Hi guys, i saw a couple of circuits that used the pin 7 of the 555 timer as the output and not the pin 3, and i was wondering how it works. My first contact was in the "EEVBLOG" of Dave that he made it work and explained how, but sadly i did not understood it completely so i came here to ask for an explanation if someone here knows.

The video is:

 

Pin 7 (Discharge) is open collector. It can sink current, and its transistor is turned on by the same flip-flop that controls the output at pin #3:

(Image from Paisley, citation below)


For explanation and examples, Rob Paisley's site is excellent: http://home.cogeco.ca/~rpaisley4/LM555.html Click on #14, using the 555 as a voltage comparator as one example. Note that a pull-up resistor is needed. Some data sheets say the current capacity of Pin7 is internally limited to 50 mA. Paisley says 200 mA, as do some others.

Edit:
Here is the output structure for Pin3


Note that Pin3 can sink (Q26) or source (Q28) current and its rating is 200 mA. Here is an image from the datasheet for the LM555:


Note the discharge pin is not characterized above about 30 mA.




John

 

The circuit in #14 position is a little weird to me, but the idea of using the pin 7 is the same as using pin 3 ?

 

Pin 7 has the same logic state as pin 3. The main difference is that pin 3 has a "strong" output buffer that can source and sink current while pin 7 can only sink current to ground and it would then require a pull-up resistor or something similar.

 

but sadly i did not understood it completely

it is equivalent to an OC output pin: it works as long as you have other means to discharge the rc network.

 

Pin 7 is an open collector output, so its collector voltage source Is not committed to VCC. You could connect it to a different voltage source and use as an output if desired.

 

So. I use pull up resistor to pull the Pin 7 to external voltage source (Vext). When Pin 7 is High, I get Vext-Vcc=0 volts. When Pin 7 is Low, I get Vext-0=Vext. Is that how it works?

 

When pin 7 is off, the pullup resistor makes it high...then when pin 7 is on the output is low.

 

The output on pin 3 is tied to its supply voltage. IF the 555 is powered by a 5 volt source then pin 3 can be either ground (zero volts) when its output is low, or 5 volts when its output is high.

Pin 7 does not have an output. It is either ground or infinite resistance (like an open wire - hence - open collector). Pin 3 can only give you zero volts or five volts because of its supply. But pin 7 gives you a solid ground or no output at all. When its output is "OPEN" - depending on how you wire it externally, it can be any voltage you like. But suppose you connect pin 7 to a resistor and the base of a PNP transistor. The emitter of the PNP can be connected to any voltage supply you like. You are not tied to the supply rail of the 555. You could control 3 volts, 12 volts, 30 volts - whatever the transistor is rated for. When pin 7 goes low, current is sinked from the base of the PNP transistor, thus turning it on. The emitter voltage is then passed through to the collector, whatever that voltage is. There are some losses involved, but for the sake of argument and understanding, if you have 30 volts on the emitter and ground the base, you'll have 30 volts at the collector.

Remember, there ARE losses when you go through a PN junction, typically 0.6 volts. But even that depends on the device you're using. That's where the spec sheet comes in handy.

Did that clear it up for you? I'm hoping it did.

 

I have a vague recollection from early 1970's of seeing a spec. on pin 7 as 200 mA max. jpanhalt's post is the
only other time that I have seen a spec. on pin 7. A few years ago I configured a 555 as one shot, Vcc 12 V,
& dumped a 33,000 uF into pin 7to see what would happen. It just kept on dumping, so I assumed that pin 7
was a constant current dump of un stated value.

 

on pin 7 as 200 mA max.

it varies but generally far less than 200ma on pin7. bjt 555s generally can do 200ma on pin3.

 

Hi guys, i saw a couple of circuits that used the pin 7 of the 555 timer as the output and not the pin 3, and i was wondering how it works. My first contact was in the "EEVBLOG" of Dave that he made it work and explained how, but sadly i did not understood it completely so i came here to ask for an explanation if someone here knows.

The video is:

The signal at pin 7 is in phase with the output at pin 3. The normal circuit shows a current limiting resistor from pin 7 to Vcc )1k min) that also feeds the CR timing network. Its pretty much as simple as swapping the two around - pin 3 directly drives the top of the CR timing network, pin 7 is available as an open collector output that can have a pull up resistor as low as 1k.

The 1k is in series with the CR network during charging - but not during discharge. If you want equal mark space ratio; the pin 7 pull up resistor should be as low as you can get away with (usually 1k). Using pin 3 to drive the LC network is an option to get improved equal ratio - you can also include a pair of steering diodes and a splitter pot to get variable mark space ratio.