tutorial all transistor light bar graph build

I will put the schematic at the very top and bottom to make it easy to navigate to when reading through the tutorial from top to bottom of the page.
also a short video of this circuit is shown at the bottom link.


from Data books LED is established as 2V @ 30 mA.

Since this is a tutorial I will try to record all the steps I took to design this circuit.
Vcc is chosen to be 7v. this will allow a 9volt battery to run it comfortably.

Now Vled is 2v @30 mA then using kirchoffs voltage law, (7v - 2v) = 5v which is the remainder of calculated voltage between the LED and the VCC. This is the drop of voltage across both the resistor R1 and the transistor from emitter to collector.

Now I arbitrarily chose to make VCE to be 500mV, which again calculating remainder voltage across the last component (R1) in the voltage loop, comes to be 4.5v.(VR1)
Now to determine the value of R1 can be done, by taking (VR1 / 30mA) where 30mA is the series current flowing through the led and R1 this current will call (IR1). Therefor R1 can now be calculated by (VR1 / IR1) = (4.5v / 30mA)=150 ohms.

now for R2 value, first assume 0.7v dropped across the base emitter junction of Q1 which puts the base voltage (VBQ1) = (4.5v - 0.7v)= 3.8v.

Now since I chose not to implement a collector resistor into Q1 circuit, means that I can assume emitter current of 30mA (IR1) will flow through collector back to the supply. Using this assumption will allow me to calculate a value of base current IBQ1, to a ball park figure, as follows,
IBQ1= (IR1 / Bmin)= (30mA / 50)=600uA.

R2 forms a voltage divider to keep the base voltage stiff under varying beta conditions, so in order to calculate the value for R2 I need to make sure that at least 10 times more current will flow through R2 than the base current (IBQ1). this makes this current = to 6mA. call it (IDR2) the current in the divider flowing through R2.

R2 now can be calculated as (VBQ1 / IDR2)= (3.8v / 6mA)=633 ohms, so 620 nom. value is chosen, for R2.
R3 is calculated as again using the voltage loop theorem, we have from ground to the top of R2 3.8v(VBQ1) and from the top of R2 to the Vcc we have another voltage drop across R3 and R4 which is (Vcc - VBQ1)= (7v - 3.8v) = 3.2v

The divider current (IDR2) flows in series through R3-R4 so the value of R3-R4 is calculated as (3.2v / 6mA) = 533, make it nom. value of 510 ohms. Now since I want this voltage to be variable using a potentiometer I will make R3 = 510 so when R4 is varied down to zero ohms the calculated value of base voltage can be attained, to drive Q1, however I will choose a pot value of around 1k or so to turn the transistor Q1 off completely until the pot is varied to a voltage input.

Now this next procedure is used consecutively throughout the rest of the design for the remaining stages.
Now 6.5v is at the emitter of Q1 (VEQ1) I will choose an arbitrary value of 500mV to be dropped across the base resistor of each transistor in the next stages.

So (VEQ1) (6.5v -0.5v) = 6v at the base of Q2. (VBQ2)
Now (VBQ2 - Vbe) = (VEQ2) so voltage at the emitter of Q2 (VEQ2) = (6v-0.7)= 5.3v.
Again using 30mA as current flow through the LED2 ant R6 and using 2v drop across LED2 than in the emitter loop the voltage drop across R6 (VR6) is (VEQ2 - Vled2)= (5.3v - 2v)= 3.2v across R6 (VR6). That will also cause the voltage (VCE) of Q2 to be (7v - 5.3v)= 1.7v.

Now R6 can be calculated as (3.2v / 30mA) = 106 ohms, make it nom. 110 ohm resistor.
R5 can now be calculated, sincwe I chose to drop 500mV across each base resistor, that means I need to establish the amount of maximum base current to flow through Q2 base (IBQ2) that's done by assuming for every stage a beta of 50.

So IBQ2 will be (30mA / 50) = 600uA
Now R5 = (500mV / 600uA)= 833 make it nom. 820ohms.
Since I want all succesive stages to have 500mV dropped across the base resistor, I can use this same 820 ohm value for every base resistor, and the only resistor that needs to be recalculated will be the emitter resistor connected to each LED.

for instance stage 3 will calculate as follows, with 5.3v at the emitter of Q2 (VEQ2) subtract the 500mV across R7, will bring the base voltage of Q3 to 4.8v (VBQ3).

Subtracting Vbe of 0.7v from this value makes the emitter voltage of Q3 (VEQ3)= 4.1v
again subtracting LED3 voltage of 2v from this value makes the voltage drop of R8 (VR8) = (4.1v - 2v)= 2.1v
now R8 is calculated as (VR8 / IR8)= (2.1v / 30mA)=70 ohms make it nom. 68ohms.
Now R9 is 820 ohms so repeat above steps to find values for the emitter resistor(R10)
and so on until the emitter voltage reaches 2v or less where no more emitter resistor is used.

light bar2.jpg

Here is a video of it working

Blog entry information

Last update

More entries in General

More entries from hobbyist

Share this entry