# How to calculate this resistor value

Discussion in 'The Projects Forum' started by masked, Aug 9, 2010.

Jul 1, 2010
48
1
Hello,

Attached is a portion of an example Ron made for me. I understand the general idea that turning Q6 on will turn Q4 off. Or, as Ron put it:
"When Q6 is saturated (Vce≈0.1V), it steals the base current from Q4. Q4 cannot be ON unless the base voltage ≈0.7V."

Now, if I want to add an LED with a voltage drop of 1.8 at point "A", I'm confused about how to calculate the values for R6 and R7 such that Q4 will be either saturated or off.

Using a meter and trial & error, I came up with an R7 of 150k and R6 of 1.8k. However, I'm missing how to reverse engineer that and determine my values mathematically beforehand.

How's that work?

(Thank you),

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2. ### whale Active Member

Dec 21, 2008
111
0
hie,
since your LED will cause voltage drop of 1.8 v, the voltage level at base - emitter of Q4 will be low when compared to the circuit without LED. the difference in voltage level at base - emitter of Q4 before and after fixing LED will be 1.8 volt.
Due to decrease in base - emitter voltage the output current of Q4 will also be decreased. Now you want to choose the value of R6, such that the output voltage of Q6 will be compensated after fixing LED. ie, the output voltage should be increased by 1.8v to compensate the LED forward voltage drop.
There will be also loss in base current of Q4 due to LED, but it will be smaller.
If you think to compensate that current level and to make the output exact, then you should change the value of R7. To do this you want to find out the current loss due to application of LED at the base of Q4 and add up your current loss with the base current of Q4 before application of LED, then you set up the R7 value to produce desired current level ( calculated current value ) with the knowledge of gain of the transistor Q6.

Jul 1, 2010
48
1
I'm not too concerned with the current at Q4$\small{_b}$, just trying to figure out how to calculate that my voltage will be above .7V.
Current draw of R6 would be .6mA (9V/15000Ω).
Current draw of Q4 will be 2.64uA. (to saturate when Q4$\small{ce}$.)
So it sounds like I would just use a total current of 2.64264mA, correct?

Normally for series resistors I'd calculate the total resistance to get current, and then multiply the total current by each resistor value to get the drop across that resistor.

With only 1 resistor connected to a battery , it has "all" the voltage dropped because the potential between both leads is the same as the potential between the battery terminal.

In this case there are basically 3 components: resistor-LED-Q6.
I guess what I'm trying to figure out is how to calculate the voltage at Q4$\small{_b}$. I know the LED will subtract 1.8V from something, but can I simply consider Q4 as another series resistor to calculate voltage?

Thanks,

Last edited: Aug 9, 2010
4. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
If Q6 is off there will be more than enough voltage at the base of Q4. Once the break over of 0.6V is reached it becomes all current.

LEDs have a similar problem as the base emitter of a transistor. You didn't say the color (it matters), so I will assume a red LED with a Vf of 2.5V.

These voltage drops are treated as constants. So you have 2.5VDC for the LED and 0.6V for the transistor. This means the resistor R6 will drop the remaining voltage.

Remaining voltage:
27V - 2.5V - 0.6V = 23.9V

The current using 15KΩ through the LED and the BE is:
23.9V ÷ 15KΩ ≈ 1.6ma

The voltage drops are constant, so the remaining voltage is constant. If you want a specific current you calculate accordingly.

You've been here long enough I'm sure you've seen the link, but just in case...

LEDs, 555s, Flashers, and Light Chasers

Jul 1, 2010
48
1
The LED is yellow, btw; I measured it's drop at 1.74.

I thought my (silicon) transistor dropped .7, and I was using 9V rather than 27V thinking to calculate the circuit: 9V - R6 - LED -Q4 - Gnd.

So I'll have a Vf of 2.4V (1.8+.6).
So the remaining is 9V-2.4V = 6.6V.
Current is:
6.6V ÷ 15KΩ ≈ .44mA

But I think the point is that my remaining voltage is above the .6 (or .7) required to run the transistor, and I don't need to worry about R6 dropping the V below .6V.
Is that right?

6. ### hobbyist Distinguished Member

Aug 10, 2008
764
56

I don't know what the collector of Q4, is connected to, so have to use a current assumption for showing this example.

You want to reverse engineer this to understand how to get resistor values without actual imperical measurements, and prototyping.

Here is another suggstion on how this can be done.

Example:
supose you want 50mA, of collector current through Q4. = (ICQ4)
According to the data sheet, baase current needs to be around 1/10th of collector current for saturation.

So you need 5mA, of base current. = (IBQ4)

LED = 1.8v.
Vbe Q4 = 0.7v.

So voltage at the base (VBQ4) needs to be at least 2.5v. to start conducting. Now this is really the voltage at the collector of Q6.

VCC = 9v. = supply voltage. as shown, in your schem.

Therefore Voltage drop across R6 = VdR6 = (VCC - VBQ4) = 9v. - 2.5v. = 6.5v. = VdR6.

earlier it was determined that IBQ4 should be around 5mA. so R6 = (VdR6 / IBQ4) = 6.5v. / 5mA. = 1.3K ohms.

Now to put Q4 into cutoff, assume Q6 is saturated with 0.3v. VCE.
That means then a current of (VCC - VCEsat.) / R6 = ICQ6sat. = 9v. - 0.3v. / 1.3K ~=6.7mA.

This will be the current Q6, will need to deliver in order to drop the VBQ4 to 0.3v. actually voltage at the collector of Q6.

Now with 6.7mA. being established as sat. current for Q6, than base current of Q6, will be 1/10th of it, or 670uA. = IBQ6.

From there you decide on what the input voltage (Vin) is to be, than again solve for the value of R7 by taking (Vin - VbeQ6) / IBQ6 = R7.

That will get you close deepending on transistor parameters to calculating resistor values.

Last edited: Aug 9, 2010