Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

lm567 positive edge pulse

Status
Not open for further replies.

neazoi

Advanced Member level 6
Joined
Jan 5, 2008
Messages
4,119
Helped
13
Reputation
26
Reaction score
15
Trophy points
1,318
Location
Greece
Activity points
36,918
Hi I have tried to convert the lm567 to output a positive pulse on signal detect but I had no success yet. I have also tried an NPN ans drive it's base directly from pin 8 but no deep pulse is output. Any ideas?
 

Attachments

  • lm567.PNG
    lm567.PNG
    3.3 KB · Views: 80
  • lm567b.PNG
    lm567b.PNG
    4.3 KB · Views: 74

You don’t have enough base drive to turn that transistor fully on, 33k is way too large. How did you arrive at those values?

And you might damage the output since you’re applying 13.5V through the 22k; it’s rated for 9V max.

Also, that divider on the power supply input to the lm567 is awful. Use a zener,or a regulator.
 

You don’t have enough base drive to turn that transistor fully on, 33k is way too large. How did you arrive at those values?

I just saw them on a circuit on the internet. But I have tried even the transistor directly connected to pin 8 and with a 2.2k to the vcc (13.5v) and again It could not be switched.
Suggested values or different circuit?
 

Again, you don’t want to connect the output to any voltage above 9V. If you connected the base of the transistor directly to the LM657 output, you probably damaged the transistor, lm657, or both.

And, also, again, use a regulator or zener to drop the voltage. Maybe use a 1k base resistor and a 10k pull-up.
 

Hi,

I just saw them on a circuit on the internet.
Don't use random schematics from the internet.
Read and keep on datasheet.

Datasheet says:
* maximum recommended VCC = 8.5V
* V8 max voltage = 9V. And this in the "absolute maximum ratings". This means any millivolt for any short microsecond beyond this value may immediately destroy your IC. This does not mean this really happens immediately, but you should not be surprised if it does.
* datasheet mentiones:
13.1 Layout Guidelines
The VCC pin of the LM567 should be decoupled to ground plane as the device can work with high switching
speeds. The decoupling capacitor should be placed as close as possible to the device. Traces length for the
timing and external filter components should be kept at minimum in order to avoid any possible interference from
other close traces.
I see no decoupling capacitor at all. Don't know about your GND plane and trace lengths.
Btw: they mention a proper GND plane. This is almost equal to: the circuit can not be tested on a breadboard.

Usually they don't mention twice the proper use of the VCC decoupling capacitor ... if it could be omitted.

There is the chance that the whole circuit is weirdly oscillating. Either continously or in certain situations.

Klaus
 

My datasheet says V8 can range up to 15V. Anyway, missing supply bypass is a problem.
 

My datasheet says V8 can range up to 15V. Anyway, missing supply bypass is a problem.

Ah, you’re right, I was looking at supply voltage. And output can sink 100ma, so probably didn’t damage that.
 

Hi,

I might be wrong but:

If you have to use a resistive divider to divide down 13.5V to more or less the recommended 8.5Vmax, make R1 (from V+) = 1k and R2 (to gnd) = 1.6k = 8.3V and at same time meet required max device supply Iq with about 13mA (datasheet says it needs 6mA to 15mA, not 34mA).

Tie PNP pull-up resistor to Vdivider voltage, not Vsupply voltage. Make Rpull-up 10k and place it from the PNP base to Vdivider voltage, put 1k from PNP base to pin 8 of the LM567.

Might work that way.
 

Hi,

My datasheet says V8 can range up to 15V
Sorry I've confused it with VCC absolute maximum value. 15V is correct.

Klaus
 

Hi,

Also, regarding resistive divider supply voltage, it might be of help to place a 10nF to 100nF capacitor across each resistor.

What is the PNP feeding - is it literally an LED, a high impedance input pin, a current-driven or current-heavy load or what?

For 13.5Vsupply down to 8.5V to 9V, a 3V dropout 7808 or 7809 would work. That or experimenting with a homemade regulator or a "poor-boy LDO" (as someone else called them). The resistive divider approach on a supply pin with fluctuating input current range is a grim method to be avoided and prone to causing problems with the LM567 functioning correctly, plus must repeat point of bypassing capacitor(s) on suppy pin, as already mentioned.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top