# Common transistor topologies [closed]

I'm trying to get my head around transistors and I had some difficulties to understand the various way we can connect a transistor.

For example, here are a few possible arrangment:

simulate this circuit – Schematic created using CircuitLab

In theory, both of those circuits should be equivalent except on the left, the led is on when the base is low and on the right the led is on when the base is high.

The difference I see is that on the left, when the transistor is in cutoff state R1 and D1 aren't affected by a voltage drop through the transistor unlike the schema on the right.

So technically something like this is:

simulate this circuit

Here the arrangement at the top is capable to supply 15v to the gate of the mosfet but the arrangment to the bottom can't barely supply more than 2v according to the simulation.

The question might be a bit vague, but what I'm really interested to know is if there is a list known pattern for transistor arrangement and their pros/cons?

For example, there is the darlington transistor which is pretty much like my last schematic but it seems it doesn't work if you want to amplify the voltage output for a mosfet using a transistor. In my examples, I used a clock but had in mind an output pin coming from a mcu.

For example, I saw on stackexchange someone recommending this kind of pattern:

simulate this circuit

To drive a Mosfet, yet in this particular example, doesn't seem to work in my case. Actually it looks like half of an H-bridge.

Also, in my particular examples, all of the collector have a 12v applied and driven with a 5v logic on/off.

• In theory, both of those circuits should be equivalent No, they're not. The left circuit is incomplete, a collector resistor is needed otherwise you short the supply and the LED will be always on (details matter in a circuit diagram!). The right circuit needs more than about 3 V at the input to make the LED light up. The left circuit only needs about 0.7 V to make the LED turn off. So no, they are very different circuits. – Bimpelrekkie Mar 2 '18 at 18:46
• What a strange arrangement of misfitting circuits and half truth dialogue. Do you have a question to ask or is this posting just for people’s entertainment? – Andy aka Mar 2 '18 at 18:47
• There are a good set of ways of connecting transistors. However, none are pictured in any of these diagrams. The best advice is to throw all of them in the bin. – Neil_UK Mar 2 '18 at 18:48
• Circuits with MOSFET: Q1 basically "does nothing", again a collector resistor is missing. Circuit with Q3 cannot work because no current can flow through Q3 as the MOSFET has a high impedance input. I suggest taking a step back and try to really understand one of these circuits and what is should look like. Now you missed some details making your circuits not work and you not understanding. – Bimpelrekkie Mar 2 '18 at 18:49
• In your edit you left V1 and V2 at 1 V defaults. Change to 12 V as in your text. – Transistor Mar 2 '18 at 21:53

This is not an answer at this stage but merely shows a few problems with your understanding.

You are missing the power supplies on your schematics. Add them in.

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) The LED is always on. When Q1 turns on it short circuits the supply and probably burns out. (b) This will work but the emitter will be about 0.7 V below whatever voltage is on the base.

simulate this circuit

Figure 2. (a) M1's gate is permanently connected to 12 V via R3. Q1 does nothing for it. (b) With a 5 V clock signal the emitter of Q3 will reach < 5 V. This might turn on M3 a little but once current flows the voltage on R8 will rise, reduce the voltage and turn it off. If the clock signal voltage was high enough you could turn on M3 but without a discharge path M3 will probably never switch off.

simulate this circuit

Figure 3. The Q1 and Q2 parts are OKish but having M1 on the high side of the load will prevent it working properly.

Also, in my particular examples, all of the collector have a 12 V applied and driven with a 5 V logic on/off.

In all except Figure 1a you are running the NPN transistors in "emitter-follower" mode the emitter will follow the base voltage minus about 0.5 to 0.7 V because of the base-emitter diode voltage drop. You will never get the emitter above about 4.5 V so most of your circuits won't work properly, if at all.

Edit after OP's update:

The difference I see is that on the left, when the transistor is in cutoff state R1 and D1 aren't affected by a voltage drop through the transistor unlike the schema on the right.

That is correct. That arrangement is referred to as shunting the current. Your arrangement is inefficient, however as high current is high current is shunted it's rather power hungry. A better arrangement is shown below in 1a.

simulate this circuit

Figure 4. (a) All that is required is to shunt the LED current. R4 then limits the current to slightly more than when the LED is on. With the transistor turned hard on (note addition of base current limiting resistor) there will be about 0.2 V across Q1.

I'm really interested to know is if there is a list known pattern for transistor arrangement and their pros/cons?

What you are asking is for transistor circuit topologies. I had a look around for a good one but drew a blank. My old The Art of Electronics by Horowitz and Hill gives many examples.

• Thank your for your answer. I modified 1a and 2a to use a voltage divider. Pretty sure it shouldn't help because if the voltage dropped completely then the LED would be always off and the Mosfet also always Off. – Loïc Faure-Lacroix Mar 2 '18 at 21:19
• I guess I understand what you mean by discharge path but I don't understand how I could put one.I guess that anything that anything that flow through the gate must flow somewhere else then, either the drain or the source. Anyway, the question was mainly if there was some kind of common transistor arrangement. Like you said 1a) was an "emitter-follower". 2b) would be a darlington transistor if we replace the mosfet by a bjt. Apparently those are called "amplifier topologies" according to wikipedia – Loïc Faure-Lacroix Mar 2 '18 at 21:31
• See the update. MOSFETs gate is insulated as indicated by the circuit symbol. It behaves more like a capacitor and there is no leakage current and if you pull the gate high and disconnect then it will stay high. You have to drive it high and drive it low but in many cases a resistor will suffice to pull one way - usually to ground. – Transistor Mar 2 '18 at 21:51
• Ah right. Thank you again. I'll have to check this book. Circuit topologies is definitely what I was looking for. As for the MOSFETs, it start to make a lot more sense now. – Loïc Faure-Lacroix Mar 2 '18 at 21:58
• @LoïcFaure-Lacroix It's a good book, finally in its 3rd edition. Make sure that if you get the 2nd edition that you also buy the Student Manual for it. If you get the 3rd edition, get also Learning the Art of Electronics to go with it. Sadly, the third book in the 3rd edition series, "The x-Chapters" hasn't yet been published. And it clearly contains material that used to be in the 2nd edition and has been pushed into this still-unavailable book. So the 2nd edition + Student Manual might be a better starter for you. – jonk Mar 2 '18 at 22:22

Since you are into patterns, let's discuss a few.

There is the pattern of an LED, where the voltage source might be anything.

simulate this circuit – Schematic created using CircuitLab

In this design pattern, you only need to calculate the value of $R$, knowing the intended LED current, intended LED voltage at that current, and the voltage of the supply (or the voltage difference between the two nodes at top and bottom.) Here the equation is:

$$R=\frac{V-V_\text{LED}}{I_\text{LED}}$$

The choice as to which is first, resistor or LED, doesn't matter. The pair makes up a functional block.

Then there is the pattern of an MCU controllable series switch.

simulate this circuit

Here, you have the options of using a BJT or MOSFET for the series switch. The use of the series switch means that either the circuit is enabled or disabled, but not bypassed (see later.) Note that the switch is either on the "low side," meaning it is attached to ground and when enabled connects the load to ground when enabled; or else it is on the "high side," meaning it is attached to the (+) power supply rail and connects the load to that supply rail, if enabled.

If the $+V$ rail is different (higher) than the MCU's own $V_\text{CC}$ rail, then the low-side pattern is usually attempted. (This is because the high-side pattern can't be turned fully off when the $+V$ rail is higher.) In other cases, either approach probably will work okay.

The value of the base resistor, when the BJT form is selected, is:

$$R_\text {B}= 10\cdot \frac{V_\text{CC}-700\:\text{mV}}{I_\text{LOAD}}$$

$I_\text{LOAD}$ is just $I_\text{LED}$ if an LED is the load.

In this pattern, the first pattern of just powering an LED properly can be applied as the load. (The only issue being that the series switch itself has a slight voltage across it. So this added voltage must be subtracted from the earlier equation's numerator (for the LED current limit resistor) before calculating the resistor value. Usually, this isn't much of a difference, so the same resistor works fine.)

Then there is the pattern of an MCU controllable bypass (parallel) switch. (For this pattern, I'll focus on the low-side switching pattern discussed above. Similar, would apply in the high-side case.)

simulate this circuit

In this case, you must have a load that can be divided in half. Pattern 1 is such a load and shows up here split into two parts.

Here, the value of the base resistor in the BJT case is:

$$R_\text {B}= 10\cdot R\:\frac{V_\text{CC}-700\:\text{mV}}{ V_\text{CC} }$$

The use of the parallel switch means that the current that would otherwise go through the object of interest (the LED here) is bypassed through the switch, instead.

Note that in this case the current isn't turned off, ever. It either is allowed to proceed through the LED or else it bypasses the LED and goes through the switch. Either way, the current takes place. (In fact, since the switch often has a lower voltage across it than the LED, the resistor shown will actually increase its current when the switch is bypassing the LED.)

The main point here is that the transistor switch (BJT or MOSFET) must have a very low voltage across it, so that the load (LED in this case) won't have sufficient voltage across it to keep operating. An LED makes this easy, since both BJTs and MOSFETs, used as switches, are easily able to essentially shut off the LED by bypassing it.

Note also that the sense of ON/OFF, relative to the MCU I/O control line is opposite to Pattern 2.

(The high-side switching pattern also applies for the parallel case here. But it also has the same limitations mentioned earlier.)

There is another pattern called a current source. (Actually, there are dozens of patterns used as current sources.) Since the current is more important when running an LED, a current source makes sense.

This only works in cases where you are operating this as a low-side switch pattern and where you have access to a voltage rail that is enough higher than $V_\text{CC}$ for the MCU. (Several volts higher, at least.) In this case, you can attempt the following:

simulate this circuit

Here, the $+V$ rail must be at least the LED voltage above $V_\text{CC}$. But in this case, the value of the resistor is computed as:

$$R_\text{SET}=\frac{V_\text{CC}-700\:\text{mV}}{I_\text{LED}}$$

There are a great many additional patterns that can be built upon these. (For example, Pattern 4 can be used in a high-side switch form if there is access to a sufficiently negative voltage rail.) And the above patterns have problems in special circumstances, such as when used at high frequencies or large currents. Each one has a limited range where it may apply. But I thought I'd provide some basic ideas here that you may be able to relate to what you've seen before.