i am a newbie in this forum,i need help from others to calculate the current flow in the current mirror circuit.the battery supply is 12V, the transistor used is 2SC1815. the led use is amber colour led.the forward voltage for the led is 2.3V and the full drive current for the led is 25mA
can anyone teach me how to calculate for I1 and I2 please?

 

The mirroring accuracy degrades as you add more output stages, because each one draws base current, but that probably won't be a problem for you. If it is, you can just adjust the Q1 current to compensate if needed.
Also, this type of current mirror requires tight matching between all the transistors, or your output currents will not be equal. You can minimize the matching errors by adding a resistor in series with each emitter (including Q1). For 25mA, 10Ω resistors should work well.
With Q1 biased off the 12V battery as shown, your resistor should be
R1=[(12-0.7)/.025]-10, where 10 is the emitter ballast resistor
R1=442Ω
Standard 5% values are 430 and 470. Take your pick.
The LED currents will be ≈25mA each. Don't forget to add the 10Ω ballast resistors in series with each emitter.

 

Thank you so much for the reply. Actually I already choose the value of R1 used, which is 240 ohm. Due to time limitation, I cannot change the value of the R1 except I have to fabricate another PCB board which will take some time to get it.


Below is my calculation, I am not sure whether it is right or not…


V=I1 * R
12V-0.7V=I1 *240ohm
I1= 47.083mA

Theoretically, current of I2 will be same as I1 which is 47.083mA right?
will the current of I2 same as I1 in my circuit?




Or can i calculate Ic in tis way?

Ib=(5-0.7)/1k=4.3mA;

Ic=βIb=100*4.3mA=0.43A
the value of Ic obtained is very large, is it correct?

Refer to the picture,

Vce(sat)=0.4V
forward voltage for the led= 2.3V
battery=12V
12-(2.3*3)-(0.4*2)=4.3V
the 3 led will take the remaining 4.3V o the 2 transistor will take the 4.3V?

 

I hope you used a 1 Watt resistor for R1, because it dissipates 532 mW.
All resistors of a given wattage are the same size. Why can't you change the value of R1?
Your I1 current analysis is correct. Assuming Q1 and Q14 are matched, IC14 will also be ≈47mA. Q2 will not saturate (if your LED forward voltages are low enough), so you will only lose base current through it. Therefore, LED current will also be ≈47mA.
As I told you in my last post, your current mirror transistors all need to be matched (for Vbe). Discrete transistors are not inherently matched, as they can be in an IC. Even a 10% mismatch in Vbe's (at the same emitter current) will cause the currents to be different by the ratio of 1.47:1. That means that, with only 10mV of mismatch, IC14 could be as high as 69mA or as low as 32mA. That's why I recommended ballast resistors.

 

i don't have time to do the whole circuit again,what can i do now is do some calculations to show the current flow in my circuit although this will show that my circuit is a bad design. Somehow i can state that my circuit can be improved by connect ballast resistors in series with each emitter.this is a good idea perhaps.

the LED require 25mA to get the brightness that I want,if the current flow is btw 32mA to 69mA,will it burn my LED?i think that larger current flow is more preferable than smaller current,if the current is too small,our eyes can easily recognize that the brightness of the LED is not consistent.But if the current is larger,we cant recognize the brightness of the LED easily.I am wondering my thinking is correct or not?

 

Did you breadboard this, or at least simulate it, before you designed the PC board?

 

i did the simulation 1st n den use breadboard to test whether it is function onot. After that, i straight away put all the components in the PCB board without any consideration of the current.so i am facing lots of problem when i do the calculation part.

 

Why can't you change the value of R1?

 

i soldered the resistor in the PCB board ady

 

i soldered the resistor in the PCB board ady

They are easy to change,

What wattage did you use? If it's too low, it will unsolder itself.

 

i try to add a resistor in series with each emitter (including Q1), but the current of I1 and I2 is not the same
I1=26mA
I2=8.8mA
is it necessary to add a resistor in series with each emitter?

 

i try to add a resistor in series with each emitter (including Q1), but the current of I1 and I2 is not the same
I1=26mA
I2=8.8mA
is it necessary to add a resistor in series with each emitter?

If you only added emitter resistors in series with Q1 and Q14, the rest of the transistors will be running at very high currents. You must add resistors between emitter and GND on all the transistors whose bases are connected together.
Why do you not answer some of my questions?

 

sorry about. actually i dunno what wattage did I use because I bought the resistor at the shop.How to know the wattage of the resistor?Honestly, I cant change the resistor R1 because i glue my PCB board on top of the polycarbonate

i done a big mistake!but then the PCB board still can function with R1=240Ω.why

 

Really sorry