Sunday, 24 November 2013
Plus – no harm for doing that right? After all, people accidentally does it once a while…
So with the power on – and closing in with index finger – I finally get closed enough – for a spark to “shoot” into my finger – and yap, it was painful – just like being poked by a needle! And it never cease to impress me what electronic can do, with couple of AA battery, this thing can boost it up to kV for zapping!
So this is where the boost converter comes in, with a boost converter, we can easily step up any voltage to the desired level, and it is pretty common in electronics (camera flash, spark plug….) And who says electronic is boring??
Monday, 7 October 2013
For a circuit guy – like myself – I think what the saying is really trying to tell you that AC-DC adaptor with transformer built in is better. For this I can’t agree more... from low level signal perspective.
For a +5V AC adaptor without transformer – each and every node inside the adaptor circuit is considered “LIVE” which means it is dangerous – the AC outlet voltage 230V for example is directly switched by silicon IC, and filtered to give a output of +5V. See schematic below:
(picture from some online forum.. sorry that can’t remember where I got it exactly, been with me for some time)
For a +5V AC adaptor with transformer – generally the transformer will scale down the line voltage to much managed able 12VAC or so, before passing it through bridge rectifier to and further drop down to +5V. So the secondary side of the transformer is not “LIVE” and much safer to mess around. Note that since the switching is done at much lower voltage level – the electrical noise will be much less. As shown below:
There are considerations on which is better – costing, how “clean” it is..., eventually it is up to the application and individuals to decide. For me, when I build something that requires mV or nA level of accuracy – those with transformer are definitely my choice of adaptor. Other than that, it does not bother me much.
I guess at the end of the day, knowing the fundamental of different AC adaptors allow an engineer to decide what is good enough for the application, and this especially important when dealing with low level analogue signals.
Friday, 6 September 2013
I like wiring – since my childhood years – guess it is just the way i am.
One of my early experience as a teenager is to extend the telephone line – essentially i just strip the wires and twist the interconnect – with my bare hand. In the middle of doing that, muscle at my upper arm started twitching involuntary – it felt kind of funny – no pain but still… seeing part of your body behaving out of your control is kind of weird…
As it turns out, telephone line is typically at 40VDC, and when it rings it can go all the way up to 90VAC.. so i guess my twitching muscle act as the incoming call detector!!
Having your body as part of current path is never a good idea, few mili ampere of current is enough to mess up your heart beat under the right condition – so give electricity some respect is what we should do..
|IEC voltage range||AC||DC||defining risk|
|High voltage (supply system)||> 1000 Vrms||> 1500 V||electrical arcing|
|Low voltage (supply system)||50–1000 Vrms||120–1500 V||electrical shock|
|Extra-low voltage (supply system)||< 50 Vrms||< 120 V||low risk|
< 120VDC is considered as low risk.
to understand this from graphical approach,
let’s start by drawing out a x-y axis – with x being voltmeter reading and y being actual value
then draw a line with y = x (implied m=1, c=0) from –1 to +1.
Let’s try to interpret this graph
now, to further discuss this, i think it is best to throw in some number – let’s consider the gain error of +/-10%. what it means is that the line will has a slope of 10% deviation from our ideal line (m=1, c=0).
if gain error is all that we have, then we have something like below:
next let’s consider the case where gain error is zero, and +/-10% of offset error (remember that we are taking about +/-10% of measurement range offset). for this we have
Now, for the real world instrument – gain error and offset error is real and cannot be ignored.
Factor in both gain and offset error, we have resulted in a series of possible lines that the real instrument behave.
From the look of it, it seem pretty bad – as the uncertainty is high for the voltmeter reading, in fact – up to 20% of measurement range. But, what you pay is what you get, gain and offset error of 10% is chosen for the sake of our discussion here. For real voltmeter with decent pricing, you can easily get 0.1% accuracy for both offset and gain. In certain case (depend of calibration, operating temperature, pricing…) you can even get better than 0.01%! Now, that is impressive.
In this example we used a voltmeter, but most measurement / sourcing instrument (source measure unit, power supply, oscilloscope, capacitance meter….) present their spec in similar form. So you can apply this method to have a sense of what the spec mean – with graphical approach.
Sometimes, the manufacturer will give un-normalized spec for offset, which is essentially the same thing. take our example here of +/-(10% reading + 10% of range) , if the range is 10V, the equivalent un-normalized spec will be +/-(10% reading + 1V)
Friday, 2 August 2013
Take an example from the figure above – the device might be promoted to have less than 25degC per watt of junction-ambient thermal resistance (with fine print indicates test condition) – but without heatsink or large enough copper area, the thermal resistance easily go up to 40degC/Watt - as shown in the graph when copper area is less than 1 inches square.
Power dissipation specification at times shows best number - but only when mounted on large copper area - not practical for actual circuit – often heatsink is needed for low thermal resistance.
But it was the standard way for the industrial to compare using this method - too fresh and naive and some said silly to take it literally :)
Friday, 5 July 2013
Follow up on previous post, a few points to note on algebra analysis - to state the obvious for those who already know, to clarify things for those with humble beginning like myself:
- acknowledge consciously that the goal is to put variables into y = mx + c
- you need to know at the end of derivation what is on the left hand side – in other words – you need to know what is going to be y, and x before you even derive it
- always substitute repeating complex terms with a symbol – it keep you mind in the bigger picture – and save you from typo – or miss-spelled mistakes. You can always re-substitute the terms back at the end of derivation.
- watch out for stuffs that can be derived from basics variable, and make use of them – for example, identify that Vdac is a function of Vref, and let Vdac = β*Vref, where 0<= β<=1.
- always good idea to drop in some number to verify nothings goes wrong in the algebra derivation
- take it easy – nobody is going to punish you if you get it wrong – have a bit of curiosity, have fun.
Saturday, 15 June 2013
When I first started doing design works, there was this particular project that required me to communicate through I2C. It was my virgin experience using Microcontroller with embedded I2C engine– and it did not works well – errors occur intermittently – I spent 2 week checked and rechecked my codes, bus waveforms – start , stop, acknowledge conditions – every single thing that I could think of…
There was so much frustration, finally I told myself that enough is enough and there must be something else going wrong instead of my code – and Google long enough – I found out that there’s something called “Errata sheet”. And true enough – looking for the errata of that silicon revision on manufacturer’s website it showed that there was a bug in the silicon – and it proposed some work-around, plugged the workaround in and everything works accordingly
Lesson learnt –
Before choosing a controller or similar devices – always read the errata (if it has one) for the stuffs that related to you, and newly released devices might be more happening that you would expect!
Monday, 3 June 2013
I remember my lucky escape from getting injured dealing with a boost converter with electrolytic capacitor. It was a brand new design – a boost that generate 100V. Something was not right – it could not go up to 100V – so I was having this board vised up ( to see what a “vise” is, go to http://en.wikipedia.org/wiki/Vise) is to ease probing on each side – with those cap facing me (bad bad choice).
While I my face was inches away from the board – one of the cap exploded and shooting up hot boiling oily substance right pass me, hitting my cubicle wall about 1 meter way – it turns out that wrong part was installed – it was a 48V part.
It could at best cause some burnt on my not so handsome face – or at worst blinded me… come to think about that – I was simply lucky enough to escape un-injured and extremely thankful. To show you what i meant, i managed to find just a similar end result from wiki (http://en.wikipedia.org/wiki/Electrolytic_capacitor) as shown below:
Lesson learnt – those weak points being built on top of the capacitor can is the “safety valve” – it is designed to open up (“explode”) under overheat – overvoltage condition – never ever face the valve toward you or anybody else.
Sunday, 26 May 2013
Lesson learnt –
There are certain amp that are not meant to be unity gain stable – mainly for fast response – read the datasheet before using them.
Generally DAC output will be from 0V to Vref, and to get bipolar output we typically use this circuit that convert [0:Vref] -> [-Vref: Vref], with Rf = Ri.
But what else can this circuit do if we are able to tweak Rf, Ri? Can it gives Vo > Vref?
There are few ways to get the transfer function of this circuit block, namely:
- circuit simulation - overkilled for such a simply circuit - don't you think so?
- algebra manipulation - going to be tough for those not doing it for a while
- graphical analysis base on circuit inspection - quick and intuitive - my personal favourite
in this example, I'm going to shows you alternative #2, and #3, have a nice read-up!
The algebra way:
Vo = Vdac + i * Rf
Vo = Vdac + [(Vdac-Vref)/Ri] * Rf
Vo = Vdac*(1+Rf/Ri) - Vref*(Rf/Ri)
but it is time consuming to write Rf/Ri, let's just replace it by α
Vo = Vdac*(1+α) - Vref*α
but Vdac is a function of Vref, let Vdac = β*Vref, where 0<= β<=1
Vo = β*Vref*(1+α) - Vref*α
Vo = Vref* [β + α*β - α]
α = Rf/Ri
β = dacData/dacFullResolution
it is always good idea to drop in some number to verify nothings goes wrong in the algebra derivation, let's do it now
let α = Rf/Ri = 1
Vo = Vref* [2β - 1]
β = 0: Vo = -Vref
β = 1: Vo = Vref
which is expected
Now that we have the final equation, we can inspect the equation as ratio of Rf/Ri changes:
Rf/Ri = 0:
- Vo = Vref*β
Rf/Ri = infinity:
- from Vo = Vref* [β + α*β - α]
- when β = 0 (Vdac = 0): Vo = negative infinity
- when β = 1 (Vdac = Vref): Vo = Vref
from general equation of y = m*x + c, this circuit configuration allows us to configure maximum m = +1 @ c = 0, minimum m = negative infinity @ c = negative infinity, and max Vout = Vref.
For example, you are not getting more that Vo = Vref regardless of what Ri, Rf values you used.
The Graphical Way:
Figuring out what is input, what is output:
- draw out XY axis
- acknowledge that we control dac output directly, hence x-axis is label as Vdac
- Vo is the output of circuit block - we want to know what happen to Vo as Vdac changes - thus Vo as label for y-axis
Draw out the xy boundaries by circuit inspection:
- from circuit inspection - dac output can only varies between 0V to Vref, thus we draw at vertical line at x = Vref.
- from circuit inspection - when Vdac = Vref:
- since inputs of op-amp is essentially the same potential - the inverting input is at Vref,
- there is no voltage drop across Ri, both side of Ri at Vref, thus i=0A, and since there is no voltage drop across Rf, Vo=Vref
- thus draw a horizontal lines at y = Vref
Acknowledge there is essentially one variable to play around - that is, the ratio of Rf/Ri :
- from circuit inspection - consider the case Rf/Ri = 0, it means Ri = open and i=0, and the amp act as voltage follower , Vo = Vdac, from y = m*x + c, this gives m =1, c =0, draw a line that cross point (x,y) = (Vref, Vref), (0,0), this is one of our boundary.
- from circuit inspection - consider the case Rf/Ri = infinity, when Vdac =Vref, Vo = Vref, when Vdac is slightly less than Vref, Vo = negative infinity, draw a vertical lines at x = Vref.
- region bounded by the 2 lines above is the possible circuit function for this circuit configuration. For example, you are not getting more that Vo = Vref regardless of what Ri, Rf values you used.
Wednesday, 22 May 2013
So tracing back on the switching waveforms – voltage, current I deduced that the inductance was not right – VL = L * (di/dt) relationship does not hold – drop in another piece – and it start working again.
Being curious as I am (luckily), I check the resistance of the inductor and compared to the new part (did not have LCR meter on hand). The resistance is much lower lower than the number stated in datasheet.
As the failure made me very un-comfortable – I went Google around – and it turns out that inductor does have voltage rating (although not mentioned in the datasheet of the one I was using), so the design was flawed in the first case – luckily it was found out earlier than later (it pays for being curious), else can’t even imagine what it would do in the field. Changing the part to high voltage part – and it work out solid!
For more details, refer to previous post: http://electroniccircuitdesignsharing.blogspot.com/2012/12/inductor-voltage-rating.html
Inductor does have voltage rating – due to the insulation coating of the windings, and the way wires being wound – and the inductor intended for high voltage operation will have working voltage stated in the datasheet – else don’t use it at high voltages.
Always look for the root-cause – it will save your ass
Saturday, 11 May 2013
So it was the time that I look at the datasheet of that amp – and found out that it was current feedback op-amp – not the assumed typical voltage amp – it turns out to be that current feedback op-amp are generally faster – and the behaviour differs as well.
for more details refer to wiki:
Textbook op-amp circuit is generally voltage feedback op-amp – and there are much more in the market then what offered in text book.
Saturday, 20 April 2013
So my humble advice to the beginners is to think about every circuit or IC as a functional block – define the boundary yourself or from a schematic – make sense of the circuit does. To state the obvious, in a typical circuit block:
- will have a function name in the schematic - this contains important info regarding what the block supposed to do
- will generally has input, output, supply and control signals
Once you know that the input/output relationship does not hold (assuming that you understand what the block supposed to do), here are some of the steps that might helps
- Bias the circuit so that the block will have known state (or consider overwrite the input from a known source), use DMM to confirm all the DC pins, starting with supplies pins.
- Confirm the input output relationship – if it is not correct, break this block into smaller block and repeat the process (take a look at binary search debugging post: http://electroniccircuitdesignsharing.blogspot.com/2013/02/debugbinary-search.html).
- If the block is already an IC, check for part number, orientation, cold solder join, missing solder joints before “accusing” it as faulty part.
- remember that sometimes certain failure mode needs more than a DMM to be used (see http://electroniccircuitdesignsharing.blogspot.com/2012/06/why-oscilloscope-is-needed-for-circuit.html) , so if DMM cannot give you convergence, you should consider to use another instrument (depending on what this block supposed to do).
Saturday, 6 April 2013
Shown below is a circuit that convert infrared light level into a voltage. Now, the photodiode convert infrared light density into current. This circuit is to provide a 0V bias across the diode (operate the diode in Photovoltaic mode) and convert the current into a low impedance voltage level for further operation.
In this circuit, the diode being used in Photovoltaic mode, and details about the diode operation can be found from http://en.wikipedia.org/wiki/Photodiode as shown below:
This is just one of many examples of the advantage being to think comfortably in term of current. Try to make sense of this circuit only thinking in voltage term will be a nightmare, since the inverting input of the op-amp supposed to be a “virtual ground”, and 0V across the diode does not make much sense.
Saturday, 23 March 2013
But as years goes by – when I got better in circuit understanding – I realize that by translating current into voltage (using resistance, impedance, equations), there are valuable circuit insights being lost in translation. For example – if you don’t think in terms of current flows – you cannot understand how decoupling works shown in previous posts:
To show you what I meant, from my earlier post about load line: http://electroniccircuitdesignsharing.blogspot.com/2012/05/load-line-bjt-signal-gain-amplification.html
There is this BJT amplifier shown as below:
In this circuit – assuming that it is in linear region, collector current changes with base current directly, in fact Icollector = hfe * Ibase, where hfe is the current gain of BJT. Yes, Vout might be our parameter of interest, but it really is just a “byproduct” of this action: Vout = Vdd – (Ic * RC).
So for beginner – I would like to say when we view or analyse circuit – it is worth the effort to try to think in terms of current – it will open a whole new world.
Friday, 1 March 2013
As i found it tough to explain how it works, i been having this idea to draw it out in 3D, so finally i did it! Since this is related to high voltage stuffs, do read the disclaimer at the bottom of this page.
Look at the picture below, we have a test pen that light up when in contact with a "LIVE" or "HOT" wire, what's the magic behind it?
See picture below for the “BIG” picture. Through the body capacitance the AC current completed its path to Earth wire or Neutral wire, and light up the neon light inside the test pen.
In the picture, note that there are wires being routed inside the wall, floor, or ceiling, and in some cases the metallic piping under the floor will provide the return path for AC current as well.
See diagram below for the equivalent circuit:
Lots more info at link below:
Saturday, 16 February 2013
The idea about debugging using binary search as follow:
- For a know series of circuit block that give un-expected result - always try to determine which half the the series that cause the problem
- Repeat the iteration till faulty block found
In the diagram shown below, let x be the input, and output of each block be yn, where n is the block number, and block 15 coloured in red is the faulty block.
start: y16/x is not expected
- confirm y8/x is expected or not
- since y8/x is expected, we know the faulty block should be some where from block 9 to block 16, evaluate y12/y8
- since y12/y8 is working, faulty block should be some where from block 13 to block 16, evaluate block y14/y12
- y14/y12 is still working, faulty block should be some where from block 15 to block 16
Sunday, 3 February 2013
Let’s say you measured Vin = –1V, but Vout way off +1V, now what can be wrong? How do you debug it? Below are some suggestions for you to consider:
- check the supply to see if the supplies are correct
- visual inspect for
- wrong polarity
- wrong part (if manual assembly involved)
- missing solder or “cold” solder
- check if wrong resistors being loaded - power down – measure resistance of R1, R2 to see if any of R1, R2 is larger than 1KOhm – since this is in-circuit measurement – we don’t expect to see 1KOhm, but we definitely expect 1KOhm or less, since parallel resistance of any resistors is the minimum of those.
- check if oscillation occurs - if everything looks good – get hold of a scope and confirmed there’s no oscillation on either pins mentioned above
- check if op-amp already sourcing too much current - lift output pin, insert a small resistance resistor from pin to pad, measure the voltage drop to see if the amp is sourcing or sinking more that datasheet stated. It is possible that the next stage is demanding more current than expected.
Saturday, 12 January 2013
read more about this fantastic characteristic on links below:
picture from http://en.wikipedia.org/wiki/Wolverine_(comics)