Signal Processing

Yes ,you can , in simply use divider voltage to process the peaks of the pulse usually the divider voltage is two resistances are connected in series the connection point between them (use those values of a connected Resistances 1KΩ and 4KΩ you will get 6 voltage over the 1KΩ and 24 volt over the 4KΩ). regarding with time of te pulse 2ms is means 1/2ms *1000 Hz =500Hz and 10ms=100Hz so you can use divider frequency logic I.C to divide the 2ms by 5 to get the 10ms time period for this process use TTL LOGIC I.C SN7490.it can division by 5

Well the problem is I have only a single pulse generated using piezoelectric igniter. Will the frequency divider circuit work with this single pulse?

from which side you are not sure i am not sure about your piezoelectric do you want to drive it by this signal or it generates this pulse. explain to me more . the voltage and the frequency of the TTL SN7490 it can work up to 50MHZ max freq,input signal and 5.5 volt max .input and output and you can see the data sheet .

The piezoelectric igniter I am using generates a single pulse of 1400 volts for 2ms. I wanted to increase the time period of it by decreasing the voltage. Is it possible?

That is not right logic in the electronics and i can't understand you why you want to do that . as i stated to you have to step down the voltage and the frequency individually to get 6 volt with 10ms period time signal . there is an liner device can convert the voltage of the signal to frequency or frequency to volt(FTV OF VTF) you may benefit from it this I.C is LM331

It may be possible to do what you want with a passive pulse shaping circuit. Voltage and time alone do not provide an energy value. Can you define the source & load currents or impedances? Also, it is important to provide consistent specifications. Is it 30V or 1400V on the input voltage?

Read for monoshot/monostable multivibrator. You can construct one with discrete components or ICs. You will need to supply power for these circuits.

Gajanan Phadte

I agree that a separate circuit would be preferred. Perhaps a retriggerable, one shot multivibrator that provides the desired pulse would be best. Taking energy from the piezoelectric pulse will not only load it down but it may blast the trigger into next year so some care must be exercised to step down the trigger voltage.

The igniter voltage is also high enough to excite gas in a fluorescent tube. The persistence of the fluorescent power, however, may be longer than 10mSec. Just another way to think about it. trigger

Have you considered adding a high voltage capacitance to store some of the charge of the output of the igniter to extend the period of the pulse when it becomes the lower voltage?

The rule for the peak impulse response P of a "no-loss" filter of bandwidth B Hz at 2:1 reduction from flat (-6 dB)[if pulse length << 1/B] is

P = 2.B.A where A is the ingoing pulse area Volts X seconds

30 volts x 2 ms gives A = 30*2/1000 = 0.06 volt-seconds

If B = 50 Hz, P = 2*50*0.06 = 6 volts.

The period of the output oscillation is about 1/B or 20 ms for 50 Hz low-pass filter - so the half period is 10 ms. Usually, filter damping is enough for the first half cycle of output to be much stronger than the next.

I do not understand why you give 30V and then 1400V peak input, but I suggest if you have 1400V in and want 6V out, a series resistor - shunt capacitor filter preceded by resistor attenuator may do the job.

As has been mentioned by other replies, if you want precision, you will have to use a monostable electronic circuit.

67model

As there is no repetitive "frequency" involved and you are generating a high voltage with a short period from an ignitor and you want to convert the high voltage to a lower voltage with a longer period it stands to reason that a passive filter would do the trick. Only the time frame from the snap of the piezo generator to the output of the signal conditioner would be delayed. Right? I have seen others suggesting an RC circuit for damping the voltage and extending the period using this concept. Do you see any problem with such an idea?

Further to my post#10, an RC of 5 ms, say 1 megohm and 0.005 microfarads, would be near enough for 50 Hz bandwidth. 1400V 2 ms would give 300V peak out however (warning, an ordinary 1/8 watt resistor is only rated for about 500V pulse, 1400V needs care). But I doubt the peizo pulse is square [a steep front with exponential tail and pulse width measured to 1/e, 37% of peak, on tail is likely] and its impedance will be high, which would change the result. Also, the load impedance would have an effect, unless it is megohms.

And thanks for the warning about time delay, K...dweller

67model

The op (Akash) said "The piezoelectric igniter I am using generates a single pulse of 1400 volts for 2ms. I wanted to increase the time period of it by decreasing the voltage. Is it possible?"

I don't understand the 50Hz reference unless you are refering to a half cycle of 50 Hz waveform which is 10 ms. 20 ms is for a half cycle of 25Hz. I don't think he can push the button that rapidly in succesion.

The pulse he wants is 6 volts at 20 ms duration and I believe that would be the time between the rise and fall of the pulse with regard to its usable characteristics.

It is a good point you made about the need for high Voltage rated components.

I do understand that an RC arrangement can reshape the pulse and the high voltage and current can be redistributed so that there will be a lower voltage and longer sustained current. Other passive devices can be incorporated to provide limiting, clipping, clamping and shaping too. I believe this is the desired effect.

Then whatever this pulse will be applied to is going to effect all of that...:-)

JB

Well I was not sure of voltage because the CRO which I am using gives a resolution of just 50 volts/division but I need a higher resolution CRO like 1000 volts/division to get the actual voltage. But when I referred certain datasheets I came to know that the actual signal would be a degenerating sine pulse with initial voltage of 5.44 kvolts. Would the passive RC filter work for this type of signal. The total time duration for the signal to reach 0 volts is 1.2ms.

A high Voltage probe for the CRO does the job of providing a high impedence Voltage divider like 1000:1 and the volts per division on the CRO is set to 1 volt per division so 5000 volts would appear in 5 divisions. A standard probe would be damaged and you would be looking at false indications and arc damage to your low Voltage probe.

If I knew more about what your project is, like testing igniters, using igniters for triggering......needing a usable pulsewidth and amplitude derived from an igniter pulse to activate a circuit or device......

JB

Well I want run a MSP430 micro-controller with the help of piezo igniter to send a packet a data, say a single byte. But I am still testing it, don't know whether it might work or not.!

Does it matter if you get the 5kV pulse, Akash?

Do you need ignition of gas? - or is the peizo just a means of getting a pulse?

If you do not need 5 kV, you can load the peizo with a capacitor C which just absorbs the charge [Q = CV, hence V = Q/C N.B. units Volts, Coulombs, Farads] and jumps to a voltage V. The shunt resistance (you adjust to suit) will then give an exponential decay, with t = RC.

The data sheet for the peizo should give its capacitance (picofarads) and the charge q or voltage produced by a specified mechanical pulse.

It will be important to protect microcontroller input from high voltage, by zener or voltage dependent resistor VDR. To protect oscilloscope and give 1000V/division, use 20 megohm metal glaze series resistor (check voltage rating), if scope is standard 1 meg input. Good thing it was just a little peizo pulse and not a jet engine ignitor box!

k...dweller please note that calculation for bandwidth B in my post #10 gave a value of 50 Hz

A piezoelectric ignitor delivers a small amount of charge (Coulombs=Volt-seconds) at a fairly high voltage. The high voltage pulse can be converted to a low voltage, but there will be significant energy loss. Even with the loss, you may still have sufficient energy to power your circuit long enough to generate and send (RF transmit?) a packet of data.

Calculating results will be difficult without knowing the exact specifications of the ignitor, your micro code, and the transmitter energy requirements. It may be easier for you assemble a simple circuit to bench test the concept.

> I suggest using a bridge rectifier composed of fast recovery diodes with the two AC terminals attached to the piezoelectric device.

> Start with a 0.047 micro-Farad capacitor attached to the +/- rectifier terminals.

> In parallel to the 0.047 micro-Farad capacitor, I suggest 10k ohm resistor (test load) AND a 3.3V Zener diode to clamp the maximum voltage.

> Activate the piezoelectric ignitor and monitor the output of the test circuit with a scope probe.

> Adjust the capacitor value to get the best results with your specific ignitor.

IF the ignitor pulse can be converted to a +3 Volt pulse with a 0.5 millisecond duration (at about 300 micro-Amps), you should be able to generate a burst of data from the micro. Whether there is enough energy left to power an RF transmitter to send the data is another matter.

Best wishes :-)