Digital Output from a Microprocessor - I need a current amplifier

[BrianRismoen]

Hello,

I am currently building a microprocessor controled project, and i will be needing a way to take the +5v outputs and make some high current outputs with them, rather than the low current that is available at the pin of the i/c. i built a circuit around a tip3055 npn transistor that works, BUT the output is not high enough - perhaps i built an inefficent circuit.

the output requirements are 1 amp or more, able to switch fairly fast. each output will be switched on at about a maxium frequency of 16 khz, but probably much slower, like 1 khz. or even less. this is for an led display board, so what i am talking about is a multiplexed display controller, and i just need the outputs to be able to drive 50, 20ma leds at a time. usually, not all the leds are lit, but it does need to handle it if they are.

for now, a circuit that just amps up the available curent would be great. to make it easier for me, its probably also better to have the output be at 12 volts. so lets assume that the main power supply is 12 volts dc, and the input is a +5,0 small digital signal.

cost is also a concern, so while it might be quick to design, it would also be better built with a 50 cent transistor, than a $5 unit. 50x50 display uses 100 of them.

idealy, a design that uses the tip3055, or the MJE3055TG (ON semiconductor corp.) would be best, since i have some samples here now.

thank you,

Brian

 

[MikeMl]

Get a logic-level gate NFET, likely in a T0220 package. Connect source to ground, gate to the PIN pin, Drain to load. Other end of load is tied to positive voltage supply.

What is the supply voltage?
What kind of load? Inductive?

There are dozens of inexpensive FETs that will work for your application.

Cheapskate solution: Make your own Darlington (two transistors)

 

[audioguru]

Since you insist on using the TIP3055 transistor, a 1A load can be connected to the + power supply (12V if you want) and its collector.
With a 100mA base current. The 2N3055 saturates with a voltage loss typically of 0.15V. Its 100mA base current can be provided through a 39 ohm resistor by a 2N4403 PNP transistor with its emitter connected to +5V. The 2N4403 has has a 10mA base current provided through a 390 ohm resistor from the microcontroller.

 

[peufeu]

> would also be better built with a 50 cent transistor, than a $5 unit. 50x50 display uses 100 of them.

Where are you going to put 100 2N3055 transistors ? lol.

You can use double MOSFETs in SO-8 package or smaller. Typically those are pretty cheap, like 30c in your quantity. Any MOSFET will do if it has a low gate charge, RdsON < 0.1 ohm, and Vds > 20V. This will use a lot less space and cost less. You can drive those with the output of a high current logic gate like 74AC if you need speed, but it's not critical since the switching losses will be spread around all the MOSFETs.

Example :

Digi-Key - QS5K2CT-ND (Manufacturer - QS5K2TR)

Darlington arrays could also do the job.

Or something like that :

https://www.electro-tech-online.com/custompdfs/2011/03/MC33996.pdf

 

[Mike - K8LH]

If it's a 50x50 multiplexed display, and assuming a 1/50th duty cycle, then lighting 50 LEDs at 20-ma "peak" current only produces the equivalent brightness of approximately 400-microamps "average" current.