How to Build a Linear Voltage Decay Circuit (9V-12V DC, 3-5s Ramp Down)

Okay.

I need a simple but accurate circuit that will do the following. (I'm a novice)

When power is but onto this circuit 9V - 12V Dc the output starts to fall (Linear) until it reaches 0V.

I have found a ramp generator circuit (using a 555 timer) that does the opposite (starts at 0V rising to 12v )

This is to remain at 0V (not essential) until the power is taken away and then reapplied and the process is then repeated.

The rate of voltage decay needs to be at least 3- 5 secs

Thanks in advance

Maybe this helps...
(Attachment is incorrect. Refer to the immediate reply to this post).

Corrected version.

VM1 measures the capacitor voltage and VM2 measures the o/p voltage at the transistor's collector. Assuming the cap is discharged when the switch is thrown ON, it will have zero volts across it and therefore VM2 measures the full battery voltage at the collector. The transistor, LED and emitter resistor implement a current source. As the capacitor charges via the current through the collector, its voltage rises. This subtracts from the collector voltage and when the capacitor voltage reaches its maximum (battery voltage), the collector voltage will be zero (assuming R3 to be open). The capacitor voltage will rise linearly due to the current source. Therefore the collector voltage will also linearly ramp down.

Note: This may not necessarily be a correct circuit but only an idea: Charging a capacitor via a current source for a linear ramp. Subtract this ramp from the source voltage to get the linear ramp down voltage.

The emitter resistor R1 sets the current given by I = (Vled - Vbe)/R1. The capacitor determines the time-constant. Values shown might not give the required time constant - they are only to explain the basic operation. R3 might not be needed - it sets the lower limit to the output voltage - the lower this is the higher will be the lowest ramp-down voltage (i.e., it won't ramp down to 0V) as long as I * R3 < 9V.

Thanks for the replies so far.

To clear a few points up.

The circuit is "On" giving say, 12 V dc output, when the circuit is "Switched off" the voltage on the output terminals decays down to 0 V at a constant rate. ( not a sharp dip then a slow decline)

I wish the whole process to take say 3 seconds (10 V - 0 V)

Just to add,

If the supply voltage is re-applied (12 V Dc) the output must instantly rise back to 12 v DC.

I'm not that fussy if the voltage doesn't reach 0 V but must be as low as 1 Volt.

One last point i cannot open the circuit diagram above (will check it's not this computers setup)

I forgot to ask, does this ramp voltage need to drive anything? I'm assuming not much if the ramp voltage from the cap on a 555 circuit is sufficient.

The CCS implemented by the transistor, LED & R1 will ensure that the ramp is linear.

This is the formula, I believe for the "slope" of the ramp (http://electronics.stackexchange.com/questions/23020/capacitor-discharge-through-constant-current-source)

Voltage across the capacitor will be given by the formula,

V(t) = (I*t)/C, assuming that the capacitor is in discharged state at t=0.

And the o/p voltage is,

Vo = Vcc - V(t) = Vcc - (I*t)/C

Now, for the o/p voltage to change from Vcc to zero in 3 seconds would mean that the capacitor has to charge from 0 to Vcc in 3 seconds. Theferfore,

V(t=3) = Vcc = (I*3)/C

So, C = (3*I/Vcc)

if, I = 1mA, and Vcc = 10V

C = (3 * 1e-3)/10 = 300uF

Experts, please correct me if wrong.

The circuit i found using the 555 timer went from 0 V to 12 Volts. the wrong direction

I want the opposite direction 12 V to 0 V.

The output from my "new circuit" will feed into a 555 timer on pin 5 and alter it's output.

I have found a 566 Ic that might do (again not to sure)

You could also use variable pwm using a 555 timer to adjust your slope of the ramp such as used in a light dimmer circuit or use atmel or pic micro with analoguewrite( value ) function 0 to 255 then you can maintain your low votage or any preset voltage to which you want this to fall to using obviously a transitor and low pass filter network or fet and low pass filter plus you can vary your time delay on the slope with more acuracy If you are using a micro then you could also make this rise up or down and vary your time constant single shot depends what your using this for and how much accuracy you want and what scale of linearity you want but for 3 to 5 seconds ideally you would use a programmable pwm output for specified time pwm out (0 to 255 ) within a three to five second period i.e 0% to 100% where 0% reflects 12 volts and 100% relates to zero volts

Also the basis for an electronic voltage controlled volume control

Actually for the purpose of controlling something via PWM using a comparator, a downward ramp can be used in place of a upward ramp simply by interchanging the positions of the ramp and reference voltage inputs to the comparator.

-Here's a candidate circuit in both CircuitLAB and a PNG bitmap:- *[This circuit is incorrect in terms of the specifications supplied by Andrew -- my apologies!]*

The ramp is exposed at "V_Out". V2 simulates the application and removal of power. The ramp is generated by discharging C1 with a constant current sink (R1,D2,D1,Q2,R2). D1 and D2 form a constant voltage input to Q2. One of the two diodes overcomes the Vbe forward voltage and the other applies approx .6 volts across R2. This constant voltage across R2 results in a constant current at the Q2 collector (Q2 amplifies the resistance to an extremely high value). It is this constant current that discharges C1 at a constant rate of change dv/dt.

C3 supplies voltage to the voltage reference [R1,D2,D1] long enough for the post power-loss ramp to finish.

For best precision use a ceramic cap for C1 with applicable temperature rating [see: http://www.niccomp.com/Products/TC_Ceramics.pdf]. If not too critical, then use a tantalum. Adjusting C1 and/or R2 will change the ramp timing.

BUT, this circuit is not terribly temperature stable, so if that's an issue then you'll need a different solution -- such as what Mark is talking about.

Here's the URL to the CircuitLAB simulation:

* https://www.circuitlab.com/circuit/d2htx6/3second_pwr_off_ramp01/

Hi Rohit,

-Looks to me like the R2, D1 leg, in your circuit, will discharge C1 with a current nearly 10x as large as the Q1 collector current, thus swamping any linearity. I predict that the discharge curve will be more nearly logarithmic. This is the reason that my circuit has two capacitors, one to produce the linear discharge, and one to supply the voltage reference.- *[I made this evaluation from the incorrect assumption that Andrew wanted a linear downward ramp to occur _after_ power was removed from the circuit]*

Well done Rohit knew you would like that idea

If you want to see how this is done with schematics This is an early project I developed some years ago when I first started trying to learn how to do all of this with PWM, Visual basic , Pic Micro , proton basic , C , java , electronics, webpages etc to control light dimming circuits

The original plan was to incorporate this into someone’s car at the time A request which I made for someone but never actually came back to pick it up or have this installed

It featured led display in steps of 10, Memory and obviously pwm control for a 12 volt lamp It could have gone into many units bedside lamps , caravans , cars even act as emergency lighting or for motor speed control on these minicraft drill sets that was when I had ideas about trying to set up a training dept or create work for people who were unemployed at the time also who wanted to learn new skills the less unfortunate of people Lots of them there are in the UK which is quite sad when you think about what this country once was


As usual there wasnt any real interest Dont ask me why Typical , Normal this Must have thought like Collin

So you might as well have that one took me a few days to design and build

http://www.harrington.force9.co.uk/Projects/LightsController/LightsController.htm

Be careful because this could be advertising Mustn’t' upset Communist 555 timer Collin you know lol Such a joke this !!

Everything you need is on there including PCB layout hex code, schematics, drill plots the lot

Its No good to me I dont own a car anymore cant afford it due to the Communists

All Thatchers fault probabaly slept with that Russion President , gobalot or something cant remember his name , plus those Americans , whjats his name "The Peanut farmer Ronald Raving Reagan ", created all those macdonalds places that make everyone Fat and those beeedin Africans all over the place they are trying to take over the whole of britains must be bringing them in by the billions I reckon theyve had a secret side deal with Mandela myself

Yeah , just waiting for them to biuld the Carl Marx statue with the old red carpet

Anway Jokes aside Im sure it will come in usefull to you thats if they will ever pay you to do anything Thats another story all on its own that one

Hmm - I simulated the ckt, this time for a ramp that takes 6 seconds to go from Vcc to 0. Attached are the new schematic and the o/p voltage waveform. R1 seems to affect the time it takes to ramp down but not the linearity. With R1 = 10k, the ramp down time is about 5.5 seconds and with R1 = 560, it is about 3 seconds. I guess if R1 itself is a current source then the timing would be more accurate.

The voltage source is a 0-12V square wave of 50% duty cycle and 12 seconds period (6 ON, 6 OFF).

Thanks!!

While re-analyzing your circuit Rohit, I realized that I misread Andrews requirements. For some reason I thought that he wanted the Ramp to happen after power was removed from the circuit -- that is WRONG, thus *MY CIRCUIT IS WRONG*!!

It appears you win the prize Rohit!

Thank you!!