Rail to Rail 555

Thread Starter

Wendy

Joined Mar 24, 2008
23,396
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.

R-R 555 Osc.png

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:

xtra drive #2.png

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

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Audioguru again

Joined Oct 21, 2019
6,606
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.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,396
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
 
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crutschow

Joined Mar 14, 2008
34,012
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..

1679458108894.png
 

Ian0

Joined Aug 7, 2020
9,458
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.


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:


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.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,396
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.
 
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Ian0

Joined Aug 7, 2020
9,458
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.
 

Audioguru again

Joined Oct 21, 2019
6,606
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:
 

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Thread Starter

Wendy

Joined Mar 24, 2008
23,396
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.
 
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Alec_t

Joined Sep 17, 2013
14,206
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.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,396
Shoot through on a 555 is a well known issue, and handled (sort of) by decoupling caps. I've never seen a good solution.
 

AnalogKid

Joined Aug 1, 2013
10,943
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
 
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Ian0

Joined Aug 7, 2020
9,458
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
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,396
Would you share info on this driver IC? I would like to keep the low voltage specs 3V to 4.5V.
 

AnalogKid

Joined Aug 1, 2013
10,943
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


555-Driver-1-c.gif
 
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