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Old 8th Dec 2019, 3:14 am   #1
Martin G7MRV
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Default Question about a simple RC Integrator

Question relates to the circuit shown in attachment, (LED just represents a load for the 2nd transistor)

Simple RC integrator intended to 'extend' incoming pulses to a fixed pulse length, with the values shown of 5k and 0.1uF pulse length should be 500uS,

but - does the 1k collector resistor for the 1st transistor have to be taken into account as part of the RC time constant? This resistor is their from an earlier version where the 1st transistor was just a drive for the 2nd, connected directly emitter to base.

Everything i've read so far just deals with the RC constant of the single resistor/single capacitor, nothing about whether further resistance separated by a transistor junction has a noticeable effect. If it does, i'll need to adjust the value of the 5k resistor to compensate.


Cheers
Martin


Edited to add - I think the 5k resistor should actually be in parallel with the capacitor!
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Old 8th Dec 2019, 3:28 am   #2
Argus25
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Default Re: Question about a simple RC Integrator

Its hard to answer because the driver circuit is missing:

Whether or not the 1k will have an effect depends upon the source resistance driving the base of the first transistor. If its a high valve, say at least >10k, then the first transistor will still turn on and saturate, so you can regard the 1k then as being in series with the 5k. However, if the driving source resistance is low, say a hundred ohms or less, the current that charges the integrator capacitor will come via the B-E junction of the first transistor, so the 1k then will barely have any effect and you could regard the charging circuit is being just the 5k.

With an RC integrator, the math will only tell the truth if you are driving it from a very low source resistance, and the load across the capacitor is zero to negligible, like say the + input of an op Amp or the gate of a fet etc, so you are not drawing charge away from the capacitor into another load. An OP amp set up as a voltage follower works well, but if you don't want that you can wire up something like a MPF102 as a voltage follower.

One problem trying to get the equation to apply to your circuit is that the integrator capacitor is fed directly into a B-E junction of a transistor which has a very low resistance, suddenly, over about 0.65V and the knee where the transistor switches on & off will not be defined by the timing of the R-C filter, that is easy to calculate at least.

So the circuit for incoming pulses will have an asymmetrical attack and decay profile as the charge and discharge rate of the integrator capacitor is not matched. Circuits like the one you have posted do work to stretch a pulse out, for an actual LED too, for example as a peak audio level indicator, where a narrow peak of audio doesn't light an LED for long enough to see it easily. Sometimes the circuit was called a charge pump too. You will be able to find the pulse stretch value by experiment, and it will also depend on the width of the incoming pulse too.

One way to make a pulse stretcher (without a monostable) is just to use some cmos inverter gates, the output of one charges a capacitor quickly via a diode and that feeds into the input of the next inverter (that doesn't draw current) and you can adjust the resistor across the capacitor for the amount of stretch. Cmos gates can work with less than 5V too.

Last edited by Argus25; 8th Dec 2019 at 3:50 am. Reason: typo
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Old 8th Dec 2019, 3:41 am   #3
Martin G7MRV
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Default Re: Question about a simple RC Integrator

The drive comes from the cathode of a geiger-muller tube, 10k from cathode to base, 100k base to ground, incoming pulse is about 6V at the base. What I cant be sure of is the incoming pulse length.

The RC constant chosen for 500uS is pretty arbitrary. All im looking for is a longer pulse than the short, variable pulses from the G-M tube.

Normally, i'd mock it up and look at it on the 'scope - but im working nights...

Interestingly, the G-M tube itself has an inherent dead-time after any pulse of 190uS, so an RC constant of 200uS would be ok, as no further pulses can be detected until after the pulse plus dead-time anyway,
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Old 8th Dec 2019, 4:26 am   #4
Martin G7MRV
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Default Re: Question about a simple RC Integrator

Quote:
Originally Posted by Argus25 View Post
So the circuit for incoming pulses will have an asymmetrical attack and decay profile as the charge and discharge rate of the integrator capacitor is not matched. Circuits like the one you have posted do work to stretch a pulse out, for an actual LED too, for example as a peak audio level indicator, where a narrow peak of audio doesn't light an LED for long enough to see it easily. Sometimes the circuit was called a charge pump too. You will be able to find the pulse stretch value by experiment, and it will also depend on the width of the incoming pulse too.

One way to make a pulse stretcher (without a monostable) is just to use some cmos inverter gates, the output of one charges a capacitor quickly via a diode and that feeds into the input of the next inverter (that doesn't draw current) and you can adjust the resistor across the capacitor for the amount of stretch. Cmos gates can work with less than 5V too.
Cheers,

I think the incoming pulses are some 5-10uS, with a very fast attack but then and exponential decay. Sometime early in the coming week i'll be able to scope the input and see what the incoming pulses are like.

This particular circuit needs to use the minimum parts and be as simple as possible - its just a small pocket unit, but im providing an output to connect to an external counter (probably arduino or PIC based, my eldest lad will develop that!) so im just looking for a longer, stable pulse. The external unit can do any fancy, leading edge triggering etc stuff.

I actually have the circuit on bread-board, except the RC parts. Although there is a capacitor in the required position, 22nF, which was added to cure leakage of the HV circuits oscillator, which was causing a very faint but annoying flickering of the LED. That has resulted in a better flash from the LED, so im wondering if it is acting as a pulse stretcher anyway (im guessing with the 1k?)
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Old 8th Dec 2019, 5:43 am   #5
Terry_VK5TM
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Default Re: Question about a simple RC Integrator

Just getting a little ahead of you, if the tube is ultimately going to be interfaced to a PIC or some other micro, there is no need for pulse stretching.

The attached picture is part of a schematic from the EPE Feb 2007 Geiger Counter and the tube is connected by a couple of resistors and one transistor directly to the PIC (the circle is just to indicate where the tube is).

All the other components are part of the 500v generation circuit (also generated by the PIC).

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Old 8th Dec 2019, 5:47 am   #6
Martin G7MRV
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Default Re: Question about a simple RC Integrator

Quote:
Originally Posted by Terry_VK5TM View Post
Just getting a little ahead of you, if the tube is ultimately going to be interfaced to a PIC or some other micro, there is no need for any semiconductors.

The attached picture is part of a schematic from the EPE Feb 2007 Geiger Counter and the tube is connected by various passive components directly to the PIC (the circle is just to indicate where the tube is).

Attachment 195030
Not quite, the interface will allow it to attached to the electronics for counting on a different, more advanced unit (with a much bigger tube - 10"!), but this unit itself is a little portable 'clicker' type using a very small tube (SI-19BG, about 1cm x 2cm), so it makes sense to improve the pulse a little in the indicator circuits. I plan to add an opto-coupler and a 3.5mm jack socket to allow external connection.
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Old 8th Dec 2019, 6:09 am   #7
Martin G7MRV
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Default Re: Question about a simple RC Integrator

Returning to the basics - am I right that the 5k resistor should actually be in parallel with the capacitor?
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Old 8th Dec 2019, 9:35 am   #8
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Default Re: Question about a simple RC Integrator

TBH, relying on an RC circuit driving the base of a transistor will not be very predictable anyway. You'd need a resistor across the cap such that its load is way more than the transistor's base and even then the output rise time will be a bit slow. Really need to see the input waveform, but I think given you are driving a PIC and I'm assuming that you are driving an interrupt what you want is a fast clean square waveform with no chance of bounce. I would therefore use a schmitt trigger circuit which can be done with an op-amp or a couple of transistors. I don't know what the min pulse width is for a PIC to register an interrupt but I'd be very surprised if it were more than a uS, so you probably don't need to stretch the waveform.

TTFN,
Jon
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Old 8th Dec 2019, 10:48 am   #9
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Default Re: Question about a simple RC Integrator

The capacitor is in a place where the voltage change is very small. It isn't going to work very well at all. I'm afraid the original post circuit is destined to be disappointing.

If you want a pulse stretcher, look up 'monostable circuits'

The Art of Electronics covers them quite well.

David
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Old 8th Dec 2019, 12:10 pm   #10
Argus25
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Default Re: Question about a simple RC Integrator

Quote:
Originally Posted by Radio Wrangler View Post
The capacitor is in a place where the voltage change is very small. It isn't going to work very well at all. I'm afraid the original post circuit is destined to be disappointing.

If you want a pulse stretcher, look up 'monostable circuits'

The Art of Electronics covers them quite well.

David

It is interesting that sooner or later, everyone who attempts any kind of electronic circuit of moderate complexity comes up against the problem of how to shorten a pulse, or how to lengthen it.

The former case is usually pretty easy but the latter well, not as easy.

I'm pretty sure I read in Horowitz & Hill that if you use monostables to do it, it was a mark of a "neophyte designer". In fact I think I read that just after I concocted some circuit that was full of them and felt suitably insulted, but I did see what they were getting at. Sure enough monostables do have their issues with triggering and jitter.

Yet the digital way to do it might not be that helpful if you don't have some stable master clock that you can count for some specific delay.

Of course these dilemmas suddenly got taken out of the hands of the hardware designer, more or less, when even the simplest of projects (like a PIC micro controlled tin can opener) needed some delay, it was all about digital counting, so the delay became a software problem. This is a little like the software solution to de-bounce switch contacts (but don't be fooled, actually it works much better with an S-R flip flop as it largely throws away time domain considerations that the software cannot).

Of course if you "bodge up" a circuit with hex Schmitt trigger inverters, that are supposed to be for "digital applications" with analog RC networks and diodes jammed in between them, the world will frown upon it (you) and the designer will be accused of being a neophyte, but it still might just well be the most practical and economic solution to a problem, that works well.

Recently I saw a "kludge" in a circuit where a small surface mount part had been added, it was actually a PIC micro, that had been programmed to simply delay a pulse, really? thousands of semiconductors deployed to do it digitally.

Last edited by Argus25; 8th Dec 2019 at 12:33 pm.
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Old 8th Dec 2019, 2:04 pm   #11
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Default Re: Question about a simple RC Integrator

But when a micro can be had for pennies and easily made to do a reliable job, isn't that a saving of effort in a different direction? Yes, it's adding "valves" needlessly; but sometimes, minimum parts count isn't the overriding consideration. Good analogue designers who know what they are doing are expensive .....

The replacement of analogue signal shaping with digital alternatives is really just the next logical step after the replacement of inductors with op-amps and capacitors.
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Old 8th Dec 2019, 5:14 pm   #12
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Default Re: Question about a simple RC Integrator

Quote:
Originally Posted by julie_m View Post
Good analogue designers who know what they are doing are expensive .....
The replacement of analogue signal shaping with digital alternatives is really just the next logical step after the replacement of inductors with op-amps and capacitors.
I guess that makes the next logical step the replacement of the designers with AI.
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