For years I've been trying to figure out a way to do this for years on a standard (not CMOS) and I think I've got it, The 1.2V V+ drop on the bipolar 555 positive rail.

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Since the - rail is usually reached on a stock 555 all I really needed was the positive rail pull up, which this should nicely. It allows for much greater drives like so:

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Definitely less elegant than I was hoping for, but I'll test it out <Sigh> Yet another revision.

 

Hi Wendy. The Discharge pin goes low when the Output goes low.
You need an inverting transistor to drive the new PNP transistor to pull the output up when the Discharge pin goes open-collector.

 

Actually not, I verified this experimentally here

 

Hi Wendy. The Discharge pin goes low when the Output goes low.

Have to agree with Audioguru again on that statement.

 

I have retested this again after AG's post, Pin 7 conducts when the output is high. It is the discharge pin after all.One of the things I love about electronics is the ability to test things out in reality.

555 Tester

 

I have retested this again after AG's post, Pin 7 conducts when the output is high

Then you have a bum 555.
Pin 7 conducts to ground when at the end of the pulse to discharge the timing capacitor for the next period (simulation below).
As you can see the pin 7 (DIS) goes high (blue trace) when the output (red trace) is high, showing that pin 7 is not conducting..

 

For years I've been trying to figure out a way to do this for years on a standard (not CMOS) and I think I've got it, The 1.2V V+ drop on the bipolar 555 positive rail.

View attachment 290389​

Since the - rail is usually reached on a stock 555 all I really needed was the positive rail pull up, which this should nicely. It allows for much greater drives like so:

View attachment 290392​

I'll let y'all know if it works.

Why? What's not to like about the ICM7555, TS555 or TLC555?
It's like using a 741 when there's such a wide choice of good op-amps.

@Audioguru again is correct - Discharge and output are in Phase. I connect them in parallel if I need extra output current sinking.

 

What's not to like about the CMOS 555s is the current output.

 

Why? What's not to like about the ICM7555, TS555 or TLC555?
It's like using a 741 when there's such a wide choice of good op-amps.

@Audioguru again is correct - Discharge and output are in Phase. I connect them in parallel if I need extra output current sinking.

For the vast majority of application it is great, but now and again more is needed. such as a class D amp.
I actually tried it for another project, at higher frequencies the CMOS output becomes more and more of a sine wave. Also a CMOS 555 has lousy driving characteristics at lower voltages like 5 volts power supply. A conventional 555 has full drive even at 4.5 volts.

 

Wouldn‘t it be struggling for speed in a class-D amp? If I need more current, then I add a MCP1402.
You can even implement oscillators with just a MCP1401 (the inverting version), although they aren’t as accurate as a 555.

 

On another thread, Wendy was using a 555 that was not an NE555 or LM555. Wendy, what is your 555 part number?
Here is the schematic of an original NE555 showing that the Discharge pin 7 and the Output pin3 both go low together:

 

Look at the picture of the tester in the other thread it is clearly shown.
You are right I screwed up. And I figured out what I did.

Back to the paint board.

 

According to LTspice, adding a pull-up transistor to the 555 output introduces (or worsens?) a shoot-through pulse of several nS when the 555 output goes low. That may or may not be significant.

 

The shoot-through current on the TTL output of an NE555 is almost 400mA!

 

Shoot through on a 555 is a well known issue, and handled (sort of) by decoupling caps. I've never seen a good solution.

 

As shown in post #11, the bipolar 555 Discharge pin is an open-collector output. As such, it cannot source the base current necessary to turn on an external NPN transistor. If that transistor's base is pulled high with a resistor, then the Discharge output can turn off the transistor by "shorting" its base to GND.

Something that will work is an external NPN and PNP driver pair, like the right-most image in post #1, with one current-limiting resistor and the bases connected together through two zener diodes. The zeners prevent output stage (outside of the 555 internal circuitry) cross-conduction. The zener voltage is dependent on the circuit operating voltage; no single zener value can cover the entire 555 operating voltage range. For crisp turn-off, each transistor's base needs a turn-off resistor to the respective rail.

ak

 

Shoot through on a 555 is a well known issue, and handled (sort of) by decoupling caps. I've never seen a good solution.

There is none. Hans was really very truly great, but not perfect.

ak

 

Shoot through on a 555 is a well known issue, and handled (sort of) by decoupling caps. I've never seen a good solution.

That's another reason why I always use the CMOS version. If I need more current a complementary emitter follower is still better than the dropout on a NE555, and if I really need rail-to-rail with lots of current I use a low-side driver IC, and they can deliver up to 10A from supplies up to 18V

 

Would you share info on this driver IC? I would like to keep the low voltage specs 3V to 4.5V.

 

Here is the circuit I described in #16, adapted from something I did on another forum. I reworked it for a bipolar 555 running on 12 V producing an output from 0 v to 10 V.

R1 and R2 assure a rapid and complete turn-off of the transistors. R3 limits the base current into both transistors (only one transistor at a time is on).

Because the centerpoint of the input waveform is not equal to the centerpoint of the output, waveform, the two base currents are not equal. To make them more equal, you can increase the Q2 base current by decreasing the D2 zener voltage, or decrease the Q1 base current by increasing the D1 zener voltage.

The transistors start conducting at 0.5 Vbe or less. To prevent cross-conduction, the sum of the two Vbe's plus the two diodes should be at least 2 V greater than Vcc for the circuit. 3 V is better. For the circuit as shown, the cross-conduction margin is 4 V.

ak