# Calculating resistance to make BJT transistor act as a switch

Let's say you're given a circuit diagram as follows:

I want to find R1 and R2 so that forward voltage drop across the LED is 2V with an overdrive factor of at least 2.

Current of 25mA from a source which outputs 0 or 5V through an output resistance of 610 is given. Beta ranges from 50 - 300

So here's my approach. Since we want to find beta forced with ODF of 2,

$$\beta_{forced} = \frac{\beta_{min}}{2} = 25$$

Since source current = 25mA, we know this is the current going into the base, so Ib = 25mA.

Then I can find Ic at saturation which is given by

$$\beta_{forced} = \frac{Ic_{sat}}{I_b} \text{ which leads to } Ic_{sat} = \beta_{forced}I_b$$

$$Ic_{sat} = 25 * 25mA = 0.625A$$

Since we want a forward voltage drop of 2 across the LED, voltage at Vn can be found by

$$Vn = 5 - 2 = 3V$$

We know at saturation Vce = 0.2v, which is Vc

We can then simply find R2 by ohms law since we know the current and voltage at each end of R2

$$\frac{Vn - Vc}{R_2} = Ic_{sat}$$

$$R_2 = \frac{3 - 0.2}{0.625} = 4.48\Omega$$

That seems like a really small resistance to put at collector, but moving on, we can write a KVL in the lower loop to find R1 like so

$$5V - 610I_b - I_bR_1 - 0.7V = 0$$

Solving the above equation gives me R1 = -438 ohms.

Something is definitely not right, I can't have negative resistance here. Where did I go wrong?

• Are you asking us to answer your homework question(s) for you? This looks suspiciously like an Electronics textbook question. – Robherc KV5ROB Jan 29 '16 at 22:43
• Well in practice there should be two R2's one for each LED. Diodes shouldn't be paralleled like that. – Tom Carpenter Jan 29 '16 at 22:44
• @RobhercKV5ROB I'm asking where I went wrong in the design process of picking values of R1 and R2. Homework or not, I think the question still stands. – Xiagua Jan 29 '16 at 22:49
• What does "an overdrive factor" mean in this context? – Andy aka Jan 29 '16 at 22:59
• I would start by deciding how much current you want in the LEDs. Then work out what collector current you will need. As stated above, use separate resistors for the LEDs. Then you can calculate the base current you need. You seem to have started with a base current, that's where you went wrong. – user1582568 Jan 29 '16 at 23:14

$$5V - 610I_b - I_bR_1 - 0.7V = 0$$

Where does '5 V' come from in the last equation? It looks to me as though you have written the equation for the base circuit but used the 5 V from the collector circuit.

To get 25 mA through 610 Ω $V_{S1}$ will need to be at least 15.25 V.

Please put individual resistors on the LEDs. It's upsetting us all.

• Vs1 is 0 or 5 volts. I've written the KVL from vs1, 610 ohm resistor, R1 and Vbe. I can't do this? – Xiagua Jan 29 '16 at 23:13
• What's the max current you can get through 610 Ω when $V_{S1}$ = 5 V? (Hint: Ohm's Law.) – Transistor Jan 29 '16 at 23:17
• I = 5/610 = 8.20mA? – Xiagua Jan 29 '16 at 23:19
• Yes, and as I read your question you have assumed $I_B$ = 25 mA. Is this the problem? – Transistor Jan 29 '16 at 23:22
• "Current of 25mA from a source which outputs 0 to 5V" is what I've been told. Does this not indicate the current that runs through Vs1 into the gate? If not, what is this current they are referring to? – Xiagua Jan 29 '16 at 23:23