# help with resistor value

#### coll234

Joined Dec 14, 2021
1
I am doing a flip flop circuit on my breadboard i am using one from an online guide , i have left the link below , It uses 2 LEDS and 4 resistors and 2 BC547 transistor , the two 330 ohms are for the LEDS and the other two resistors are for the 2 BC547 , The resistors for the bc547 are 10k for each , Can anyone explain how i calculate the resister value for the bc547 specs .
Flip Flop LED Flasher Using BC547 Transistors | Lindevs

#### AnalogKid

Joined Aug 1, 2013
10,065
To be clear, that is a multivibrator, which is a modified form of a flip-flop.

R2 and R3, working with C1 and C2, set the frequency of oscillation; each resistor controls one of the half-cycles. This is covered by the first equation in the text. You rearrange it to solve for the resistor values.

Things are much more simple if you want a symmetrical square wave (50% duty cycle). In that case, R2 = R3 = R, and C1 = C2 = C. This is covered by the second equation in the text. The equation has three variables: f, R, and C. You select values for any two of the three variables, and use the equation to solve for the third one. For example, you can state that you want a frequency of 440 Hz, using 10 K resistors, and solve for the required capacitor value.

ak

#### Audioguru again

Joined Oct 21, 2019
4,693
The maximum value for R2 and R3 are calculated from the BC547 spec in its datasheet of "Collector-Emitter Saturation Voltage" because you want each transistor to saturate when it turns on. The spec says that the base current should be 1/20th the collector current. The collector current is (5V - 2V)/330 ohms = 9.1mA when the LED is 2V red. Then the base current must be 0.45mA.
The base voltage is about 0.65V and the resistor is (5V - 0.65V)/0.45mA= 9666 ohms, use 10k.

Note that most 2Nxxxx transistors specs say to use a base current that is 1/10th the collector current for saturation.

#### Alec_t

Joined Sep 17, 2013
12,818
Can anyone explain how i calculate the resister value for the bc547 specs .
As I said on another forum:
A red LED drops about 2V, so for the linked circuit the collector current will be a ~(5V-2V)/330Ω = 0.01A = 10mA.
For circuit design purposes a bipolar transistor used as a switch is assumed (by a rule of thumb) to have a current gain in the range 10-20. Provided the transistor used has an actual gain considerably greater than that it should make a good switch.
That rule of thumb implies a base current of 5mA-10mA 0.5-1mA here. The base current is actually about (5V-0.7V)/10kΩ = 4.3 0.43mA, which is in the right ball park.
The time constant RC defined by the base resistor and the connected capacitor is what sets the time for which one LED is 'off' while the other is 'on'.

Edit; current values corrected ^^

Last edited:

#### Audioguru again

Joined Oct 21, 2019
4,693
As I said on another forum:
A red LED drops about 2V, so for the linked circuit the collector current will be a ~(5V-2V)/330Ω = 0.01A = 10mA.
For circuit design purposes a bipolar transistor used as a switch is assumed (by a rule of thumb) to have a current gain in the range 10-20. Provided the transistor used has an actual gain considerably greater than that it should make a good switch.
That rule of thumb implies a base current of 5mA-10mA here. The base current is actually about (5V-0.7V)/10kΩ = 4.3mA, which is in the right ball park.
The time constant RC defined by the base resistor and the connected capacitor is what sets the time for which one LED is 'off' while the other is 'on'.
Nope, your decimal point is in the wrong place.
The base current should be (5V - 0.7V)/10k= 0.43mA.

#### Alec_t

Joined Sep 17, 2013
12,818
Oops. You're quite right. I've edited my post ^^.