Transistor Biasing help (long post)

Hello all, i have my end of year exam in a few weeks and i'm running through a transistor biasing circuit that i made up just to see if i get it. The schematic is included, i just wanna run through this and make sure i'm right, and ask some questions. I'v breadboarded it and it works mostly.

Ok,

Assuming that the blue led = 4v, green led = 2v,
Ie=Ic,
The phototransistor has a measured range of 10k ohms at bright light and 1M ohms plus in darkness.

The circuit is supposed to be off during light, and light the leds (turn on) at night.

Vb = Q2/(R1 + Q2) All x Vcc
= 1M ohm/(2.2kohm + 1Mohm) all x 4.8v
Vb = 4.7v

Ve = VB - 0.6v
Ve = 4.7v - 0.6v = 4.1v

Ie = assuming i want 60ma to distribute through 3 leds (R2 for current control, i think?)

R2 = Ve/Ie
= 4.1v/0.060A
R2 = 68ohms

ie=ic so ic = 60ma

R3 = vcc-blue led/20ma(desired)
=4.8v-4v/0.020A
=40 ohms

R4 = vcc-green led/20ma(desired)
=4.8v-2v/0.020A
= 140ohms

R5 = R4 = 140ohms

Ok, I breadboarded the circuit and simulated it in livewire. The circuit simulates fine, led's go off when Q2 is reduced, on when Q2 increases.

Led's reduce slightly on breadboard in full light, dark causes led's to be on full.

Questions:

1. Does the biasing set up seem correct? I seem to think i'm missing something.

2. For R2, i assumed i needed 60ma, so is it right to do this for the value of R2? for example, say i needed ie=ic=20ma, would i just do ve/0.020 = R2?

3. How would i improve the circuit to have a better 'snap on' time? for example, on the breadboard the leds are on in decent light, how would i modify ciruit so that they would be off unless Q2 was 1Mohm ish?

Thanx for taking the time to read that.

Ross

Myrmidon said:

Questions:

1. Does the biasing set up seem correct? I seem to think i'm missing something.

2. For R2, i assumed i needed 60ma, so is it right to do this for the value of R2? for example, say i needed ie=ic=20ma, would i just do ve/0.020 = R2?
Wont the current thru R2 be a direct function of the voltage at the base of the transistor.???
Example: say when 'dark' the voltage on the base is 2V, then the emitter voltage will be 0.7V lower ie: 1.3V.
So if R2 was 50R then Ie would be 1.3V/50R =26mA.

If when 'light' the Vbase is say 1V then Ve would be 0.3V and the current Ie would be 0.3/50 = 6mA.
These values are for explanation only.


3. How would i improve the circuit to have a better 'snap on' time? for example, on the breadboard the leds are on in decent light, how would i modify ciruit so that they would be off unless Q2 was 1Mohm ish?
To get a 'snap' action its usual to provide positive feedback from the output to the input.
A Schmitt trigger would do just that, a second transistor is used.
Thanx for taking the time to read that.

Ross

Click to expand...

hi,
Why do you think you need an emitter resistance to control the LED current, also look at the effect the emitter resistance has on the input impedance of the transistor.
Do you follow.?

Thanx for replying!

I think i follow you. Bare with me:

Isn't R2 used to set up the collector current though? If it was taken out, surely you would just be relying on the hfe(gain), which isn't the most reliable?

My thinking of R2 was to set up a collector current of around 60ma to direct 20ma through each led.

I'm not clear on what you mean here:

Wont the current thru R2 be a direct function of the voltage at the base of the transistor.???
Example: say when 'dark' the voltage on the base is 2V, then the emitter voltage will be 0.7V lower ie: 1.3V.
So if R2 was 50R then Ie would be 1.3V/50R =26mA.

If when 'light' the Vbase is say 1V then Ve would be 0.3V and the current Ie would be 0.3/50 = 6mA.
These values are for explanation only.

Click to expand...

I did the calculations basing Q2 at 1M ohm, and worked out the base voltage to be 4.7v at nite(dark) so 4.1v at the emitter. Dividing that by a desired 60ma to get R2.

Doing the same but taking Q2 to be at light = 10k ohm, R2 ends up being 55 ohms. I did this to attempt to get ic = 60ma, hence 20ma per led.

Was i right? it just seems to be the snap on i don't know how to add which brings me on to:

To get a 'snap' action its usual to provide positive feedback from the output to the input.
A Schmitt trigger would do just that, a second transistor is used.

Click to expand...

How would i do this?

also look at the effect the emitter resistance has on the input impedance of the transistor.
Do you follow.?

Click to expand...

Not really, i haven't really covered this at college yet, i think that's next year tho.

hi,
Have a look at this link for a Schmitt.
https://www.daycounter.com/Circuits/Schmitt-Trigger/Transistor-Schmitt-Trigger.phtml

This circuit uses the emitter coupling for positive feedback by sharing the common resistor in the two emitters.

I did the calculations basing Q2 at 1M ohm, and worked out the base voltage to be 4.7v at nite(dark) so 4.1v at the emitter. Dividing that by a desired 60ma to get R2.
I suggest you measure the Base voltage when the LDR is dark ie: 1M0 and light 10K0

Doing the same but taking Q2 to be at light = 10k ohm, R2 ends up being 55 ohms. I did this to attempt to get ic = 60ma, hence 20ma per led.

Click to expand...

Using your calculations gives transistor Ie currents of 57mA and 45mA.!

Hello again eric,

Using your calculations gives transistor Ie currents of 57mA and 45mA.!

Click to expand...

Oops, yeah, i rounded off the R2 value to 70 ohms as the result was 68.33 ohms, i should have mentioned that.

Hmm, the base voltage just seems to stay at a constant 2.06v, which isn't the 4.1v i expected. Checked my wiring and it seems ok.

Any ideas?

Myrmidon said:

Hello again eric,



Oops, yeah, i rounded off the R2 value to 70 ohms as the result was 68.33 ohms, i should have mentioned that.

Hmm, the base voltage just seems to stay at a constant 2.06v, which isn't the 4.1v i expected. Checked my wiring and it seems ok.

Any ideas?

Click to expand...

Hi,
Its the input impedance of the transistor which is MUCH smaller than the 1M0 [dark] So effectively you have the 1M0 and the transistor input resistance in parallel.
That means the voltages at the base will be lower than you expected.

If they were as you said, there would be not sufficient voltage across the LED's.!


EDIT:

Look here.
https://www.electro-tech-online.com/custompdfs/2008/05/Experiment7.pdf

Its the input impedance of the transistor which is MUCH smaller than the 1M0 [dark] So effectively you have the 1M0 and the transistor input resistance in parallel.
That means the voltages at the base will be lower than you expected.

Click to expand...

Ok, err...so how do i know the input impedence? Is it on the datasheet or is there a formula?

How would i amend the circuit to take this into account? I'm struggling a bit.

Just reading that pdf.

Myrmidon said:

Ok, err...so how do i know the input impedence? Is it on the datasheet or is there a formula?

How would i amend the circuit to take this into account? I'm struggling a bit.

Just reading that pdf.

Click to expand...

hi,
Approx, the input impedance is Hfe[transistor gain] * Re

I sense i'm losing this battle, would that mean that R2 (emitter resistor) should be of a much higher value then? But that would then lower the overall current at the collector?