H-Bridge Circuit Design Help

Hello,
First of all, I would like to say that I am relatively new to electronics. I have read a couple books now, but I don’t know how to determine what values of components to use in my own circuits.

I attached a schematic of an H-Bridge circuit I came up with. I was wondering if this design would work, and how can I determine what values to use (particularly with the resistors.) Is there a good book on this or some sort of free/low cost software that could be used?

- Nick

Hi Stoggs & welcome.

There are a quite a few different simulators, evaluation versions are free.

Try www.tina.com I think their evaluation version won’t allow you to save files, I only suggest this one because its so easy to use.

www.kpsec.freeuk.com/trancirc.htm

looks like a good place to figure out how to calculate the resistor values.

Hope that’s helpful, do let us know how it goes, if you include more details (transistor, motor) I’m sure you’ll find lots of useful help on this forum (quite a friendly bunch)

Regards Bob

I’m not sure what motor I’m going to use, I know that it will be a DC brush motor, probably between 6 and 12 volts. I have 2 rechargeable 7.5 volt batteries, so those are most likely going to be the power supply. I’m also not sure about the transistors, but I think that 2907A (PNP) and 2222A (NPN) are good choices. Could someone explain the process that they would use to determine the values for this circuit? Thanks

- Nick

You may use L298 from ST. I have seen a simple schematic **broken link removed**

Thanks, but I'm trying to learn how to make my own circuits.

Its just a case of experience, so all you have to do is start putting a few circuit together. Try to pin down the parameters that you know (like the voltage supply) and work from there.

The easiest place to start is to grab a transistor and a handful of resistors and start to play! The fun part is blowing components off the board after all (it sort of helps you remember what not to do too)

You know that when you have a transistor in the current path, there will be a voltage drop of around 1.4V (collector – emitter) in this case you will have three transistors so you will have a voltage drop of

3 * 1.4v = 4.2v

With your batteries you have a choice of 7.5v or 15v so realistically you need a motor that will run at 3.3v or one that will run at 10.3v (7.5v – 4.2v = 3.3v) or (15v – 4.2v = 10.3v).

A couple of things to think about here:

A 3.3v motor that’s going to leave a small margin for voltage drop when the batteries get tired.
On the upside you get twice the current.
On the downside high current low voltage is a bad because the power in any component is proportional to the voltage, but proportional to the square of the current, so the lower the current the better.

So ok, lets assume your going to choose a 9v motor and a 15v supply and you can get a maximum of 10.3V across the motor. If you run the motor continuously at this voltage it will probably burn out. If you drive it with a pulse it will be fine.

Plug your motor into a 9v supply and measure the current it draws with no load, hmm lets say 0.5 A.

You now know that you will have to be able to supply at least 0.5A through the transistors. You know you get a voltage drop of 1.4 volts across the transistor, this tells you that the power of the transistor you need has to be better than (p=v*I) 0.5A* 1.4v=0.7w so your looking for a transistor that can cope at 700mW. If you use a transistor close to it’s power rating it will get really hot, so try to use one a little bigger.

A BFY51 transistor can cope with this level of power (and its quite s sturdy metal cased transistor) so lets say you chose this one.

Now you need a base resistor, the data sheet for the transistor gives a figure called the hfe, this tells you how much current you need to drive into the base to allow for a current in the collector emitter path.

You know your motor needs 0.5a in the collector emitter path, the hfe is at least 40 so the base current is 0.5a/40= 12.5mA. The hfe can vary, and you want the transistor to switch on so lets double that current. (technically this is called an overdrive factor of 2 ). Now you check back with the data sheet and find that the max base current is 100mA. So driving 12.5ma*2 (25mA is no problem).

Ok you want 25mA into the base and you have a 15v supply, so 15v-0.7v(0.7v is the junction voltage) = 14.3v. the resistor you need to put in line with the base is v/I, 14.3/25mA=572ohms. You can get a 560ohm resistor so use that. (that’s a 25.5mA base current)

Plug it in and test it, have fun and let us know how it goes.

Regards Bob

Thanks a ton, Engineer_Bob, for taking the time to explain that. It really helps to have an idea of what to start with. Now I'm going to get those components and similar ones to play around with. I'll be sure to get back to yall to post the results.