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In that case I would change the 22 ohm resistor to a 47 ohm 1w, to drop the dissipation across the 2N2222. Otherwise you will exceed the device rating of the transistor when using the 33 ohm load at 5 volts. But since the pulses are only 1ms the resistors won't even know the difference. So you can ...

1. The actual value needed for 1ms pulse is close to 4K. A 20K will work but a 5K might be easier to adjust. Or you can connect a 6.2K resistor across the 20K pot. That will set the control near the middle of it's range to get close to 4K, if using a linear pot. Steve G :D That is great. 1. I have ...

Nice! I like it. :grin: Steve G I have made the following changes to my circuit as the availability of components. Does it affect the circuit performance ? 1. 20K tuning resistor at 4047 2. HCF 4047 BE as monoshot 3. LM 317 T as regulator 4. HCF 40106 BE as debouncer 5. 620 ohm instead of 820 ohm a...

Can I use 7805 with the selected values of resistances to get the adjustable voltage around 5-7V. You can but I would use a LM317L with the values chosen to give a more finite adjustment Steve G I think 820R of LM317T should be replaced with 620R. From the datasheet of LM317T (I'm ignoring Iadj): V...

final circuit using 4047

LM317 should have a load of at least 10mA to ensure regulation. The 555 doesn't draw much current.

Just add a 1k from the LM317 output to ground. That will give you 5mA through the 1KΩ resistor, and another ≈5mA through the 220Ω, for a total of slightly more than 10mA.

Here is the final circuit...

sghioto wrote:As requested.

Steve G

But for your circuit the pressing time would affect the pulse width of 555 output ?

I think latch output should be subjected to some type of signal conditioning like Differentiating ! 1nF C2 is doing that function.

Since the cmos 555 canot source much current I modified the circuit again using another LM324 stage to buffer the resistor ladder. This will eliminate the load on the 555 to give 4.95 volt output with 5 volt supply Steve G 1. U3(now you renamed as U1 LM324) is there so that the output current of 55...

sghioto wrote:Use an adjustable regulator with the 555 so you can set the output at exactly 5 volts

Steve G

Can I use 7805 with the selected values of resistances to get the adjustable voltage around 5-7V.

How can this latch configuration will avoid bounces? It works because the inverters are much faster then the mechanical switch. As soon as the NC position is opened the output of IC3A will go low and make the input of IC3B low before the switch gets to the NO position and vice versa when released. ...

Eliminate the 50K pot. For the sake of example, lets say you had a 10 volt output at 555 pin 3. That would require a 1K series resistor in series with the 1K ladder resulting in 5V at the top. Use a precision resistor close in value but somewhat lower than that, maybe 950 ohm. Now add a 100 ohm pot...

Robert Reed wrote: One question though, do you have the quality instrumentation to confirm the accuracy you are after?

Keithley 2700 multimeter
TDK Lambda ZUP 60-7 420W supply

Vinod, Have you had a chance to test the circuit using the 74LS14 debouncer? You mention you were using a SPDT switch. I came across this circuit in my design library that uses two inverters to debounce the switch wired as in the attachment. I have not tried this circuit but thought it looked inter...

If you want more accurate steps then you have to clamp the 5v going to the divider. A way to do this is through another op amp set up to clamp the output at 5v, or if a zener diode is good enough you can use a resistor plus zener. Im wondering how did you expect to get 5v out of the 555 when it is ...