Learning about H-bridges and transistors

Question

I'm trying to learn about transistors and H bridges towards a final goal of controlling a bipolar stepper motor with Arduino.

I've done the following half H-bridge in the process of learning:

My first question: Why can't I use a single resistor to connect the base of both transistors to ground?

not2xs · Accepted Answer

If you want to get on to efficiently and reliably driving the stepper motor then as others have said a chip will be easiest; otherwise you will need to do everything the chip designer did. If you want to learn about H-bridges, the links supplied by mcb1 and Clem show good explanations. If you want to understand how transistors work, here is a quick but somewhat thorough description:

The base to emitter connection works just like a semiconductor diode. Perhaps easiest to think of it as a variable resistor where the resistance changes with the voltage. Like a diode, if you try to drive it "backwards" it behaves like a near infinite resistance, and near no current flows. If you try to drive it "forwards", like a semiconductor diode if the voltage is less than about a half a volt it still behaves like a near infinite resistance, but somewhere a bit above that voltage the resistance starts to drop, until by around 0.7V (for silicon devices) the resistance is very low, nearing zero-ohms (except for that 0.7V lurking inside). The exact voltage between base and emitter where this happens will vary, mostly with temperature. That is why you typically connect an external resistor to the base, to control how much current flows there, otherwise the transistor will allow more than it can handle.

The next thing to know about a transistor is it has a parameter (properly specified as h-subscript-fe if your fonts will allow that) I will abbreviate as h-fe here. This is the ratio of how much current will flow in the collector relative to how much in the base, if the rest of the circuit will allow it. Typically the h-fe will be somewhere around 1000 to 100000, and it will vary wildly with temperature and from individual transistor to transistor (even if they have the same part number). This means that the current in the collector will be 1000 to 100000 times the amount of current in the base - IF THE REST OF THE CIRCUIT WILL ALLOW IT. The current in the transistor's emitter is the sum of the current in the base and collector. However, when using transistors as switches, you usually want to drive them into "saturation". This is where the circuit cannot manage the collector to base current ratio and the collector to emitter voltage goes really low (below half a volt) while the transistor is trying really hard. With very low collector to emitter voltage, not much power gets consumed inside the transistor meaning it does not get as hot and more power is available to go into the load.

There are two common configurations for connecting up a transistor, and you have both in your circuit above. One frequently used way is called "common emitter", here the emitter is wired to some known voltage, the base is wired to have some current flow into it (or not) while considering that 0.7V drop, and the "load" is connected to the collector where a higher amount of current is switched. Another common way is called "emitter follower", where the base is driven to a known voltage, the "load" is connected to the emitter, and the emitter will stay around 0.7V away from the base voltage with any needed extra current to achieve that coming through the collector. To get an emitter-follower into saturation, the base will need to be driven to a voltage beyond the battery voltage for the rest of the circuit. This is why it is most common to suggest combining NPN and PNP transistors into the H-bridge, they can all be wired common-emitter and saturated without using multiple supply voltages.

If you want to use transistors as amplifiers instead of switches it gets more difficult due to having to deal with the variations mentioned above. When dealing with motors or coils you should be aware of the "kickback" voltage spike they produce, hence all the extra diodes (and sometimes capacitors) in well-designed H-bridges.

If this is what you were interested in learning, then good for you.

strb · Answer

I'm not sure that this schematic works... the led isn't reverse polarized? Anyway use a pnp as pullup transistor and a npn as pulldown, it's a good thing to put also a limiter resistor to control the current that flow into the led. In your schematic you can't wire together bases because they are at different potential (b1 has a voltage of about 4.3V, b2 5-Vec2- Vled-0.7) I suggest you to take (from internet or a book) a schematic of a full bridge and study how it works from there.

mcb1 · Answer

What I'm having issues understanding is how transistors work You can consider a transistor as a switch (in this case they are either ON or OFF) The control part of the switch is the base to emitter (NPN) or emitter to base (PNP) junction. This usually requires 0.6v across it to 'switch' on the Collector to Emitter (NPN) or Emitter to Collector (PNP) junction. When in the 'ON' state the C-E/E-C is as low as 0.2v or less in larger devices. In your half bridge circuit, when you switch S2 on, there is current flow from the +ve through the E-B of Q3 thru R3 and S2 to ground. ... Q3 is switched ON and current is able to flow through the E-C of Q3, thru LED2 and appears at the Emitter of Q2. .... Current will flow via Q2's E-B, R2 and via Switch2 to ground BUT .... as soon as Q2 turns ON the E-C of Q2 will reduce the voltage to 0.2v at the emitter.... Q2 will then be OFF. It is likely that the circuit will adjust itself to end up with 0.6v at the Emitter which is not ideal. BUT the real biggy here is you have effectively dumped 5v across a poor led with no series limiting resistor, which tends to make them very bright for a millisecond. You really need to add a 220 ohm resistor in series to stop them being destroyed. The only thing saving your leds and giving the illusion of the circuit working is that Q2 is not turning on and the led current is flowing through the E-B and 330 ohm resistors. Please note I have only described one side, but the same could be considered for the other half. We all start learning somewhere so trying to understand is a great way to learn ...well done. Mark BTW It's not your fault but generally positve voltages are at the top and ground is at the bottom. It tends to come from our right to left and top to bottom reading ... plus water flows downwards.

clem57 · Answer

strb is right. You need to learn the what before the how! Read https://en.wikipedia.org/wiki/H_bridge for the what is. For the how to, read nathandumont.com : H-Bridge Tutorial that explains the how. Here the H bridge is used with a motor in the middle. The opposite switches open and close to induce the motor to turn one way or the other(reverse). In the diagram the top has voltage applied with the bottom as a drain of the current. Think this is like water and it makes more sense way you have the drain. This works for high current motors. Clem

ovidiub13 · Answer

I fixed it. I think. I've added a resistor for the LEDs, and 2 bigger resistors for the transistors bases to + to make sure the gates are closed when the button is not pressed.

Problemchild · Answer

of course the Vf of the Blue diode is more like 3.2-3.8 and we have 2? junctions worth of drop between 5V and GND so even less current all told

Problemchild · Answer

Yes except they are rarely identical unless you are limiting the gain with resistors and the switching logic and drive for the bridge is often not considered so you end up with a system that can short out or not have enough gain to turn on properly or have too low voltage to do the same. Failing that just dealing with back EMFs etc is just an annoyance. Beyond proving you can do it you are better off with the chip unless you are switching mega-amps or something

mcb1 · Answer

How can I fix this? How can I make them work? prove my point by shorting out the E-B first to confirm the suspicion. A quick check using Google shows this. H Bridge motor driver theory & practical circuit using transistors- animation/ simulation: Getting started with Robotics… This also explains how it works. We already pointed out that you don't require a bridge to drive a stepper. HOWEVER unless you advise what stepper you intend using, then it's hard to help. shabaz is quite right that your circuit is not designed for inductive loads, and there are plenty of simple IC's that take all the hard work out of it for you. Mark edit 3x spelling mistakes .. I blame the keyboard not keeping up.

Robert Peter Oakes · Answer

If you are new to electronics then your diagrams are not that bad... wrong perhaps but pretty good start for a newcomer and you will only get better over time In the diagram above i would suggest a small resistor in series with the LEDs as they are in parallel with the coils and so will limit the volts to the coil and conversely when the transistors switch off there will be a back EMF that could destroy them. You should have a protection diode (NOT an LED) across each transistors load to minimize the impact of this The diagram you have above btw is not an H Bridge, the stepper you're using in the diagram has 5 or 6 wires with the center taps of each coil going to +Vcc, the rest are sequenced to ground by the controller, you can also simplify the components by using an LM2003 darlington driver chip, it will handle steppers of this kind when they're not too big, it will also handle motors of more than 5V if needed, LEDs with series resistors can easily be added to give visual indication of operation, you could also put them on the input of the chip rather than the output, then just use a 330R to 1K Ohm resistor in series with an LED to 5V and it should not load the output of the arduino too much btw, here is a more readable version of the H bridge schematic (The bit with the transistors anyway, you just need to add your IC back to it. Any NPN transistors that have the required specifications will work, it all depends on the load your switching and the drive your using (The chip in the case of the diagram above). This diagram is also the more traditional way of showing the H Bridge, yours is not wrong, just harder to follow. I would also use 1N4001 ot better diodes than a 1n4148 as those are just signalling diodes and may not last long with any but a tiny motor

Learning about H-bridges and transistors

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