# BJT theory and practice - please help me understand

I've never built a circuit using BJTs before but I'm more inclined to go this route over FETs for this task.

I have 3x 12V 200mA (approx.) loads in a vehicle which I want to switch on with only a 12V 20mA (max.) signal. I also have a standard 12V car battery as a supply, which we can assume could give anything from 12-14V.

From what I've learnt so far this is my intended schematic

R3 = 12V 600mA load

Ideally I'd like to know what I could use for R1, R2, Q1, D1 and, I suppose, RY1 too.

From my research into transistor theory I should be using the NPN transistor in common-collector configuration. Assuming beta = 100 (approx.) I should have more than enough signal current to saturate my transistor for it to essentially work as a switch to my relay, which I assume at this stage will need 100mA (approx.) coil current.

What I am confused about is calculating resistor values based on the BE, BC and CE voltage (assuming a 0.7V drop on BE?) and if R2 is even needed at all (I guess to act as a voltage divider?)

Any guidance that can be offered here will be extremely appreciated.

With a $600\:\text{mA}$ load (three devices to be simultaneously activated), a relay may be a good choice. There are plenty of relays available rated for vehicle use. Some of the really cheap small ones that are more often used with regulated $12\:\text{V}$ power supplies are rated to present a transistor load of about $320\:\Omega$. So this would mean a collector current (coil current) that is less than $40\:\text{mA}$. You'll have to decide on a specific relay, though, to know for certain because these things vary from one device to another.

You've indicated no more than $20\:\text{mA}$ as a switching current, but I think falsely concluded that you could rely on a $\beta=100$ for the transistor acting as a switch. There are two basic kinds of bipolar-based transistor switches commonly used for this application: a simple BJT or else a Darlington. The Darlington will drop about one volt when switched on, so your relay would lose that much. But most relays are rated to switch just fine with that loss. The Darlington choice would likely exhibit $\beta\ge 1000$ in some cases but generally not for switching due to some resistors that they usually include inside. So while this is a fine choice, I wouldn't consider more than $\beta=200$ for actual switching. A regular BJT (NPN) would likely exhibit $\beta=10$ or maybe $\beta=20$ for switching. But I wouldn't count on $\beta=100$ for a simple NPN -- especially one that can handle that much collector current.

Either way, if the relay coil current is no more than $100\:\text{mA}$ and given that your switching signal can comply with as much as $20\:\text{mA}$, I think both options (NPN or Darlington NPN) are still on the table. So that's good.

Your circuit looks fine. $R_2$ in either case doesn't need to pull down much current. So I'd just set $R_2=10\:\text{k}\Omega$. It's mostly there to make sure the transistor stays off unless intended to be on and to quickly remove any stored charge in the transistor. But it's value isn't critical and a lot of times you won't even see it present, at all. I think it's fine you keep it, though. The value of $R_1$ will depend on which transistor type. For the NPN, you'll want $R_1\approx\frac{12\:\text{V}-1\:\text{V}}{10\:\text{mA}+\frac{1\:\text{V}}{10\:\text{k}\Omega}}$ or something from about $1\:\text{k}\Omega$ to no higher than $1.5\:\text{k}\Omega$. For the Darlington, I'd probably use $R_1\approx\frac{12\:\text{V}-2\:\text{V}}{\ge 500\:\mu\text{A}+\frac{2\:\text{V}}{10\:\text{k}\Omega}}$ or something from about $10\:\text{k}\Omega$ to perhaps no higher than $15\:\text{k}\Omega$.

Either way, your switching signal appears capable of handling the situation. Just make sure that your transistor can handle the dissipation. This will be on the order of $\frac{1}{4}\:\text{W}$. A TO-220 packaged device would be totally safe. A smaller TO-92 device may get hot.

• This was so helpful, thank you very much! It did make me realise my knowledge of BJTs is rusty so I did some further reading which leads me to some more questions: For an NPN to operate as a switch, the β value is no longer relevant, correct? To operate as a switch V(B) > V(C) > V(E)? With R(1) dropping 11V, V(B) = 1. With RL1 dropping 12V, V(C) = 0. And V(C) = 0 also, as it's connected to ground. But then there is no allowance for the voltage drop between collector and emitter, so will this even work? – Thirtyniner Jun 3 '18 at 9:12
• @Thirtyniner For an NPN switching this much current, expect $V_\text{BE}\ge 800\:\text{mV}$. I had rounded that up. Operated as a switch at this much current, there also still will be a small voltage drop of perhaps $V_\text{CE}\approx 300\:\text{mV}$. So the load will not get the entire $12\:\text{V}$. It's not a perfect switch. But relays have wide tolerances on this requirement. So this won't be a problem here. – jonk Jun 3 '18 at 13:18
• @Thirtyniner BJTs will be operated in one of two modes: active (wide range of applications) and saturated (usually as a switch.) For switching operation, you want to be sure that there is many times more base current than you would otherwise expect to need so that the collector current into the load drives down (forces) the collector close to the emitter voltage. Flooding the base with current is a good way to go. Here, $\beta=10$ is a common figure because this is usually about 10X more than might otherwise be needed. It makes sure the goal is met, this way. – jonk Jun 3 '18 at 13:27
• @Thirtyniner However, you can often get by with higher values. So this is a judgment call of sorts. There may be times when compromises in the situation may push you into $\beta=20$, for example. And that might be more than enough for the situation. But BJTs vary one to another and you have to be careful here, since as you move towards the nominal $\beta$ (but lower-end) value specified for active mode operation, you then become increasingly susceptible to variations in the BJTs you may use. – jonk Jun 3 '18 at 13:31
• thanks again for taking the time to reply, you do this community a great service! FYI I'm really struggling getting inline TeX to work so apologies for the formatting. I think this is indicative of my lack of understanding (despite many many hours of study) of basic circuits. I assume the relay, first in the circuit, would "take" all of the 12V, leaving nothing for the transistor. Instead the transistor seemingly takes it's share of the voltage first, and the relay takes what remains. – Thirtyniner Jun 3 '18 at 17:35

Ok, let me give you some advice.

First of all, the relay RY1 is not really needed, you can connect each load to a BJT directly.

You know that your BJT must switch 200mA load current and your beta is 100, so you can calculate the base current as 200mA/100=2mA.

Next, calculate R1 = 12V/2mA = 6k Ohm. You do not need R2.

Build it up and test it :-)