Charge and Discharge Timing

https://www.electronicshub.org/astable-multivibrator-using-555-timer/

https://www.electronics-tutorials.ws/rc/rc_2.html

Dear Guru,

The charge and discharge were at constant current, in my case, 5.96 and 5.92 mA respectively. I believe the curve should be linear; not exponential.

Kind regards,

Job.

That is a false belief, then, because it is not borne out by experience.

Dear Guru,

You are correct. I have two separate links for charging and discharging. With a constant current of 5.96 & 5.2 mA the voltage rise and fall were linear from 5 to 10 V and 10 to 5 V. With a 0.1 uF capacitor, the timing was 75 uS. I expected the time to fall to around 7.5 with a 0.01 uF capacitor; instead, the charge time was 30 uS. The discharge time was 8 uS instead of 7.5 which is reasonable. The ramps were perfectly linear on both charge and discharge. What is puzzling me is why it took about four times longer that what it took to discharge at the same rate?.

A small oscillation (with a few mV amplitude) for about 1 uS at the beginning of the on ramp.

Kind regards,

Job.

"with 1 uf & .01 uF, the timings were 760 and 75 uS"

Are you sure, or is that a typo: only a ten fold decrease in time for a 100 fold decrease in capacitance.

Depending on the way it's prototyped, down at 1 nF, stray capacitance may be having a significant effect.

Can you show us a circuit diagram; a picture of the physical circuit, and, any waveforms (have you got a scope?)?

Hi Randall,

You are correct; it was a typo. The values were 1.0 and 0.1 uF.

Let me see if I can tale a snapshot of the waveform and attach.

Kind regards,

Job.

Can you give us the circuit, draw it around Rixter's 555 Layout. Thank you.

Ho Oomborrie,

Please see below.

Kind regards,

Job

This is what LTspice says. Notice 10 fold reduction in caps. and time scales of 4ms, 400µs, 40µs and 10µs

1µf

100nf

10nf

1n

Very cool! In my day Spice was just getting started and I never had the opportunity to use it. I guess I'm only seeing the dust of the current generation. Still, I think what you've done is, as I said, very cool!

LTspice is free and easy to use. I am retired now, but, when I was at work the company paid for another spice package which included a several day course for myself and several others. It probably would have been more powerful, but, I always went back to LTspice for ease of use.

Hi Randall,

Does "LT" stand for Linear Technologies?

I am a retired mechanical engineer, who graduated 51 years ago. In my university days, we had the opportunity to learn about only diodes and bipolar transistors. Had known about integrated circuits but never seen them. Whatever I know is just referring the data books and never worked with a company which manufactured electronic gadgets. This is my background.

For a person with my background how difficult is to learn and use this LT spice? how can I install this software and use this? Are there ant training manual available?

Kind regards,

Job.

Yes: Linear Tech.

Just follow the link

and, follow the instructions. Lots of tutorial stuff available there.

Here are some clues to get you started (there will be mistakes in here so expect a little fiddling).

from left to right: add wire, ground symbol, resistor, capacitor, inductor, diode, "component"**, move, and drag (keeps connectivity).

**component brings up the library

The NE555 is in [Misc]

The LT317 is in [Power Products]

I always forget how to place a power rail: you need to use the Voltage component in the top level library. This component can also be used to create an AC signal source.

Use the escape key or choose another function to stop doing something.

When you have placed all the components and wired them together: hover over each component and RMC (Right Mouse Click) to enter details.

You only need the values (210, 1K etc) for resistors.

Same for capacitors

15 V for voltage supply.

Obviously you can enter other details if you want to get other information from the tool.

Finally click on simulate

Edit Simulation Cmd

leave it on transient and enter the end time

and click OK

Then just click on the running man (for RUN)

Hover over the signal your interested in (say Out) and a probe appears click and the signal is shown in the output screen.

Followed by pin2/6

Hover over D1 and click and you get the current in D1

Good luck.

I Appreciate it very much Randall;

I will try and see. Since you have already wished me good luck, am not asking for one.

Kind regards,

Job.

Hi Randal,

Thank you very much for the help. It appears that I will not be able to achieve a 50 5 duty cycle, triangular wave with linear ramps if the frequency is greater than 60 to 70 K Hz.

Kind regards,

Job.

Bob Pease always recommends this constant current circuit

I've used the inverse for the sink

Randall, In his 317 circuit he also use a diode like, D1 in the charge circuit, in the discharge circuit.

Definitely not needed for this circuit, and I don't think it makes much difference in the original.

Thank you.

Hi Randall,

Learned something new. What is the equation for output current?

Kind regards,

Job.

P.S.

No offence; it is not "Bob". Just to let you know, "Job" is a common Christian name in Kerala state (in India).

Job, not unknown over here, it occurs in the bible: my brother in law's middle name is Job; when he was young he was always known as Joby.

The drop across D4 matches the drop across the emitter in Q1, and, that fixes voltage at the "bottom" of R3 equal to the voltage at the junction of D3 and D4. So the current in R3 is

I_R3=V_D3/R_R3

and of course most of that current (>99% depending on the H_fe of the transistor) goes down the collector.

I was aiming at 6 mA, and assuming about 0.6V drop across D3: hence 100 Ohms

Thank you Randall;

Kind regards,

Job.

I think you are running into the lower limits of operation of the LM555. Rounding off your constant current sources to 6ma and the capacitor at 10^-3 μF your charge discharge rate is is 6V/μS. A 5V swing, 5 to 10 charging and 10 to 5 discharging is around 800 nS. I assume you are using the output (pin 3) to steer the current sources, if that is the case then the LM555 spec on fig. 8 shows a propagation delay of 300-800 nS @ 25°C, when switch between charge and discharge. 1V @ 6V/μS is around 180 nS which might account of the shift you see from 5V to 6V. It looks to me as if you are operating at the margins of the device. The fact you see 30 μS and 8 μS seems odd considering the capacitance and charge/discharge currents involved and the expected cycle time of 800 nS; this too makes me think the input characteristics of the comparators are coming into play as well as the propagation delay of the output.

Hi Rich,

Please see the circuit that I posted just a few minutes ago. The charge current is provided by the main supply not from pin 3. The output was left floating. The output swings between 1/3 and 2/3 VCC as expected.

We have a steady supply voltage and a constant current source which supplies a charging current of 5.96 mA.

The expected charge time for a 10 uF capacitor is approximately = 5 X .0000001 / 0.00596 seconds, which is 8389 uS, and the observed one was 8,100 uS. for 1 uF, the observed time was 760 us and for 0.1 uF it was 75. If the pattern is followed, why should it take 30 uS instead of around 8 micro seconds to charge the 0.01 uF capacitor unless the capacitor is a leaky one? A few capacitors of different makes and voltages ( More than 25 v) were tried but gave the same result indicating that leakage was not the factor.

kind regards,

Job.

There might be a delay in the ramp-up of the constant current source.....

You are correct; there is a slight delay; the discharge transistor unclamps only after one micro second after the output goes high which does not account for the total ramp up time of 30 micro seconds.

Job

Thankan, the output on pin 7 never goes high. It only switch inside the 555 to pull pin 7 to ground, that is after all its purpose. The voltage rise on pin 7 is an exsternal action and depends on your 317 between Vcc and pin 7. The rise in Voltage on pin 7 charge the capacitor through the diode between pin 7 and 2/6. The voltage at pin 7 is also the control voltage for the 317 and the 317 need to proses the rise in voltage to adjust the voltage it clamps pin 7 at, (regulate) for sure that will add time to the rise in voltage on pin 7 and therefore the rise in voltage on pin 6/2.

Hi Oomborrie,

I was talking about switching of pin 3, which is the totem-pole output; not the open collector pin # 7 of the IC.

Upon rising the voltage to 2/3 Vcc (10 V in my case), the discharge transistor turns ON discharging the capacitor through return branch. The function of this transistor is only to discharge the capacitor. Upon reaching the voltage at 1/3 Vcc (5V in my case) the discharge transistor turns OFF, unclamping the source (5.96 mA) allowing the current to flow to the capacitor until the voltage reaches 2/3 again. The voltage on pin 7 will always depended on the voltage on pin 6. During the charge cycle, this voltage will ride 0.7V (Which is the forward voltage of the diode) above the capacitor voltage.

There is a slight delay between the output (Pin 3) going high and un-clamping of the discharge transistor (Pin 7). The delay is only 560 nS. This delay, in my opinion should not make any appreciable difference

Another strange thing is that it did not make any difference in charge time when the charge current was increased to 10 mA.

Kind regards,

Job

What in effect happens is that the control voltage on the adjust leg of the 317 depends on the rise in voltage on the charging capacitor and therefore the action of the 317 gets throttled by the charging capacitor. There will always only be 0,6 volt more on the Control leg of the 317 than the voltage on the chargeing Capacitor.

Hi Omborrie,

If it is the case, Why is it not happening when the capacitance is 0.10 uF and above?

kind regards,

Job.

Look at Rich's answer again and remember that every that happen is directly dependent on what happens at pin 6 and 2. Can you post your complete test circuit with test instruments. A measuring instrument with internal resistance will always subtract time from discharging circuit and add time to a charging circuit if working with short time periods.

Hi Oomborrie,

You are correct in saying that the resistance of the measuring instrument will have a loading effect and extend the charge time and reduce the discharge time. All of us can agree. The resistance of the probe remaining the same, why the charge time remained with 5.96 mA and 10 mA charge current?

The instrument that I am using is an old one, a Tetronix TDX220 with a P6112 probe.

Kind regards,

Job

I've used the LM317 in the past for a current source for trouble shooting but the way you've used it is rather unique. I think the problem may be your running out of voltage as the capacitor is discharging to keep the LM317 operating as you expected it too. The specification for the input voltage range for the LM317L is 3.75-40V. You need to add the 1.25V needed for the resistor to keep it operating as a current device. You are now at 5V needed across the LM317L with the Resistor (3.75+1.25). Now add the diode drop of 0.75 to 1V and the saturation voltage drop across the open collector capacitor and you're at 6V. When the LM555 capacitor is charged to 10V and the pin 7 switches to low, the capacitor discharges down 4V leaving 6V across the LM317L. The discharge current has dropped to whatever leakage current can get through the device, it appears to continue to discharge at a slower rate so instead of 8 μS you are at 30 μS. This problem was always there but the larger capacitor values masked it out because the extra 22 μS wasn't significant enough to notice. I suspect if you had a scope across the capacitor you would see a funny looking waveform at the 6 to 5 volt level where the linearity starts looking more like an exponential waveform.

Hi R_i_c_h,

The linearity is perfect on both charge and discharge as shown below on the snap-shot. The upper trace is the voltage at discharge pin (#7) and the lower one is the voltage of the timing capacitor, which is one diode drop below the upper one on the charging ramp.

Kind regards

Job

Job

I agree the waveform shows linearity in charge and discharge and it is apparent the LM 317 discharge side I was concerned about stays in control so it doesn't stop regulating at 5V. So what to think about the anomaly? The theoretical rise time for 10, 1, .1 and .01 capacitors is 8445 μS for 10 μF to 84.45 μS for .1 μF. The measured values are low by 6%, 10%, and 11%. The .01 μF is high by 355%. With the linearity displayed on charging it would appear the charging current dropped to less than 2mA or the charging current is being split into two parts, the latter doesn't seem viable because I don't see a path to accomplish it. The discharge side doesn't have a similar problem so you have a real head scratcher. I would suggest you put a 10 ohm resistor between Vcc and the charging LM317 and use a scope to monitor the waveform across the resistor. As you switch from between charging and discharging the waveform across the resistor would give you an idea of the stability of the charging regulator current. I will be very interested to what you discover.

Thanks for your patience,

Take care,

Job

It just occurred to me why your charging time is extended out to 30 μS. I think the charging current is being split between pin 7 and the capacitor when the output transistor is turned off. It just occurred to me when the output is turned off there is a residual charge on the base of the NPN transistor. This allows the transistor to be in a conducting mode for a short time as the base charge is dissipated. During this period the collector remains in a conduction mode for a short period of time which allows the constant current from the charging side to split until the transistor has turned off. This is affecting the other times as well but, again, it doesn't register because it doesn't have much affect on the longer charge times. This affect wouldn't have as much affect on the discharge cycle as charge/discharge modes are switching from one to the other. The specs indicate a rise time of typically 0.1 μS but the rise and fall times are affected by the load current. The schematic shows a 3.3K resistor to shunt the base charge to common. You might be able to overcome this with a 1K resistor between Vcc and pin 7 and then another diode, cathode to pin 7 and anode to the junction pin 7 was previously connected to. Just a thought and something relatively easy to try. Your little issue has occupied way to much time in my mind but it's been fun...

Take care.

Hi R_i_c_h,

I did not follow. Can you please attach a circuit?

Kind regards,

Job.

It's a pretty pathetic drawing but it is what I have at the moment to work with. The thought is leaving the collector open to follow the charge on the capacitor leaves a leakage path for some of the current through the open collector until the charge is dissipated from the base-emitter junction. The time is affected by the LM555 base-emitter resistor, in this case 3.3K, to drain the base charge off. I haven't enough info to make any calculations but your cycle time is short enough that it might be affecting the charging rate. The pull-up resistor helps pull up to Vcc and the diode blocks the pull-up resistor from affecting the charge/discharge timing. Hope this shows improvement, something is taking current away from the charge cycle, so it's a thought. There also may be something in the LM555 that is allowing the output to maintain some conduction longer than otherwise expected. The schematic is involved and not necessarily exact. Happy hunting...

Take care,

It seems to me that you could also get rid of D2 (in my diagram)

It has improved the discharge time but has left the charge time as it was.

Thankan, my opinion is that the heart of your problem is around the way you employ the charging 317. The control on a 317 is dependent on the supply Voltage on the input leg and the regulating voltage on the control leg. Trying to run it effective from a near 0volt base is totally opposite to what it was designed for. It works on the voltage difference between input and control to determine and clamp the output voltage. Your circuit want it to control the rise in voltage from the output pin. I asked you in a previous answer for the complete circuit with measuring devises. You want us to read the speed on the speedo while you obstruct the view. Try to pull the control voltage up with a control resistor from Supply. In the present circuit the control of the 317 has to wait for the output to rise before it can react and clamp the output. How much time that add is just a guess.

Hi Anonymous/Thankan/Job,

What you observe has not been explained by anyone yet!!

I will use your diagram with numbers added to the diodes.

Suppose you have reached time just before discharge changes to charge.

1. Pin 7 is connected to pin1 [zero volts, common] by a "single pole" switch. Pin 7 is open circuit when not connected to pin1. U2 is discharging the timing capacitor through diode 2. U1 is delivering its set current through pin 7 to common. Diode 1 is reverse biased because pins 2 & 6 are at +5 volts.
2. When capacitor is below 5 volts, NE555 switches, pin 7 becomes open-circuit. Pin7 voltage will move positive at a rate set by the circuit capacity and sum of U1 & U2 currents. Note that U2 is still discharging the timing capacitor, so pins 2,6 will fall below 5 volts.
3. As voltage across U2 falls its current may fall but it will still discharge the timing capacitor more before the rising voltage at pin 7 forward biases diode 1 and pin 2,6 voltage rises as expected.
4. So part of charge Q required to raise pin 7 circuit capacity by 5 volts is removed from the timing capacitor. This will increase the voltage swing at the timing capacitor, but its effect will not be noticeable until the timing capacitor is small e.g. Q = CV ; 10 volts on 100 pF becomes 1 volt on 1000 pF = 1nF.

You could add 1000 pF from pin 7 to pin 1 to exaggerate & verify the effect.

This does not explain why the charge time is increased from expected 8 μs to 30μs.

It seems most likely that U1 is giving less than its set current initially with fast voltage steps.

The LM317 data sheet does not give any transient performance as a current source e.g. when the voltage from IN to ADJ terminals is reduced quickly. Again this would not be noticeable until frequency is high. Since U2 seems to discharge at near the expected rate with 1 nF, current behaviour is different for sudden voltage step-up.

You could change the NE555 to a conventional RC oscillator @ 10 kHz with e.g. a 6 volt supply, then apply its pin 3 output via 3 x 1.5V manganese dry cells in series to an LM317 current source; while observing its output current on oscilloscope. This would give a 4.5 to 9.5V step test [bypass with single 10μF tantalum capacitor across the 3 cells]

The LM317 circuit does have some internal capacitors, which affect its transient response......

Regards,

67model

"Note that U2 is still discharging the timing capacitor, so pins 2,6 will fall below 5 volts."

I don't think so: pin 7 is behaving as expected; I'm not sure why you think it would take a long time to rise when its transistor is switched off.

First this is the wave forms with 100nf; the blue trace is pins 6/2 and the red trace is pin 7.

Note that the current scale goes down to -5mA (both diodes are fully off when they're supposed to be).

Now if we switch to 1nf

The charging 317 is not turning on properly (quickly enough), which I think is what you said in the second half of your post.

Hi Randall,

1) What is the red trace seen below the red charge ramp?

2) I could not agree with guru 67model since the charge IC (U1) is always conducting (Not turned OFF) irrespective of the state of the discharge transistor. While the transistor is ON, the current will be directed to the common ground and as soon as the clamp is released, the current will be flowing to the capacitor.

Randall, is it possible to simulate the current through U1 and see how the trace look like?

Kind regards,

Job.

"1) What is the red trace seen below the red charge ramp?"

Not sure what you mean. The traces are

Green: out; Blue: 2/6; Red: 7; Pale Green: I D2; Pink I D1

In this shot the green trace is the voltage on pins 2 and 6, and, the blue trace is the current into the chargeing 317 (unfortunately it's U2 in the LTspice circuit because the 555 is U1). The Adj. pin is designed to not source or sink current so apart from the very brief spikey bits the input current is representative of the current supplied to the rest of the circuit

These are traces of the voltages either side of R1.

You can see that in the second charge cycle the voltage never gets close to the required 1.25V.

This is why

As the voltage on the Adj pin rises that rise is seen on the other side of C2 slowing down the correcting amplifier. I believe this is called the Miller effect.

They used this as reference.

Thanks for using the model to analyse, Randall!

It must actually be a good simulation of LM317, it does not always follow that a computer model actually reproduces all the (about 20) transistors in an LM317.

I started wrong on the reasons for the circuit behaviour, but did note that the problem could be the LM317 cannot be stepped from one in-out voltage to another instantly and still regulate.

The circuit below is from an article by Bob Pease, the designer of LM117/317.

The important point is that the two transistors marked E and 10E in Fig 4 [Q16 & Q17 in your post #51] and the two above them are the 1.25 volt reference. The E and 10E indicate they operate at 10 to 1 difference in current & the 1.25 volt comes from the emitter base voltage of the lower pair (and difference across R6] plus R5 volt drop. The transistor with base connected to two capacitors works as the error amplifier with the pair at top right carrying the power current to the current setting resistor from OUT to ADJ.

But with the capacitors putting current into the reference with fast voltage changes, its balance is upset & it delivers a reduced voltage.

The U2 stepping from off to 5V seems to work better than U1 stepping +10V to +5V.

As I noted, Thankan/Job could apply volt step cycles to U1 or U2 LM317 samples with the aid of the NE555 [as an R-C oscillator] and observe how they behave as current regulators separate from the timing/oscillator circuit.

Thanks: I looked for that circuit, and found another even more simplified circuit schematic which just helps understanding the basic operation.

In this pdf

https://www.pearl-hifi.com/06_Lit_Archive/02_PEARL_Arch/Vol_16/Sec_53/Pease_Lab_Notes/Bob_Pease_Lab_Notes_Part_9.pdf

Which also includes the circuit you found

Loads of good stuff in that article

AND there's more in the directory above

https://www.pearl-hifi.com/06_Lit_Archive/02_PEARL_Arch/Vol_16/Sec_53/Pease_Lab_Notes/

Dear Guru Randall,

I have made an inverter, a low power one, as per the circuit shown below. A ringing occurs at the drain terminals (junction of primary windings and the drains of the MOSFETs). Can you help me with this problem, as you have with some other problems in the past, in understanding why this happens and guide me to rectify this problem? The amplitude of the first wave is 50 volts above the Vcc with a duration of 100 n Seconds. The amplitude of the second one is only 10 V above the Vcc. The total duration of the ringing is about 400 n S. Cycle time approximately = 34 u S.

Kind regards,

job Thykkoottathil

I am sure Randall will put the same message in another way.

1/ Ringing is normal when you suddenly turn off current through an inductive load.

2/ A classic case is contact & coil ignition on petrol engines before electronics. A contact connected the coil [a transformer] to battery. While it was closed, current rose in the coil primary until limited by resistance or closed time too short to reach resistance limit at higher rev/min. A capacitor [condenser] across the contact stopped arcing causing short contact life. When the contact opened, the stored energy in the inductor caused the current to keep flowing - it had to flow into the capacitor with contact open. The stored energy in coil was enough to raise the voltage on the capacitor & primary to several hundred volts - transformed to 20000 volts or more on secondary connected to spark plug. The coil time constant is milliseconds and the "ringing" time tens of microseconds. Side note - When Mr Royce designed his first motor car, the first thing he did was design a good ignition coil - he was already a proven manufacturer of electric cranes.

3/ Your circuit has a transformer and its winding capacitance + capacitance of FET drain + "wiring" capacitance [mostly due to the thin insulating washer between FET drain & grounded heatsink if present] make the capacitor.

4/ Since you have a 650V FET, & your Vcc of 15V + 50V overshoot is much less, there is probably not a problem of FET failure.

5/ The FET can take turn-off at 650V peak for 10 microseconds at 9 amps.

6/ Being careful, one should know that the drain current of the FET is < 9 amp when it turns off.

7/ Maybe the DC resistance of the transfo half-primary is 15/9 = 1.66 ohms or more, which satisfies 6/.

However, you could blot out Long/Medium Wave AM & shortwave broadcast locally with a "spike" maker of this power - just try a radio near your inverter!!

The ringing may be bigger than necessary because the transformer is saturating which makes the stored energy high, this is inefficient (and the wasted energy heats the FET) - How many turns on half-primary, what DC resistance?

You need to add quench diodes across the two halves of the primary.

Same as you would across a relay coil:-

https://en.wikipedia.org/wiki/Flyback_diode

If you're still having problems start a new thread.

Dear Guru,

Diodes D3 & D4 were inserted as shown below with no success. The diodes were getting very hot indicating that a large amount of current was flowing through the diodes. For some reason, the voltage at the junction of the winding, diode and the drain was clamping down to 5 volts; why?

Kind regards;

Diodes do not work in "push- me - pull - me" - even with Dr Doolittle's help.

With switch closed, imitating FET, you get positive voltage on upper diode - near short circuit.

As I wrote in my first reply, the 50V +15V is nothing to worry about, so long as current limits of FET are observed.

As Randall suggests, another post is best.

67model

1N4007s should be fine.

The circuit looks Ok, are you sure that the diodes cathodes are connected to +Vcc?

OK: I'm sure you're right.

Looking at your diagram I'm trying to work out what happens to the current after it goes through the diode. If for example you put another rectifier in the output of the power supply: wouldn't the extra current from the second half of the primary just boost the current in the first half. (Just thinking aloud really; haven't got a clue.)

_{I've removed the off topic, so that if you reply I can give you a proper GA.}

In reply to post #70 by Randall.

1. The optimum way to put a centre-tapped winding on a core is to twist two wires together, then wind the bifilar pair into a coil on the core. Since the two wires are as near co-located as possible and the twisting serves to equalise capacitance turn-turn they are as equal as possible, share the same magnetic flux and have the same end-end e.m.f. Bifilar is necessary for balanced high frequency windings. Externally, wires from opposite ends of the coil are interconnected to make the centre tap, leaving two ends to connect to the FETs. To get a good 1:1:1 three winding transformer or a good 1:2 ratio (1:1+1), trifilar windings are used.
2. So if you apply 15 volts to the lower primary half, you will get 15 volts on the upper half. Unfortunately, this forward biases the upper diode and current flows from the winding, through the diode and back into the winding at the centre tap.
3. The circulating current in the upper winding must come from the lower winding by auto-transformer action. So the lower winding must carry an equal current to the upper winding [plus a magnetising current for the core, which provides a back-emf to oppose the DC power voltage less voltage drop across winding resistances].
4. If the core had a third winding, with a load, there would be the expected current in the lower primary to get this , so the current in the lower primary can be described as "boosted" by the unwanted current in the upper diode.

It is not necessary to add a rectifier in the power supply side for the above to happen; since current is always flowing the same way from power supply, a diode would just add a [non-linear] volt drop.

Actually, there is an unmentioned ghost in the machine. To get e.g. +15V supply + 17V = +32V at the high end of the winding you must get +15 -17 = -2V volts at the low end; but this would forward bias the inverse source-drain diode in the FET and act as a clamp.

Spikes of 50V seen are due to FET breaking the current in the leakage inductance of the primary which is due to the primary flux which does not link to another winding. The leakage inductance shows up if you short circuit the secondary and measure the primary - rather than the short circuit for an ideal transformer, you get a small inductance plus the winding resistance. In a real circuit you will get the effect of the DC supply wiring inductance etc..

I goofed in the calculation in post #69 - the mean value of an 0.2 amp peak triangular sawtooth current would be 0.1 amp.

Thanks Job,

Thanks is appreciated - it does not happen much but when it happens it tells folk their efforts have helped to answer a question.

You could compare the current regulator to a mechanical bottling plant, bottles have to be loaded, filled & a cap fitted. There is a limit to how fast you can start or stop the line and the bottles per minute. Too fast & you have part-filled bottles, contents clogging the machine and less good bottles per minute!

Apologies for final posts about details difficult to understand. Got diverted into the detail.

The LM117/137 is an electronic integrated circuit which has been made with minor change for 50 years - like a water level control ball valve there are probably one or more in millions of houses.

67model

Pranam Guru!

"You could compare the current regulator to a mechanical bottling plant, bottles have to be loaded, filled & a cap fitted. There is a limit to how fast you can start or stop the line and the bottles per minute. Too fast & you have part-filled bottles, contents clogging the machine and less good bottles per minute!"

For someone of my understanding the analogy is good.

Kind regards,

Job Thykkoottathil.

Hi Guru,

1. "When capacitor is below 5 volts, NE555 switches, pin 7 becomes open-circuit. Pin7 voltage will move positive at a rate set by the circuit capacity and sum of U1 & U2 currents. Note that U2 is still discharging the timing capacitor, so pins 2,6 will fall below 5 volts."

I am just wondering how the capacitor can continue to discharge after the discharge transistor is turned off.

I think that the charging and discharging take place independent of each other. In the charge cycle, the current is blocked by D2 and during discharge cycle the current supplied by U1 is passed to the common ground by the discharge transistor preventing the capacitor to receive any current.

U2 will remain in conduction independent of the state of the discharge transistor. Therefore, the capacitor must get charging current as soon as the "clamp" is removed. The voltage at pin 7 should jump from 0.7 (Collector emitter voltage of discharge transistor) to 1/3 VCC and start rising. There should not be any question of any delay in supplying the charge current by U1

Clearly, we can see that at lower frequencies, the charging and discharging takes place in a linear fashion with equal timing, but why the charge time increased on charge, while still at a linear fashion, at higher frequencies? Why it did not affect the discharge cycle?

I think that the voltage at pin 2/6 should fall below 5 volts (1/3 VCC) only if the control voltage at pin # 5 is less than 2/3 VCC.

Kind regards,

Job.

1st, the 0,7V, collector emitter voltage you talk about is a miss conception, the only place with forward bias voltage is the Base Emitter. If the base emitter current is high enough which it is on a 555 pin 7 goes to hard ground 0V. U2 circuit can only reverse bias the discharge diode if the voltage on pin 7 (the cathode side of the diode) rise to higher than the Anode voltage and that depends on the switching on of the U1 317 of which the control voltage on the control pin tells it to be 0V. The way you use the 317 in the charging circuit is the problem with your circuit at higher frequencies. Put the U1 and U2 circuits in parallel but opposite ways around. Pull pin 7 up with a 100 Ohm resistor. Now you will have exactly the same current circuit for charging and Discharging between Pin 7 and the Capacitor. You will pull pin 7 down to 0V with a short circuit when the 555 pulls it low and you will pull pin 7 up to Supply with the 100 Ohm resistor when the 555 allow pin 7 to go high. Use a 10 Ohm if the 555 internal circuit allows it. Make the 2 circuits as equal as possible.

See that 1k pull up resistor. But for higher frequencies the 1 k will drop the voltage to much and limit the current you need for your experiment. Do not add the diode on pin 7.

Drive, supply U1 from pin 7 in my suggestion. u1 and U2 is connected to pin 7 and pin 6/2. 0V on pin 7 comes from the switching of 555 to discharge time and pull up of pin 7 comes through the 100E or 10E pull up resistor between Supply voltage and pin 7.

Thankan,

You are correct in how you describe the operation of the circuit if all the parts are theoretically ideal & they are, for practical purposes, for long timings. If you add a capacitor Cx (U2 self capacitance) from pin7 to GND or V+ to circuit you will see that U2 can can continue to discharge the timing capacitor while drawing current through Cx & charging it +ve.

Real circuits have capacitances which are not on the block diagram and these affect operation for microsecond periods when timing capacitor voltage charges/discharges at about 5 volts in 8 microseconds [about 0.7 volts/μs]. The detectors which switch at the 1/3 & 2/3 supply volts cannot switch instantly & the bistable in the 555 which controls pin 7 will not switch instantly nor will the transistor whose collector pulls pin 7 to 0 volts. It is usual for turn-off times to be much longer than turn-on times.

If the pin 2/6 detectors actually takes 0.5 microsecond to switch, the voltage at pin 2/6 will actually overshoot by 0.5 μs x 0.7 volt/μs = ~0.35 volts, at both 1/3 & 2/3 supply - you report ideal 5 volt increases to 6 volts at expected half periods of ~ 8μs; which is more.

Another commentator has simulated your circuit with 0.001 μF timing capacitor and found a charge time greatly exceeding expected.

Although U1 seems to have an easy task, it has to conduct at the same current before and after pin 7 switching, it actually sees a 5 volt step in the voltage across it. Some LM317 data sheets show response graphs for 1 volt step of 10 volt input on output voltage and it fluctuates - for a load resistance R this means current through LM317 must fluctuate in same proportion, as a current source it has same transients.

It appears a 5 volt step down can initiate a step drop to about 25% of the set current for 30 μs. As I pointed out, you can test this by using 4.5V dry cells and the NE555 to apply 5V step cycles to your actual LM317 "U1" and observe the response on your oscilloscope.

67model