Peak Detection and Hold Query

I think you have the solution, but you don't know it!
Effecively you are saying you have the 2 voltages one of which moves relative to the other, but they start at the same level. Right? (or maybe I've missunderstood and you need to simplify the question)

The answer is to change the level of one of them, a simple potential divider will shift the voltage a little or, it can be done with an op amp (unit gain and a pot to adjust).
Sounds like you are having fun...this sort of thing is a great way to learn electronics, which is of course easy in theory...but much trickier in practice.
If you've made a peak detect & hold circuit which works you are doing fine.

I hope that makes sense.... What I'm really saying is that as long as the switching occurs at the right voltage, the 'actual absolute' value of either voltage doesn't matter, so you can shift one of them up or down to get your switching.
(Also you won't get it to switch positive going and negative going at the same voltage, there is always some hysteresis, especially if you have some positive feedback. Another solution may be to have two comparators switching at different levels one for positive going and one for negative going)

Have fun
Del

"I think you have the solution, but you don't know it"

tend to agree, staring at me but cannot see it. Have thought of changing level of one side to an Op-Amp to shift the voltage a little. A pot is a good idea for fine tuning and I think I may to do this for each circuit connected to its assigned battery.

Hysteresis might be a problem, but will only know once I have completed the final half of the circuit after a few weeks of trying this and that.

"Effecively you are saying you have the 2 voltages one of which moves relative to the other, but they start at the same level."

Very much the case, I was worried I might have shared some of my confusion in my question, but glad I (mostly) did not

Thanks for the advice.

You have your peak detector which (in a charging example) follows an increasing voltage from 1.0 V to somewhere around 1.5 V. When the battery voltage reaches the peak and has that "droop" that you have mentioned of 30 mV, the peak detector still has a memory of that peak voltage (I don't say 1.5 V because this is what you wanted, a not the same value at the total charge).

You cannot compare (for a fixed difference of 30 mV) because the peak fluctuates.

What to do? Repeat the peak value (whatever that it is) from which you subtract 30 mV. Compare this "new peak" voltage with the battery voltage. When the difference is zero, you have reached a voltage 30 mV lower that the maximum charging peak.

Good reply....

basically to re-iterate what indel said, you need to have the output of the peak detector compared with the input voltage using a seperate comparator...

If the 30mV is needed just use a potential divider from the peak detector, so when the input voltage goes down below Vp - 30mV the output switches...

Simples!

Electroman,

"If the 30mV is needed just use a potential divider from the peak detector, so when the input voltage goes down below Vp - 30mV the output switches..."

Potential divider from peak voltage can not give fixed 30mV for different peak voltages. Potential divider forms a ratio based divider.

A differential amplifier or instrumentation amplifier can subtract voltage (Peak-30mV) is a much better choice.

Peak detect and hold circuits have hold signal drop rate and to avoid this either digital peak detect and hold can be used or double peak detect and hold can be used.

Easiest way is to use ADC and uC to take decisions. 10-bit internal ADC may be all right.

Thank you. I have read some of your previous posts (prior to posting my query) regarding peak detection and found them informative.

With regards to my application in theory and practice, I do not have an issue with the hold signal drop rate. The voltage drop that occurs once the battery reaches a charged state, takes place before any real losses from the storeage capacitor in the peak detect circuit.

"A differential amplifier or instrumentation amplifier can subtract voltage (Peak-30mV) is a much better choice."

I will once again look into the option of a differential amplifier. But in my clouded ignorance, cannot for the life of me, understand how to get a seperate amplifier (in relation to the peak detect circuit) to switch when a voltage drop relative to the peak occurs.

Due to my lack of equipment, I have must avoid the use PIC's and the like.

Thanks for your time and thoughts.

If your voltage source is only 1.5V or less then first thing you need to do is to boost this voltage using DC-DC booster circuit (oscillator and voltage multiplier rectifier etc and you may need an transistors, inductor, capacitors and diodes to help this happen). Once you have 3V-6V or about that then most of the circuit will work easily.

You can also use amplifiers that work on 1V but what you have works above 2V. I am not sure if there is anything that can boost voltage source below 0.5V and is also powered on that voltage.

There is a component named Zener-diode which has a defined voltage drop and can be used as a reference. Attention its voltage depends on temperature but not much.

Depending on its voltage you can via a divider compare the variable voltage to the diode voltage and obtain the signal you need when the variable voltage has a drop as you define. The schematic is simple: from the main line you supply the diode via a resistor and the diode is connected to ground. The voltage between resistor and diode will be constant which ever variations the supply in front of the resistor has. You connect the voltage now to a divider dimensioned so that if the voltage is 1.5 V the reduced voltage is slightly over the diode voltage and if main voltage goes XmV lower the reduced voltage is under the diode voltage. You connect your comparator (OP amp) between the 2 voltages and have your signal when main decreases.

If your main voltage is under the diode voltage then you use a divider to reduce the difference.

Please read the post nick_name he doesn't want a reference voltage.... D'oh!

I read it but without a reference it can recognize a drop only if it is a dynamic one, and for that he can use a capacitance to save it for a limited time. If not both voltages move same way since in fact it is only one voltage which has to be compared with a value at origin which does not any more exist! This at least what I understood but as usual I never am sure to be right.

The OP states: "I have had no problem with constructing a peak-detect and hold circuit".

I will try this option out. Have thought of something similar, but did not like the fact that diodes can vary their specs due to temperature shifts, both circuit and climate induced.

For a shift of something around 30mv, I might get around any sensitivity issues. Will try it out and see what results.

If it was me, I'd use an A/D and a small micro-controller.

What a waste of time and space.... a dual op-amp and a few resistors do the job in a fraction of the time....

One 8 pin PIC is all you need. Hardly a waste of space, or time.

You still have to code the thing = time

CrazziestOzzy,

Could you fool around with this kind of comparator circuit.

With the Trip point set above the reference Voltage the Capacitor across R2 will prevent the comparator from toggling when there is a rising voltage. A drop in voltage at the trip point will lead the drop in voltage in the capacitor causing the output to change state until the capacitor has equalized.

The trip and ref voltages should be lower than the Supply Voltage (V+) for the IC. Then V Batt can be pretty much whatever you want if the components are selected for the V Batt range and sensitivity.

Jon

A small PIC with the right coding is the most accurate way to perform the negative delta-V test when charging NiCads. Considering you can implement negative delta-V, peak voltage, min voltage, pulse charge, burp pause (for gas recombination), temperature monitor, and high charge rate timer functions all in one little 8 pin micro, it is really no contest.

The negative delta-V test is performed after the charging current is briefly halted and the open cell voltage stabilizes. The present cell voltage is compared to the previous one (or several values are integrated to remove noise) to find the voltage drop indicating a "full charge" state has been reached. Trying to perform this test WHILE charge current is still flowing is usually worthless.

It is fairly easy to configure a single op-amp to work as a "negative slope differentiator". This would theoretically detect the negative delta-V and provide you with the "full charge" signal (I believe) you seek. However, op-amp differentiators can be very difficult to use if there is ANY noise present. While this circuit could detect a negative delta-V, trying to use it while charge current is still flowing will probably not work.

Review these or google "op-amp differentiator circuit"

http://www.ecircuitcenter.com/Circuits/opdfr/opdfr.htm

http://www.ecelab.com/circuit-differentiators.htm

I'm not say it can't work. But, by the time you get something that "kind-of" works, you could have learned and written several hundred lines of simple PIC code to do it more accurately. I specialize in analog circuit design and am just an AMATEUR PIC programmer. While I'd prefer to go analog for a project like this, the PIC micro (or one of the ASIC's) wins this one.

Good Luck!

Well spoken. I envision that I will indeed be using a PWM to "burp" charge (will be pulsing at the optimum rather than using a constant current) with measurements for peak detection and voltage drop being done during the brief offstate between "burps". At the moment though, I am aiming towards a working conceptual circuit that can be later expanded into a more reliable real-life system...wihout a PIC or AVR