= V_in)

Due to regulation of second Opamp the voltage across the third resistor is identical, too. (Still V_in)

If you say V_CEsat of output BJT to be (close to) zero, then the equation becomes true.

Btw: replace the BJTs with Mosfets and get much better accuracy (due to almost zero gate current and almost zero V_DSsat.)

Klaus

Hi!

is the transistor Q1 saturated? It seems to me that it is on the verge of saturation. Am I correct?
Also, does the first opamp function as voltage follower?

 

[danadakk]

Ignoring Vbe/Aol error and Vcesat errors of the transistors

Vin is impressed across Re of Q1, so Ic ~= Vin/R

The V at collector of Q1 ~= Vcc - [R x (Vin/R)] ~= Vcc-Vin

The I at collector Q2 ~= (Vcc-Vin) / R

So Vcc ~=Vrq2 + Vcesat +Vload ~= Vrq2 + R x (Vcc-Vin)/R ~= Vrq2 +Vcc - Vin

Vrq2 ~= R x (Vcc-Vin)/R ~= Vcc-Vin so

Vload + Vin + Vcesat ~= Vcc ~= Vload + Vin

or Vload ~= Vcc - Vin

Sim seems to confirm -

Regards, Dana.

 

[danadakk]

I changed load R from 1 k, like all the rest, to 2k and got the following -

Based on sims looks like Rload <= R in rest of schematic for Vload ~= Vcc - Vin

Regards, Dana.

 

[crutschow]

The left opamp and transistor are just to translate the signal, so you can have a grounded load (R4).
As such, R3 can be changed to provide a different constant.-current gain.
Thus for R1=R2, the constant current would be approximately V1 / (R3/R2).

If you can have a high-side load (not grounded) then you just need the left opamp and transistor, with R1 being the load.

Note that using MOSFETs instead of BJTs will eliminate the error in the constant-current due to the BJT base-emitter current (for typical BJT betas that error is about 1%).

 

[KlausST]

Hi,

is the transistor Q1 saturated?

Impossible to answer ... without voltages, currents and part values.
But generally this situation should be avoided.

Also, does the first opamp function as voltage follower?

No. (But almost)
It regulates the voltage across the R to be the same as V_in.

Klaus