Base resistors needed for saturated transistors? 8

[mrkenneth]

Are resistors needed between the transistors and the PIC microcontroller in the diagram below?

Each transistor will be driving a variable load (actually, multiplexed LEDs) from 0 A to 1.25 A in saturation. The LEDs are connected in series to current-limiting resistors, so the current between the collector and the emitter terminals will be not be exceeded. I am worried about whether the base current would destroy the transistor, since there are no resistors... Does the base draw current when the transistor is "off" (Vc = Ve = +5V, while Vb = 0V)?

Thanks in advance!

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[cbarn24050]

Hi, you allways need a base resistor to limit current. In your circuit there is not enough output voltage from the pic to turn the transistors off, so your circuit wont work anyway.

 

[IRstuff]

Why would you think that you could run 5V forward bias across a base-emitter junction?

TTFN

 

[mrkenneth]

To turn the transistor off, can I TRIS the port pins to inputs, so that the pins would present a high-impedance to the transistor and thus not sink any current?

The datasheet says the maximum Vbe is 5V... I guess the low output of the PIC would be around 0.6V because of the high current? How much current would the base source?

Thanks in advance!

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[IRstuff]

Max Vbe rating is REVERSE breakdown. Base current is proportional to e^Vbe/(kT/q). Vbe is normally around 0.7V; you can do the math.

I suggest that you review some basic transistor theory before wiping out a cadre of innocent transistors.

TTFN

 

[cbarn24050]

Hi, well you could do that but first check the datasheet to make sure its allowed and that the input leakage isn't too high. As you want to switch 1.5A you will find your allready at the upper limit of a PIC/single transistor output stage so you'll probably have to use a 2 transistor output anyway. You also could do with a base-emitter resistor as well unless you are prepared to tolerate dimly lit "off" leds.

 

[BrianR]

The only way you will turn those transistors off in that circuit is to apply +5V to the base. (Reduce Vbe to 0). The data sheet should not say the max Vbe is 5V it should say the max Veb is 5V. The max Vbe is always less than 1V as typically only 0.6V is needed to turn the transistor on.

 

[mrkenneth]

Thank you all for the information. I have not had to work with transistors much. (IRstuff, I have learnt to use electronic components through their destruction. )

The reason I thought the transistors could be connected directly to the microcontroller pins was because the emittor of the first transistor in a Darlington pair is directly connected to the base of the second without any resistors.

I shall add the resistors between the PIC and the transistors. From the datasheet, I can get close to 1.25 A with around 25 mA of base current at a Vce-sat of around 0.3 V.

Will there be a current between the collector and the emittor terminals if the base is left floating (when the PIC pins are set as inputs)? How much current will be sourced or sunk by the PIC in that state?

Thanks in advance!

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[BrianR]

This is a bad circuit and is not going to work.

You will never get those transistors to switch off until you reduce Vbe to 0, ie raise the base to 5V the same as the emitter.

It is also bad design to ever have an open circuit base as leakage current (Icbo) and noise can cause the transistor to conduct.

A practical way to make this work is to use NPN transistors, connect the emitters to ground, use a base limiting resistor and one to ground and put the collector load resistors to +5V.

 

[itsmoked]

Brain is correct mrkenneth (like everyone else 4 that matter) Remember just because you can draw 25mA from one PIC Pin doesn't mean you can do it from a truckload of them. Look at the PIC specs you will see a TOTAL limit per port and per chip.

Also as you drag more current out of a pin the voltage on the pin drops rapidly. If the PIC is sucking the current into a pin then the voltage on the pin rises ...a lot...

Just get a handful of these. Put them in Source to ground.
Hook the Gate directly to the PICs use the Drains on your LEDs. Add a current limit in series with the LED. Have a blast while not thrashing the PIC.

A lot of other FETs would do to..

 

[itsmoked]

Oops forgot the example:

http://www.irf.com/product-info/datasheets/data/irf3706.pdf

 

[felixc]

Yup, to make it easy and simple, follow itsmoked advice, otherwise it will smoke! P-Mosfets will enable you to do the job easily. Search for logic-level P-Mos.

 

[logbook]

I hate to be a dissident, but if the PIC is run from +5V then it should be workable (provided resistors are used in series with the PIC outputs). The PIC outputs will go all the way up to the power rail, definitely turning the transistors off. The 25mA base current is presumably only on one transtor at a time because this is a multiplexed LED drive circuit.

 

[cbarn24050]

Your right logbook, I don't think anyone thinks differently. He's still going to have trouble getting the needed output current unless he can find a high gain power transistor.

 

[mrkenneth]

Thank you for all the advice. I am not able to run the PIC off 5 V, unfortunately, because all of the pins are used so I am using the regulated 3.3-V rails as the reference for the onboard A/D convertor.

I should have been more detailed and clear in describing my application in the original post. I have created the following schematic that I hope will clarify the circuit:

The PIC will be controlling an 8 x 8 LED matrix, with eight pins driving an 8-channel NPN Darlington IC and another eight pins driving a set of PNP transistors. (I have only shown one PNP transistor simplify the schematic.) Only one PNP transistor will be driven at any time.

Can I add pull-up resistors at the base of the PNP transistors so that when the PIC pin is set as an input, the pull-up resistor would raise Vb to +5 V? I can just use a bussed SIP resistor for all eight pins.

I am planning to use the MPS751 PNP transistor, which should be able to output 1 A from 25 mA with a Vce of under 1 V.

If the bipolar transistors do not work out, I guess I will try itsmoked suggestion to use FETs. I would prefer to use the smaller TO-92 packages and reduce the cost at the same time.

Thanks in advance!

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[cbarn24050]

Hi, 8 leds at 160ma each is 1.2A (ish). Take the datasheet for your transistor print out figure 6 and draw in the 1.2A saturation curve. Notice at 25mA base current you are on the corner of the line, you really need to be on the flatish part to the right, they are typical values imagine how much worse it is for a worst case device. Have you checked the PIC datasheet for the input leakage current for 5v input on 3.3v supply? Do you really want to run your PIC outputs at their datasheet maximums? You will need a base-emitter resistor.

 

[IRstuff]

You need to at least 3 diodes in series with the base to come close to being able to shut off the PNP. A pullup alone will not work. The high-side drive in CMOS would be a P-channel device, whose drain will forward bias above about 4V on the output. This is bad for a number of reasons.

Alternately, you need to run from an open collector driver with a pullup to 5V.

TTFN

 

[itsmoked]

I sure understand the fixation on BJTs... and all the resistors and diodes, etc. etc.

 

[mrkenneth]

Thanks for the analysis cbarn24050. I was interpolating Figure 6 for the Vce value too... Perhaps I will have to reduce the LED current... The LEDs will be off most of the time, so I guess they would dim when all of them turn on?

The datasheet for the PIC says that the leakage current is +/-1 uA, but only when the voltage at the pin is between Vss and Vdd...

Do you think I can place a ULN2803A "sink driver" between the PIC and the PNP transistors as an extra gain stage? The leakage current of the device at 85 degrees Celcuis is 100 uA maximum.

IRstuff, doesn't a PIC pin act like an open collector when it is set as an input?

itsmoked, sorry, I have never used FETs before... Not that I have used BJTs much either. :-( I just looked at Digikey and all of the FETs that can handle 1.2 A are more than three times more expensive than BJTs. :-(

I have added a pull-up resistor to the schematic below:

Thanks again to everybody!

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[cbarn24050]

Putting the 2803 to drive the pnp transistor will work well but any small npn transistor would do equally well.