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A strange 555 Timer problem

Hi, Ive built a LF ramp generator using a CMOS 555, where the timing cap (and pins 2,6,7) is fed by a constant current (single transistor shunt type) source. All goes well using a 100uF RBLL electro and a range of currents from the source from 1uA to 50uA. The waveform is picked off the cap, and is buffered by a unity gain follower using an LF353 Op Amp. Amplitude of the waveform is as expected and ramps between 4.2V and 8.2V (using 12V supply). However, when I change the cap to a 4.7uF electro or tantulum, the waveform changes, and the waveform now only ramps between 6V and 8.2V. The trailing reset edge does however drop instantaneously to 4.2V, but immediately goes to 6V where the ramp waveform then begins. When I was using an ordinary 4.7uF electro, it seemed the negative edge spike was like a differentiated one (sharp drop, then slightly curved recovery). Changing to the tantalum stopped this, and now is only a momentary negative going spike. Does anyone know why the ramp doesnt reset and start at 4.2V as per the

100uF version? regards, JEFF
Reply to
Jeff
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Perhaps an effect of the parasitic series resistance of the capacitor?

Piotr

Reply to
Piotr Piatek

Show us your current source... it may not be constant.

The discharge pin will not stop discharging the cap until the voltage at trig gets down to 1/3 Vsupply. There is often some rebound in caps, but it gets worse (not better) as they get bigger.

Is the 12V supply regulated?

The electrolytic cap should not be recharging with a curved path unless the charging rate is proportional to the voltage across it (not a constant current).

One idea is that the common mode voltage range of the LF353 is +-11V minimum, given a +-15V supply voltage. That means it can't get closer than 4V to the negative rail (GND in your case), worst case. If you go below that, the JFETs on the input could start sourcing current. If this is the case, measuring it without the opamp in the circuit should change the results.

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Regards,
  Bob Monsen

Cantorism (set theory) is a disease from which mathematics will have to
recover.
- Henri Poincare
Reply to
Bob Monsen

Thanks. The constant current source design used is here.

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Uses set resistor of 470K for approx 1uA, and 12K for approx 47uA. The main ramp portion of the waveform is quite linear. There was only the slight charge curve from 4.2V to the 6V start point of the ramp. Using a tantalum removed this curve, and is now only a brief negative spike down to

4.2V (from 8V2) then immediately up to the 6V start of the linear ramp. 12V supply for 555 and LF353 is regulated with 7812. LF353 is standard strapped unity buffer. I th>

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Reply to
Jeff

. ----+---- +V . | . R1 . | . ,----+ . | | . | R2 . V\| | . |--+---o out . /| . | . gnd

That simple "current source" is not especially constant. Following the Ebers-Moll equation, for each factor of two change in R1 current the output current will change by about 18mV/500mV or 3.6%, which can amount to an 11% change as Vout approaches 90% of +V. A further 2% effect comes from the Ic/beta factor that's at work in this circuit, if R2 = 15k for Iout = 33uA, but this error can grow to a sobering 500%, for R2 = 470k for 1uA with R1 = 10k. Awwrrkk!!

Try this improved circuit instead, described in our dozen pages of current-source and current mirror discussions, e.g. see AoE page 76.

. ,-----+---- +V . | | . | Q1 R2 . V\| | . |---+ . /| | . | |/V . +---| . | |\ Q2 . R1 | . | '----o Iout . gnd

R1 sets a steady current through Q1 and thereby a steady Q1 Vbe bias (independent of Vout), which establishes a steady emitter current for Q2, thanks to a Q1 Q2 feedback loop. You'll find there's very little change in Iout vs Vout, which means the current source has a very high output resistance, and can therefore generate highly-linear ramps.

Use 1mA of current in Q1, and 1uA, 47uA, in Q2, as you choose. Enjoy and go in peace. If you like the circuit, you can vote against Bush.

--
 Thanks,
    - Win
Reply to
Winfield Hill

Almost like a resistor in series with the 4u7. If the Idss of the Discharge Fet is (say) 100mA at 4V, then a value of 20 ohms would cause the effect you see.

That LF353 unity gain follower might cause trouble. An Opamp UGF can do weird things when taken outside its CMV range. An npn emitter follower would work ok.

12v-------------+--------------+----->

| | | [R] \| | npn|------+-----+ +Vout e\| | | pnp|---+---> Vc to 555 | /| | | | | [R] | ===C | | | 0v--------------+--------+-----+----->

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Tony Williams.
Reply to
Tony Williams

You are ignoring the base current required to support the bulk of the emitter current through the 10K to 12V. The voltage across the 10K varies approximately from 4V to 8V for 400uA to 800ua emitter current. The beta of the 2N3906 is not that good, so you will have at least 4uA to 8uA base current in addition to VBE/RT charging timing CT. By going with something like shown below, the 2N3906 base current requirement is reduced to supplying 2N3904 base current which puts you somewhere around IB in the 8uA/50 - | +----------+ | . | ^ CBP| | VCC/RST | | . +-----+-[RT]--+ | === | | | . | | | | 1U| | 555 | LF353 | . | >|IB-> | | | | | |\ | . | 2N3906|-----+--+---------|THR/ DIS|-[1K]-+-[1M]---|+\ . | /| | | | |TRIG | | | >--+-OUT . | | | | | | | | +-|-/ | . \| | CT |+ | | | | | | |/ | [1M] . |---+ === | | | GND | | | | | .

Reply to
Fred Bloggs

Best one yet-

Reply to
Fred Bloggs

Tony's trick of using the PNP as both an EF and a current source is cute, but we've established the CS performance is poor. Here's a true current source, for better ramps.

. ,-----+---- +V . | | . | Q1 R2 . V\| | . |---+ . /| | . | |/V +V . +---| | . | |\ Q2 | Q3 . R1 | |/ . | '---+-----| ------- Vc to 555 . | _|_ |\V . gnd C --- | . | +----- linear ramp out . gnd | . R3 . | . gnd

An additional pnp emitter-follower stage may be added if the single Vbe downward dc level shift is unacceptable.

--
 Thanks,
    - Win
Reply to
Winfield Hill

Tnx Fred, appreciated.

I know the CC-source is crappy, but the OP has said that he is satisfied with the results at C= 100uF. So the problem is really one of imagining what he not said about his circuit.

A low value R in series with the 4u7 is one possibility.

Another possibility is a charge-dump, onto the timing C, after the end of the discharge.... say for example a 1uF connected between the 555 output and the top of the C. When the discharge reaches 1/3Vcc, the Vout goes to 12V.

That would dump Q= 1uF*12V into the timing C. A 100uF would +step by only 100mV, but a 4u7 would +step by 2.5V.

--
Tony Williams.
Reply to
Tony Williams

On the rare occasion when a CC-source is needed I've found the 2/3Vcc available at the CONTROL input pin to be useful.... as the base voltage for a single pnp CC. Your drawing modified below.

+V | R2 | +---------------
Reply to
Tony Williams

Hi Guys, Thank you all very much for your ideas and solutions on this. I spent a lot of time on the problem last night. The type of tant or electro made a big difference to the spike size. Got an absolutely perfect waveform by using a Polypropylene cap. Is the variation something to do with dielectric absorption? The PP I used is huge (250VAC type), so going to try a smaller Polyester when I get some. Failing that, will glue the PP somewhere off the PCB :-)) Have printed out your replies, and yes, I will improve the basic CC circuit from what it is now, according to your advice, as under close scrutiny, the ramp has a slight upwards bow in it. regards, JEFF New Zealand.

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waveform

Reply to
Jeff

I don't think Jeb's announced his candidacy yet, has he?

Reply to
Walter Harley

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