No, you control the LED current by changing the value of R2 - the transistor should be saturated when switched on.

It shouldn't do!.

You might find my PIC tutorials helpful, the "Hardware Extras" section shows various ways of connecting LED's etc.

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PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Which transistor are you using?
R2 adjusts the LED current. If R1 is less that 100 ohms, you are taking more than 32mA from the 74HC595. Its absolute max is 35mA.
The transistor's emitter voltage is supposed to be the supply voltage of the 74HC595. If they are different then you blow-up the 74HC595. The recommended max supply voltage of the 74HC595 is only 6V.

Big problem. It has much more than 0.8V drop across it. Your 5V supply and even 6V is too low for two LEDs in series when you add the voltage drops.
Use Mosfets to sink the voltage down to zero.

Attached Images

uln2003a.png (16.8 KB, 236 views)

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Uncle $crooge

 

Hi AudioGuru

Thank you for your reply.

I am using the 2N4403 PNP with R1=120ohms going from the 595 to the Base and 22 ohms going from the collector to the LEDs.
I am supplying the 595 with regulated 5V.

I was wondering if I can put say 7.5V on the Emitter of the PNP? (I have tried this and this has a much more dramatic effect on current through the LED's than changing the R2 (=22ohms)).

I am using the ULN2003 right now, it "works". I drive the ULN2003 with a PIC, and it is grounded (but the COM port is left unconnected). Only sometimes does no LED light up. At this point I just switch off the power supply, give it a couple of minutes and try again and it works again. I am not sure what I am doing really, any comments would be greately appreciated.

Once again thank you for all your help.
Regards
a.

 

Hi AudioGuru

Sorry it did not work as I reported before. What I see happen is the entire matrix lights up (quite brightly). If I put the Emitter back to 5V then the right sequence of LED's are on.

When I measure the V at the Base with the 5V supply and the 120Ohm R1 I see that V = 4.1V and 5V depending what the 595 is driving. Why is the 4.1V when 595 is showing 0 (actually its showing 0.6V or so)?

Now I am really confused.
Any comments or suggesionts would be greatley appreciated!

Regards
a.

 

The emitter is at +7.5V. When the 74HC595 is supposed to turn off the LEDs then its output to R1 is its supply voltage of +5V. The transistor has current through R1 so turns on the LEDs continuously. Actually, the current in the base resistor is trying to pull the output of the 74HC595 above its supply voltage, that's very bad for it.

If the display is multiplexed then each voltage is present only for a moment. A 'scope will show the transistor's base voltage at +5V when it is off, and at about +4.1V when it is on. A multimeter might average the voltage at 0.6V across the transistor's base-emitter.

The 32mA from the 74HC595 causes a voltage drop of about 0.6V across its output resistance. Its output voltage is 0V for a low without any current.
The 32mA in R1 is created by the remainder of the voltage across it.
Actually, with R1 at 120 ohms and the base at 4.1V and the output of the 74HC595 at 0.6V, the 3.5V across it limits the base current to 29.2mA.

The 2N4403 is nearly the same as a BC327, but has a different pinout.
They work very well up to about 100mA then above that their current gain drops and their saturation voltage rises.

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Uncle $crooge

 

Greetings everyone, I have been playing with the transistor switch circuit in Uncle $crooge's post for exactly the same application - switching segments of a 7 segment display (multiplexed) - from a 74HC595. The only difference between this circuit and mine is that I have put a 'pull down' resistor from the output of the 595 to ground to prevent the base of the transistor from floating.
I have a couple of questions;
Why do we need to try and pull almost max current from the 595? Surely this will eventually damage the chip.
I have experimented with various values for the resistances and found that what seems to work best is R(base) = 5.1k , with R(pull down) = 1K, R(load) = 510. I am using 3V to simulate the signal from the 595 and a different 3V source for load. I have found that using any R(base) value less than 5.1 K, causes the LED to light without the 3v for the load being connected (must be getting a leak through the transistor). Obviously I will have to change these values for the 5v that will be eventually used. I am just thinking that the 100 and 10 ohm resistors will cause to great a current, or does the transistor require this base current to switch properly.
Last one.......
It does not seem to matter if I use a BC337 (NPN) or a BC327 (PNP). I get exactly the same outputs. Why would that be?

Sorry for the 'noob' style questions, but that is what I am Well not really, its just been over 20 years since I played with electronics!

Regards Jon

 

Anyone got any ideas?

 

No one able to help?

Guess I shall have to keep looking elswhere..........

Cheers

 

Post a schematic. I don't think your question has much if any relevance to the OP's question.

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Ron

 

Well, the schematic is here......

http://www.electro-tech-online.com/a...f?d=1135185273

The only difference between this circuit and mine is that I have put a 'pull down' resistor from the output of the 595 to ground to prevent the base of the transistor from floating.

Q1. Why do we need to try and pull almost max current from the 595? Surely this will eventually damage the chip. (32ma / 35ma)

Q2. It does not seem to matter if I use a BC337 (NPN) or a BC327 (PNP). I get exactly the same outputs. Why would that be?

Jon

 

Transistors saturate best with up to 1/10th of their collector current as the base current.

Impossible, the transistors are completely different.

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Uncle $crooge

 

If you substitute NPN for PNP in the schematic, the LEDs get driven directly through the forward-biased B-C junction, with no transistor action. However, the logic will be reversed (high level from IC --> LED on).

Jon, are you using the same resistor values as in the schematic? You don't need 32mA of base current if you are reducing the collector current. As Audioguru pointed out, a good rule of thumb for driving a transistor into saturation is to set Ib=Ic/10.

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Ron

 

Thanks for the replies, that explains the NPN / PNP difference to me.

No I am not using the same resistance values as the diagram.

For the base, I am using 5.1k and for the collector 510 ohms. Both having 5 volts (app) creates the 10:1 ratio you mentioned. These values were decided upon using trial and error on a test circuit and worked on the test.

Jon