Hi everyone,

With reference to the circuit shown below, could someone please suggest what happens when Q1 and Q2 are switched on?

I can see that a N/A is listed in the truth table, but I would like to understand the physical effect on the motor when A=1 and B=1 condition occurs.

Thanks!

 

All the transistors conduct and create a short circuit from +V to GND.

 

The N/A on that truth table should be labeled "Smoke".

 

Oh crap...i made a mistake by not looking at the circuit closely. My question is basically about Q4 and Q2 as ON, and Q3 and Q1 as OFF (as shown below). What happens in that case?

Please refer to the following circuit for the question,

 

Nothing, you'll have equal potentials of V+ on each of the motor leads.

I notice the circuit has changed, the new one uses both NPN and PNP transistors.

 

In all H bridges (both schematics) having both Q3 and Q4 or Q1 and Q2 on at the same time is a disaster. Having Q2 and Q4 or Q1 and Q3 both on is effectively off, and it is probably a form of dynamic breaking, since shorting the leads of a motor will cause the back EMF of the motor to stop faster.

 

I know that this is an old thread, but I have a question.

Is there an advantage to the H bridge that uses NPN's and PNP's over the circuit that just uses NPN's? I notice the NPN only circuit is much simpler, and easier to understand, and also has less components (no diodes). So why not just use it?

 

In the all-NPN circuit, the low side transistors are saturated switches, but the high side transistors are emitter followers. The saturated switches will have a low Vce, for low power dissipation in the transistor itself. The emitter followers will have a minimum Vce of ~0.62v, which can increase considerably depending on motor current.

So, the power dissipation in the high side NPN transistors is considerably greater than the low side transistors.

This is different from using all N-ch MOSFETs in an H-bridge, but that's another story.

 

In the all-NPN circuit, the low side transistors are saturated switches, but the high side transistors are emitter followers. The saturated switches will have a low Vce, for low power dissipation in the transistor itself. The emitter followers will have a minimum Vce of ~0.62v, which can increase considerably depending on motor current.

So, the power dissipation in the high side NPN transistors is considerably greater than the low side transistors.

This is different from using all N-ch MOSFETs in an H-bridge, but that's another story.

Its true that the power dissipation in the emitter followers will be big, but I thought that other than being just switches to turn the motor on and off, the transistors in the H bridge are also supposed to drive the motor. Don't you want the current to be as high as the motor or load requires? The motor won't take more current than it needs.

I guess its probably because I did not take the time to figure out how the H bridge with the PNP's works. How does it decrease the power consumption?

Arghh why do professors always skip PNP's and concentrate on NPN's? So a "0" at the base turns a PNP on, and a "1" at the base turns the PNP off when being used in saturation and cutoff mode, right? Reverse is true for NPN.

 

Its true that the power dissipation in the emitter followers will be big, but I thought that other than being just switches to turn the motor on and off, the transistors in the H bridge are also supposed to drive the motor. Don't you want the current to be as high as the motor or load requires? The motor won't take more current than it needs.

Undesired power dissipation is a big problem nowadays. It didn't matter so much 30-50 years ago when energy was relatively cheap and plentiful, but that's no longer the case.

If you're using an emitter follower configuration, then you have at a minimum an 0.62v drop from collector to emitter (Vce) before you start to get any useful amount of current out of the transistor. Depending on the gain of the transistor, base current and transistor temperature, the Vce increases along with increases in the load current even up to the point where the transistor is destroyed due to excessive power dissipation.

Conversely, if a properly-rated transistor connected as common emitter is in saturation, the Vce will typically be < 0.3v, so the power dissipation in the transistor will be far less than the emitter follower; at least 50% less.

Since less power is being wasted in the transistor as heat, there is more power available to deliver to the load. This becomes very significant when operating low-voltage circuits.

I guess its probably because I did not take the time to figure out how the H bridge with the PNP's works. How does it decrease the power consumption?

It's not that total power consumption is decreased; it's that more power is delivered to the load instead of being wasted as heat in the transistors.

Arghh why do professors always skip PNP's and concentrate on NPN's?

H-bridges can be difficult to grasp if one is not already familiar with the concept. Keeping it as simple as possible has big advantages to decrease the learning curve. Of course, once you get the basics down, it's time to move on to more advanced techniques.

With the all-NPN H-bridge, if you had a way to raise the voltage on the bases higher than the motor supply voltage so that Ib=Ic/10, you would get the transistors to saturate. However, having an extra current supply for the bases would not really be very efficient. MOSFETs are used because once the gate is charged or discharged, no further current is required to maintain the state of the switch.

So a "0" at the base turns a PNP on, and a "1" at the base turns the PNP off when being used in saturation and cutoff mode, right? Reverse is true for NPN.

Cutoff is usually when Vbe is around 500mV for NPN, and -500mV for PNP.
Collector current in the mA range usually begins when Vbe is around 620mV for NPN, and -620mV for PNP.
To get a transistor into saturation, the usual formula is Ib=Ic/10, where:
Ib = required base current
Ic = required collector current
That's because on almost every transistor datasheet you'll look at, Vce(sat) is specified when Ic/Ib = 10 (a gain of 10).

So, to calculate the base resistor required to obtain the needed base current, you'd use:
Rbase = (Vin - Vbe) / (Ic / 10)
where:
Vin = the voltage applied to the resistor terminal that is away from the base.
Vbe = the voltage on the base required with the desired base current flowing
Ic = the required collector current.

So, let's say you have a desired Ic of 200mA; you'd start off dividing the collector current by 10 for an Ib of 20mA. You have an imaginary NPN transistor that has a Vbe(sat) of 0.9v when base current is 20mA. Your Vin is 10v.
Rbase = (10v-0.9v) / 20mA = 455 Ohms would be the ideal value.
Look at a standard decade table of resistance values:
http://www.logwell.com/tech/components/resistor_values.html (bookmark this page!)
Examine the E12 and E24 columns; they are pretty commonly available and relatively inexpensive. E48 and higher get expensive.
In E12, you have 390 and 470 for the closest value.
In E24, 430 and 470 are the closest values.
Now, 470 Ohms is just 3.3% higher than 455, and that's what I'd go with.
If you don't mind using a pair of resistors, here's a very handy online series/parallel resistance calculator:
http://www.qsl.net/in3otd/parallr.html (bookmark this page!!)
Just select what E-series you want to use, type in the value you need, and it'll give you a variety of combinations you can use. For example, two E-24 910 Ohm resistors in parallel = 455 Ohms exactly.

Don't forget to calculate the wattage requirement.
P=IE, so 20mA * (10-0.9v) = 182mW. For reliability's sake, multiply that result by 1.6 to get 292mW, and choose a resistor rated for that or higher. 1/4 Watt is not enough, so you'd need a 1/2 Watt resistor. If you used a 1/4 Watt resistor, it would tend to run significantly hotter than the 1/2 Watt resistor, and would not be as reliable over time.

Here's a link to a datasheet for an MPS2222A transistor: http://www.onsemi.com/pub_link/Collateral/MPS2222-D.PDF
It's a nice one because it has lots of plots documenting how a typical transistor might respond to a number of variables.

 

Thank you, very detailed explanation. This will help others.

I am going to make my first robot in a few weeks. I bought a used RC monster truck, took all the insides out, and now can control either the forward/backward motor, or the wheel turning motor.

I want to use a pic microcontroller to control the robot. So the input to the H bridge will be 5V. The robot will be able to go forward/backward and turn in either direction. So I will need two H bridges then. I want to try the NPN only H bridge and see what the difference is. I don't know the specs for the motor since its in a used toy, and nothing is written on either motor.

So far I have been doing it by trial and error. My current meter in the DMM is busted, so I still don't know how much current is drawn LOL, but I noticed that anything higher than 3V applied to the motor is too much. It goes too fast for what I want it to do.

I want to attach a few photoresistors to the robot, and program the pic so that the robot follows a flashlight that I shine on it. Kind of like a line follower, but more like a flashlight follower.

I will probably make my own thread about this soon, but for now I need to design the H bridge before anything else.

Will keep you posted.