Driving tri-state LED matrix

My link disappeared for some reason. Here it is:

Try redrawing this in a conventional manner, with Vcc at the top. I started looking at this diagram but I lost interest the minute I saw that I was going to have to unravel the circuit before considering it. There's only so much time people are willing to invest in considering a news post.

Are you talking about the way the emitter on the PNP is the bottom pin rather than the top pin? If so:

Are you talking about the way the emitter on the PNP is the bottom pin rather than the top pin? If so:

Tomas -

There is a reason for transistor symbols looking a bit like a diode between the base and emitter. In your circuit a big current will flow from VCC through the emitter and out of the base of the first (leftmost) PNP transistor, into the base and out of the emitter of the NPN transistor. The second PNP transistor will turn on, the first two transistors will be destroyed.

Why not download SwitcheCAD from

Michael Kellett

See this is exactly what I was thinking... but miraculously this worked perfectly in my college project. Even if my board mixed up the collector with the emitter for one of the transistors, it still doesn't explain why both transistors weren't always turned on... ?!

Transistors work with emitter/collector exchanged. Their efficiency, Vcc, etc. are sharply curtailed. They are not characterized for such operation.

Is there anyway I can get this to work... ? I'm thinking of using one BJT and one FET, and also a pull-up/down resistor. I'll make up a circuit and post a picture.

OK here's what I've got:

A few points about it:

• Because the source of the NMOS isn't going directly to ground, I have to get a FET that has a 4th pin for the bulk.

• The 10k is a suitable resistor value to put the transistor into saturation.

• The 200k is a pull-down resistor to ensure that the NMOS gate is kept low when the uC pin is set to high-impedence. This resistor should be as high as possible to ensure that it doesn't provide a current path that will turn on the PNP transistor all the time, but at the same time it has to be significantly smaller than the output resistance of the uC pin to ensure that the entire 5 V is dropped within uC, leaving close enough to 0 V across the 200k resistor.

Any thoughts? (..other than the repetitive "get yourself a uC with more pins")

I have already suggested the MOSFET drivers that preserve the

Much lower parts count, and no gymnastics.

-jg

At this point you can put on a shift register and actually use fewer pins on the micro.

Robert

Sorry I can't find your post where you mentioned those drivers. Are you saying that there's a driver chip out there that will preserve the tri-stateness of the uC pin?

Yes, Devices like this , Drive HL, LH, and LL on Tristate, 2-3A drive spec.

It is an emerging design trend in SPMS, to allow the flywheel FET to not discharge the load in low current situations.

Such devices can also level shift, allowing the LED drive to come from before the regulator (saves power in the regulator, improves LED load line)

-jg

I don't see how this circuit will work any better than your other one. In fact, I think it will work very poorly. I don't think you are considering the voltages in the circuit. If you have thought about that, what will the voltage be at the source of the NMOS FET when it is turned on? What will be the voltage at the emitter of the BJT when it is turned on? What will be the current in the LED in either case? What is the voltage on the base of the BJT when the MCU output is tristate? What will be the current in the two LEDs when the MCU output is tristate?

Instead of going with exotic parts, why don't you try using a practical circuit arrangement? Is there a reason that you used an NMOS FET with a PNP BJT instead of a PMOS FET and an NPN BJT? If you consider the fact that the BJT has a BE voltage of 0.7 volts when on and the FET has anywhere from 2 to 5 volts when on, I think you will find that the BJT is the part you want to use in the emitter follower configuration, not the FET.

Do you know what an emitter follower is? I believe you are looking at the components with too simplistic of a model. You need to understand the difference between current gain and voltage gain and how a FET vs. a BJT achieve each. The look at the circuit in smaller parts. Just consider what it will take to turn an LED on at a particular drive level using both types of FETs and then do the same with BJTs. The consider what happen with each circuit when the input is undriven. Only then are you ready to consider how to combine the circuits to do what you want.

BTW, someone already suggested that you download the Linear Technology SwitcherCAD (Spice) program to simulate your circuits. This is a very easy program to use, there is an excellent support group on Yahoo and you will be able to see *exactly* what is going on without building anything and burning it up.

The typical design process is to first analyze new circuits using paper and pencil based on theory. Then simulate the circuit to see if it behaves as you expect. Only then does an engineer bother to build a new circuit to test it. Testing is the worst way to verify if something works because it is *very* hard to test under all conditions. So you have only proven that the thing works on your bench, not in real world conditions (which include more than a week running time). Testing is really done to verify that nothing stupid was done in the earlier parts. After all, simulation is limited by the GIGO phenomenon (Garbage In - Garbage Out).

To be honest, your posts here are getting a little old. Not that you are asking questions, but that you aren't putting much effort into answering them yourself. Once you have figured out what will work and what won't, then I would be happy to see you post your circuit and ask for comments.

Rick