Looking for a good book on Transistors in digital logic-to-power circuits

[jerryk]

Hi All,
I've been reading a bit and was wondering if there are good books specific to transistor applications when using TTL logic to drive motors etc. I'm a programmer, and understand digital logic (and/or gates etc) but I don't know how to select the proper transistor to digitally control a power load.

Assuming I have a low-voltage logic ciruit and higher voltage power circuit, I'm not sure what to measure to ensure I've got the right transistor to handle the jump.

Similarly, the power jump may be big enough to require a darlington set. I understand darlingtons permit driving greater power, but I don't know how to tell what the 'cutoff' is between a single transistior and the need for darlingtons.

Are there books that directly address digital-to-power transistor selection? If you've learned this from a college text, I'd be happy to know the ISBN etc.

If the only way to learn it is piecemeal via google, I can do that...but I'd prefer to add a good book to my library if one exists.


Many thanks,
Jerry

 

[crutschow]

Don't know of any good books on the subject, although I'm sure there are.

Google will give you a lot of references, but it may take some digging to determine which are really appropriate to your needs. For example, here's a short tutorial on transistor drivers from a digital signal.

As far as Darlington's, they have much higher gain than a single transistor but are slower and have a higher "on" voltage, so that's a trade-off you have to make for the particular load you are trying to drive.

Logic-level MOSFETs are often a good choice since they require no steady-state drive current. But if you require fast switching, you need a gate driver that can rapidly charge and discharge the large gate capacitance of power MOSFETs.

 

[jerryk]

Thanks Carl, I'll keep looking...but that tutorial is a great help. I have a handful of TIP120 and TIP125's and wasn't sure if 5v (digital logic) was enough to feed the collector. Looks like it can. And 24V is double the loads I'm trying to switch, so I may just start experimenting.

 

[Jon Wilder]

5V logic does not go to the collector. It goes to the base.

Bipolar transistors are CURRENT devices. An NPN transistor only needs 0.7V between the base/emitter junction to turn on. The important thing is to know how much base CURRENT is needed to saturate the collector/emitter junction fully on. This is calculated via the expected load current (i.e. the amount of current that the load you're switching on/off is rated to draw) and the current gain of the transistor.

Let's take your TIP120 for example. This can handle a continuous collector current of 5 amps and exhibits a current gain (i.e. "hfe") of 1,000. This means it will take 1,000x less the load current through the emitter base junction to fully saturate the transistor on.

Now let's say your load draws 3 amps @ 24V -

3 amps / 1,000 = 3ma

This means you will need AT LEAST 3ma of emitter base current to fully saturate the transistor on. Usually you'll want more emitter base current than is actually needed to ensure the transistor fully saturates on. 10mA would be acceptable. Since this transistor is rated for a maximum base current of 120mA, 10mA is perfect.

This being said...now we calculate the required value of the base resistor to give us 10mA at our logic +5V -

5V - 0.7V Vbe drop = 4.3V

4.3V / 0.010 Amps (10mA) = 430 ohms

470 ohm is the closest standard value. This will give us about 9mA of base current with a 5 volt logic supply, which is more than enough. You'll want to make sure your logic device is rated to source this much current. If it can't, you can use a pull up resistor and have the logic device output source a ground to the base to turn it off. If the logic device is "open collector" or "open drain", a pull up resistor is a must.

So the circuit would look like this -

**broken link removed**

With a pull up resistor, it would look like this -

**broken link removed**

Arrows indicate direction of electron flow.

From the diagram you can see that emitter current is equal to the sum of base current and collector current.

So from this we can conclude -

Require load current and transistor current gain rating (hfe) determine how much base current is required to saturate the transistor fully on.

The value of the base resistor sets how much base current can flow with a given voltage supply.

The voltage drop across the base emitter junction will always equal 0.7V.

 

[jerryk]

Thanks a ton Jon. I needed a concise method of what to measure on the load current and how to scale the transistor/resistors properly for that current. This will help a lot.

 

[Jon Wilder]

For load current you would just connect it to a supply of the proper voltage with your DVM in series with the load set to measure amps.

Then you would divide the measured load current by the hfe rating of the transistor you're using to figure out the minimum required base current to saturate the transistor on with that load.

Now when calculating for the resistor, use an mA value that is a few mA greater than the minimum required base current to ensure that the transistor is pulled well into saturation rather than having it just on the borderline of saturation. If you calculate 3-4mA, use 9-10mA. Then divide 4.3V by that mA figure to get the base resistor value and use the closest standard value. I say 4.3V because 0.7V of your 5V logic signal gets dropped across the base/emitter junction.

This will ensure that you will have a little more than the required base current to saturate the transistor on with your 5V logic signal.

 

[crutschow]

Let's take your TIP120 for example. This can handle a continuous collector current of 5 amps and exhibits a current gain (i.e. "hfe") of 1,000. This means it will take 1,000x less the load current through the emitter base junction to fully saturate the transistor on.

Using the small signal gain value to calculate base current for a saturated switch configuration seems to be a common error made by many in this forum. That gain will not fully saturate the transistor.

If you look at the data sheet you will see that the minimum gain (hfe) of 1000 for the TIP120 is given for a Vce of 3V, not saturated. To properly saturate the transistor under all conditions you will need to use an hfe value several times lower than that. The data sheet uses a value of 250 (5A / 20mA or 3A / 12ma) for their Vce(sat) limits.