Pull-down resistor for MOSFETs required?

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Hello Hardware freaks,

as an output driver I want to use a (dual) N-channel MOSFETs, a IRF7530, see http://www.irf.com/product-info/....

I have connected the gates of the MOSFET to general I/O pins of an ATtiny26. Datasheet for ATtiny26 is at http://www.atmel.com/dyn/resourc....
The drains should switch the loads to ground (source potential). The loads are connected on the other side to a higher voltage (12V) than the AVR (5V).

Since the gates of the MOSFETs are very high-impedant, I want to ensure that the MOSFETs are not conducting all at the same time, if
(1) there is no Vcc voltage for the AVR, but on the load
(2) if the AVR is in reset state.
(3) if the AVR starts up.

I tried, what happens, when there is no pull-down resistor:
in case (1) the MOSFETs are not conducting,
in case (2) the MOSFETs hold its last states.
in case (3) the MOSFETs are conducting for a short time.

Case (1) and (2) are both OK for me.
But can I know it for sure, that it will be like this under all conditions?

Case (3) can be avoided by a pull-down-resistor (I used 10kOhm). (Maybe also by reducing the start-up time specified by the fuse bits.)

It is said in the datasheet that the general I/O pins are set to inputs and pull-ups off after a reset => high-impedant.

But what happens during start-up with the general I/O pins (when the voltage rises to Vcc)?

If possible, I would like to omit the additional pull-down resistors between the gates and the sources / GND.

But I want to avoid any latch-up-problems (12V before 5V present), but also spent not more parts than necessary.

Is there a way to safe the resistor?

Regards,
Michael

P.S: By the way: what is the correct adjective: high-impedance (http://www.leo.org) or high-impedant (datasheet)?

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I would add the resistor just to be sure. Resistors costs almost nothing. Unless you're going to make a million of these, then every cent counts.

Also I would add a small (100ohm) resistor in the gate to limit switching current.

Remember there is a small parasitic capacitor between the gate and drain so glitches on the drain shoot through to the gate.

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Skotti, don't try to save a few pfennig ( :) ) on such an essential resistor. And the (by Jayjay) suggested 100 ohm: I second that. Not just for the small parasitic, but to limit the AVR-pin current as well. It depends of course on the used FET, but the heavier ones: 1000 - 1200 pF gate-capacitance. And that's hard work for the AVR-pin

Nard

Btw: it's a good idea to post simular questions in the General Electronics Forum. Makes them easier to find.

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and the adjective is impedance.

Go electric!
Happy electric car owner / builder

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How quickly will you be switching the MOSFET? If it's going to be rather fast like say if you're doing PWM then you might want to add some transistors driving the gate for you to help it switch on and off faster, like so;

Edward

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Skotti,

During the start-up phase the state of the pins is usually open circuit. But can, for a short time, be indeterminate.

As some one else said, if their is a transient on the drain and no resistor from the gate to ground, this could turn the FET hard on, possibly destroying it in the process. For low power switching application their are such things as logic-FETs that have built in protective resistors (Fairchild make a variety of them). But for higher speed switching you need to make sure the FET is protected yourself.

A small series resistor (100 ohm may in fact be too large) can limit the inrush current intop the gate at turn-on, but you sometimes need a diode to allow the gate to discharge quickly enough when it is being turned off.

Saving a penny here and there won't help you. After all, resistors are only what, $20 or $30 for a reel of 4000 pieces these days! Worry more about the product warranty claims resulting from unexpected and unprotected transients - if you can process a warranty claim for $20 please let me know how!

Dean

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Thank you very much for your suggestions and your schematic example (Edward)!!!

You are completely right:
The costs are nearly nothing in relation to the costs which can arise, if something goes wrong...
In my case it was more a question about space on the PCB. The circuit board has the size of a DIL-package: it's a replacement for an obsolete part. Even the room needed for a 0402-SMD resistor is a lot. That's also the cause why I need the tiny devices.

I managed it to squeeze the pull-down resistors on the PCB.

Series resistor between Output pin and gate of MOSFET:

The frequency on the gate-input will be not higher than approximately 220Hz, i.e. no high-frequency PWM.

I looked in both datasheets:
MOSFET, Input capacitance: 1310pF (typical)
ATtiny26, DC current per I/O pin: 40mA (max.)
So, let's see the worst case:
If the output pin rises from 0 to 5V (or falls from 5 to 0V) "infinitely" fast, each capacitor (also a small one) is a short-circuit in this time.
To limit the current I need a R = U / I = 5 V / 40mA = 125Ohm resistor.

How would your calculation look like?
I did not find anything about the maximum capacitance, an output pin can drive directly.

Regards,
Michael

P.S: Thanks, sgomes ("high-impedance"); we have also Euro-Cent, no "Pfennige" anymore, Ploms. ;-)

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Hi,

Your calculation is almost right.
If you want it more precise you have to look into the "I/O Pin source charts"
Here you can see that @40mA the output voltage is about 4V. So for DC calculations you have to calculate:
R(ELim) = 4V/40mA = 100R
This is for DC, but the pulse is very short (capacitive load; don´t forget the miller capacitance of a FET)
therfore you may lower your external resistor even more. The datasheet says nothing about it.
In my opinion you should not draw more than twice the max. DC current. And calculate the dissipated power
in the AVR´s output stages. Don´t draw more than in the DC specs. (40mA x (5V-4V) = 40mW per Pin.

Dissipated power rises with switching frequency and depend on voltage waveform at the Pin. (use an oscilloscope)
But with 200Hz you will not get into trouble.

The benifit of the resistor is
* improved EMC because of smaller dU/dt
* better analog performance (AComp, ADC) of the AVR because of lower ground bounce

Klaus
********************************
Look at: www.megausb.de (German)
********************************

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125 Ohm is OK. A quick BPF learned that, even when you drive the Fet continuous with 220 Hz, the dissipation in it, due to the switch-on - switch-off delay is just some 10's of mW.

Nard

Edit: Klaus was FAST :!: :)

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BPF?

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Ruwe schatting :)

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Oooh... an educated guestimate :)

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Kinda off topic but it is related to small PCBs. I have seen capacitors (and I suppose resistors could be done too) installed INSIDE via holes connecting from topside to bottom side. You have to use an unplated via but I have seen it done. It was actually done for RF reasons more than board space but it's an option to use if you have too!

Go electric!
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Hello again,

I got another idea to save the series resistors to the gate, but probably not for my application (only for low frequency switching applications).
What about, if I do not switch directly from high to low and vice versa?
What, if I always use an intermediate state, where the "output" is configured as input with activated pull-up resistor?

The external pull-down resistor must have approximately the same size like the internal resistor.

0V => 5V transition
becomes
high impedance state => pull-up state

I have an RC network, the gate would be slowly charged to the half of the voltage (Rint = Rext).
After a certain time I switch to 5V output or leave it only with pull-up activated at nearly 2.5V, if the MOSFET is already conducting.

5V => 0V transition
becomes
pull-up state => high impedance state

The gate would be slowly discharged by the external pull-down resistor.

But, due to the time constant tau = R * C = 30kOhm * 1nF = 0.03s it would probably take too long, if I want to reach frequencies of 220Hz.

But maybe for slow switching procedures this would be a possibility to safe the gate resistors...

Michael

P.S: SMDs in holes is a nice idea, but it should be machine production...

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Quote:
But, due to the time constant tau = R * C = 30kOhm * 1nF = 0.03s it would probably take too long, if I want to reach frequencies of 220Hz.
Yes, you would probably end up with Mosfet's smoking .... one more BPF showed that ...

A ;) to jayjay1974 .... http://www.usingenglish.com/reference/idioms/ballpark+figure.html

Nard

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Why is it so important to save that single resistor? :D

In mass production you'd better find a MOSFET that's cheaper and still have a proper gate resistor.

And the pull up resistor in the AVR varies with temperature and voltage.

At some companies you get fat bonuses for finding cheaper components :shock: Recently I heard someone at got a few K $ because he - by accident - found a manufacturer that sells a particular connector for 2 cents less than the manufacturer they were planning to use :D

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The gate series resistor is there for more than just current limiting.
A real FET has inductance in its pins, there's a capacitance between drain and gate, there's inductance in the gate, and if you tie the gate to something low impedance, you have an oscillator. The frequency depends on PCB layout and FET type but is usually between 50 and 500MHz. The worst I've seen was one that emitted 5V RMS at 450MHz for 250 ns when switched, which is enough to make any device fail EMC testing, or in this case commit attempted murder on the spectrum analyzer input.

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Hello KKP,

thank you very much for the information about EMC.
I managed it, to squeeze some more SMD resistors on the PCB in the 0402 package size (now I reached the limit for the minimum trace width / distance).
So, if I understand it correctly, with the base resistor I can not only reduce the current in the gate path but also reduce a fast current change in the source drain path?
This is also important for me, because I found out, that there is a high current peak in the free wheeling diode, when I open an inductive load.
Maybe I can lower this peak current by increasing the gate resistor.
OK, the MOSFET will probably heat up, if I increase the gate resistor too much, due to a longer rise and fall time (soft switching mode with an exponentiol curve).

Any further ideas (beside frequency) what to think about, when choosing appropriate values for the resistors?

Gate series resistor: R1 = 100Ohm
Gate Pull-Down resistor: R2 = 10kOhm

Regards,
Michael

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The high current peak is logical ;) When you suddenly interrupt the current in a coil it induces a voltage (depends on how quickly you stop the current). If the energy can't go anywhere you can easily reach hunderds of volts. So you really need that diode ;)

Edit

Even long wires have enough inductance to induce several hunderd volts. A simple wire of 2 meters in length already has 0.4uH of inductance. Induces about 48V on the test board I'm working on this very moment. My FET under test has a built in clamping diode, otherwise the peak would be way higher :) Together with stray capacitance this also gives a ringing effect.

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Hello JayJay,

yes, this is clear; the diode was in my circuit from the very beginning.

What I intend to do:
Opening the MOSFET-'switch' not too fast (by the help of the gate series resistor and the gate capacity), so that the peak current is not so high, kind of slowly commutation from the MOSFET to the external Diode.
(I did not shoot it, but in the datasheet there's a maximum allowed peak value, though the effective current in the diode is not reached.)

Regards,
Michael

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Is the current spike outside of your diodes specs? If it isn't then there is no need to worry, it's just doing what it's supposed to.

Edward

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Yes, sometimes the peak current is higher than allowed by the manufacturer, provided that I undestood the datasheet correctly (very strict).

I'm using a BAT54C with the following absolute maximum ratings:

Quote:

Average Rectified Forward Current: If(AV)=200mA
Repetitive Forward Current: Ifrm=300mA
Non-repetitive Forward Surge Current (pulse width=1s):
Ifsc = 600mA

In my application I measured short peaks (maybe 1ms or less) with 2A or 3A.

But a quite interesting thing:
Another manufacturer writes as unit 'mAdc' instead of 'mA'.
What does this mean?
=> Is it allright to put once 1.2A with a pulse width of 0.5s on it?
Or once 2.4A with a pulse width of 0.25s?
Or once 4.8A with a pulse width of 0.125s?
(If I'm going on I've got a Dirac pulse with a infintely high current, which is infinitely short, but which has a finitely energy W=Uf*If*t)

Michael

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skotti: You can but, only to a degree; 100 usec maximum is my WAG.

For a large inductive load you need some sort of clamp that can handle the forward current. If you drive 5A in an inductive load, you'll get the stored energy back, and it starts at 5A.

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How about a RS1AB-13? 1A rms, 30A peak. 20cents from digikey.

Edward

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Hallo Edward,

yes, the http://pdfdata.datasheetsite.com...
looks fine, but it is too big for my application (you know, I had even problems to place some additional resistors on it...)

At KKP:
yes, at first sight this seems logical.
But what happens if the diode is too slow and there are other effects in the inductivity (coil with stray capacitance)?

Theoretically I can only have around 400mA in the diode.
When I look on the scope I can measure a peak current of up to Ipeak = 1.2V / 0.4Ohm = 3A (only for a few nano-seconds).

The effective current is rather low ...

Michael

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How do you measure the current ?

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Since he says he gets his 3A figure from 1.2V / 0.4Ohm I assume it's from the voltage drop across a current sense resistor.

Edward

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Yes, exactly: I used a resistor with a low resistance in the current path, a so called shunt resistor. In my case it was a very simple one, with wires.
But for more information:
http://www.rc-electronics-usa.co...

Michael

P.S: You have to take care with an oszilloscope where to put its GND/PE clamp to have no short circuit over a the PE line, when the circuit is also connected to earth, e.g. via a PC port, where normally GND=PE is.

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I think it is very difficult to make reliable current
measurements in the ns timescale.

Suppose your current rises from 0 to 3A linearly
in 3ns. So its a dI/dT of 1A/ns. If you have
a series inductance of L=1nH in within your shunt,
this will give additional voltage drop of
U=L * dI/dt = 1 Volt

I also found wiring of shunts and common-mode rejection
in the wires from shunt to scope often a problem.

Also crosstalk of some magnetic fields into some
measurement-loop may cause additional problems.

If you try to relate phases of voltage and current
also dont forget to account for different length of
cables. If your voltage-probe has a cable 1m longer
than the current-probe, that accounts for additional 3ns. (or more).

The damned thing is, that I have no simple solutions
to this problems. I only can say: Try to validate
and cross-check any measurements before drawing
the wrong conclusions.

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Hi scotti,

I took a peek at the data sheet and conclude the following:

Resistance of the diode pair @ 3A is approximately:

Delta V/Delta I = (800mV - 500mV)/(.1 A - .03 A) = 4ohms at the highest values given in the datasheet.

Power dissipation = I^2 * R = 9 * 4 = 36 Watts

Thermal Resistance = 500 K/W = 500 C/W, so your junction temperature would be 18000 degrees C (your part would have been a puddle long ago!).

Now, that's assuming steady state dissipation. If the duration of the dissipations is much less than the thermal time constant of the device, you can de-rate the dissipation via the following expression...

Pd pulsed = Pd CW * (thermal time constant/pulsewidth)

Assuming a thermal time constant of 10ms and a pulsewidth of 3ns, along with a max CW dissipation of 230mW from the data sheet,

Pd pulsed = .230 W * (10e-3 sec/3e-9 sec) = 766kW

So, it would seem that you wouldn't have a problem with the 3A as long as the thermal time constant of this part isn't faster than 3ns (of course heat-sinking comes into the picture and all that other wonderful stuff.)

I'd apply a couple different voltages and read the current you get from the diode, let it run for a week at full load (longer the better), and then check to make sure nothing has changed. Of course, check that my values are accurate and replace your own where necessary, but this should give you a good idea on the real limit of diodes in fast transient enviornments.

-Bryce

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to Ossi:
thank you for the hint about the shunt resistor's inductance. I did not thought about it. So the 'real' peak current through the diode may be less..

to Bryce:
thank you very much for the rough calculation, though I'm not too familiar with this thermal formulas.
So, as I understand you correctly, you calculated with Pd pulsed the maximum power dissipation for the case, that the 3ns-pulses occur in 10ms intervals. Or is the 'thermal time constant' a fixed value of the part or package?!
Since we calculated a power dissipation of 36Watt, which is less than 766kW I am probably on the safe side.
What is CW in this relation?

If I also remember correctly, the semiconductors get broken by
a) overvoltage
b) too much heat
and not by a short overcurrent (only by too much heat due to a high current).
So, if I want to get sure, maybe it is also a good idea to measure the temperature of the part, if it is in action for a long period.

Bye
Michael

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The current shouldn't get higher than the maximum pulse current specified for the device, ever :)

I believe you get micro hotspots when you overload the device, even for short times, which can lead to small cracks and device failure in the long run.

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For the Original Poster: Include the Gate resistors, one small from driver to gate, to limit current, switching speed & to absorb the D-G capacitance on turn-off. One larger resistor from Gate to Source for protection during startup etc.

@KKP
"For a large inductive load you need some sort of clamp that can handle the forward current. If you drive 5A in an inductive load, you'll get the stored energy back, and it starts at 5A."

Yes, agreed, but I think its the stored energy thats important ... if you drive 12V @ 5A into an inductive load (2.4Ohm) and switch this load off, the current will still want to flow, BUT, if you clamp the flyback into a 12V clamp of 2.4 Ohm, then I agree, the current will be about 5A, but if you clamp this energy into a diode, lets say 0.6V (0.8V+ in reality) then the current will be 12/0.6 or 20 times the current flowing in the load! 40* if 5A driven from a 24V source.

The stored energy must be absorbed somewhere.

Dont underestimate the currents involved in the clamp diode.

Also dont worry about the speed of the clamp diode too much unless this is a PWM output. The diode will always turn ON (or act as a capacitive Short Circuit) fast enough. The problem with diodes is the time it takes them to turn OFF, from conducting to reversed biased.

Ron.

 

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No, the current will start at 5A and linearly decay down to zero current. What voltage is developed depends on what voltage it's clamped too. The amount of energy is 0.5IL^2.

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rberger wrote:

Quote:
The stored energy must be absorbed somewhere.
Indeed, ..... in TIME, not in current. If 5A was flowing through the load right before turning the FET off, that 5A will keep on flowing through the clamp-diode, exponentionally going down, until the energy is gone (i.e. been converted to heat in the clamp-diode and the resistive "part" of the load)
It is for this reason that in stepper-motor-controllers the clamp-mechanism is implemented in a different way: the load is allowed to raise the voltage, but to a safe level only. The advantage is that the current will rapidly go to zero, and therefor the stepper is ready for the next step. :)

Nard

Edit: Jayjay posted while I was thinking about it, and typing. I extended my explanation a bit more.

Edit2: the current is exponentially decreasing, not linear.

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Last Edited: Mon. Jun 25, 2007 - 08:04 PM
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jayjay1974,

Consider a simple buck converter: (assuming no losses)
25Vin, 5Vout @ 5A.

Input and output power=25W

Switch current about 1A
Diode current about 5A

neglecting on/off times and all the other necessities, same for a flyback clamp.

Ron.

 

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Hmmm, ..... thinking real hard now .... :)

Nard

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Quote:
No, the current will start at 5A and linearly decay down to zero current. What voltage is developed depends on what voltage it's clamped too. The amount of energy is 0.5IL^2.

I agree that "The amount of energy is 0.5IL^2" but P=VI and from that
if the power in the load is at a voltage greater thad the clamp voltage
then the current in the clamp device is defined by the following:

1. As you stated: 0.5IL^2
2. P=VI

Lets assume the load is 12V @ 5A .... 60W

3. From 2, I=P/V ....> 60/0.6=100A ... as I said before, 12V/0.6V is 20* load current.

Ron.

 

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It's dinner-time at this side of the globe .... so I have other things to do, and you Ron: it must be 2.00 AM over there ....

I think we should be aware of the diiference between an inductor in a buck-converter and an inductive load.
In the buck-converter the coil is not driven into saturation, where an inductive load is.

And now: dinner

Nard

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First I would like to know, what load is driven.
Is it a relay, so that waiting a short time after
turn on, a steady DC current flows, that can simply
be computed by applied voltage/DC_coil resistance ?

If this is the case, we know the current trough the
coil before switch-off.

Then I assume that we switch off for a long time, so that the coil completely de-energizes before next
turn-on.

Now I try to sketch the turn-off of the transistor.
The event is triggered by lowering the gate-voltage.
So gate-capacitor discharges. At the threshold-voltage
the MOSFET begins to loose conductivity. The coil
wants to keep its current. The only escape is, that
the voltage at the drain of the transistor rises.

(This rise of the drain-voltage couples back to the
gate (miller effect) slowing down the voltage-rise
on the gate-capacitance for some nano-seconds.
we neglect this). Suppose we are able to switch
off the transistor very fast.

Then the voltage-rise at the drain (caused by the inductor-current) is limited by parasitic capacitances
(inside the coil, inside the diode, inside the
MOSFET).

After a short time the voltage at the drain becomes higher than Vsupply (here 12 Volts). If it is high enough, the diode starts to conduct. The diode takes
over the current. Now we have the conducting diode
accross the coil. So the coil (neglecting its DC-resistance) sees the diode-voltage drop. This
(low voltage) now discharges the coil.

During all that times the inductor-current had no
chance to get bigger. It only could get smaller.
So (as KKP said) the diode will start with the load current (approximately).

So the maximum diode-current (transient) is
given by the load current. The coil-energy is
dumped nearly totally into the diode. This happens
at a voltage=diode-voltage-drop. For power-diodes
there is often a diagram in the datasheet how much energy you can dump at which frequency.

This neglects stray-inductances. So the diode
must be laced near the transistor to protect
the transistor (drain-source voltage).

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rberger: I assume that you are either very green, or trolling.

As to magically getting 20A: The current in the inductor is proportional to the magnetic flux in the inductor, and the change in flux is proportional to the voltage. The energy stored in the inductor is stored in the flux. The math goes:

v=d(L*i)/d(t)
or
i=integral(v/L)dt.

This tells us that if the voltage across the inductor is 0 (the series resistance is a separate component to the inductive part), then the current doesn't change. Also, to change the current quickly, the d(i)/d(t) term becomes large, and thus the voltage really large.

Show us your perpetual motion machine. Changing the current instantaneously from 5A to 20A would mean that you instantaneously increase the energy stored in the inductor by 16 times _WITHOUT_ putting energy into it. That's a rather astonishing achievement. You just disproved thermodynamics and solved the world's energy problems! $DEITY!

Plons: he's trying to be funny with the ideal switcher. His numbers are also wrong.
If the average current in the output is 5A at 20% duty cycle (5/25), the average current is split 20% in the switch(1A) and 80% in the diode(4A). Since to get an average current of 1A in a switch that is on 20% of the time requires 5A when it is on, and to get 4A average in a diode that is conducting only 80% of the time requires 5A, the sun doesn't implode.

/Kasper

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Kasper, I don't think Ron (rberger) is trolling or green.
Confused a bit maybe .... ;)

The good thing is: this thread made me think again, about inductors and their behaviour.

Ron, I found this in Wiki: http://en.wikipedia.org/wiki/Buck_converter
Hope it helps.

Cheers

Nard

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Inductors and magnetism are quite complex, mostly because of the number of different units for the same property :)

An inductor is basically the same as a capacitor but with U and I reversed.

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Duality in theory, but problems in practice:

We have many good instruments for voltage, and it is
easy to attach them to a circuit. To measure current,
we have to "break" loops and use shunts, or good
clamps. (Do you own a DC to 50MHz current-clamp ?)

Most people (including me !) better think in
voltages. We dont use currents so often in arguments.
(Anytime thought about the question, where the current out of an OpAmp flows back ?)

Its easy to carry a charged capacitor around in a lab
for some minutes. Its hard to carry a charged inductor
with you. (Not everybody can use superconductivity).

Back to the thread:
But if it comes to power-electronics or
electromagnetic compatitibilty (EMC) we often
need to take currents into consideration.

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We are getting off topic but I must clarify what I am trying to say.

Quote:
KKP - rberger: I assume that you are either very green, or trolling.

Neither!
Quote:
Nard (Plons) - Kasper, I don't think Ron (rberger) is trolling or green.
Confused a bit maybe ....

Not confused, just trying to illustrate (badly it seems) that inductor/inductive load charge/dishcharge currents can be different.

Same for a Capacitor.

Quote:
KKP - Show us your perpetual motion machine. Changing the current instantaneously from 5A to 20A would mean that you instantaneously increase the energy stored in the inductor by 16 times _WITHOUT_ putting energy into it. That's a rather astonishing achievement. You just disproved thermodynamics and solved the world's energy problems! $DEITY!

Plons: he's trying to be funny with the ideal switcher. His numbers are also wrong.
If the average current in the output is 5A at 20% duty cycle (5/25), the average current is split 20% in the switch(1A) and 80% in the diode(4A). Since to get an average current of 1A in a switch that is on 20% of the time requires 5A when it is on, and to get 4A average in a diode that is conducting only 80% of the time requires 5A, the sun doesn't implode.

/Kasper

Haven't solved anything here, but if you do let me know as my electricity bill is too high.

So perhaps the maths was wrong in my buck example, but Kasper, you assumed output time of 20% in the diode. Its more like 5A average in the diode continuously, with a delta I of 2A. 4-6A swing. (20%)

But what I was trying to say is that the currents depend on the load.

Any SMPS with a voltage ratio of 20:1 will see a potential current increase in the load of that amount, less efficiencies.

What I am trying to say, is this:

If an inductive load is driven by a voltage, lets say 12V, and is clamped into the same voltage, lets say 12V, the on and off state currents are the same except for polarity.

If you clamp this energy into a lower voltage clamp, the current will be greater.

Consider a DC motor as an inductive load, Ratings: 40W output(not input) 24VDC in, 2.1A in, 0.6Ohm DC resistance. It draws 450mA unloaded. Clamp diode current is HEAPS!!!!! more than 450mA if you were to give it a short on pulse.

On the other hand, For a simple relay the current is limited by its coil resistance.

But, Why do you think that the flyback current would be the same if the inductive load / relay was terminated into a diode (0.6V) or into a diode/zener clamp equal to the source voltage of 12V?

Ron.

 

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No, the current does NOT get greater, the current simply decays slower when the clamp voltage is lower...

edit:

I attached a few screen shots of a simulation proving my point ;)

Attachment(s): 

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Quote:

If you clamp this energy into a lower voltage clamp, the current will be greater.

This is not true.
jayjay is right.

With a lower clamp voltage the discharge time will be longer to get the same ammount of energy.

WQith a 25W switch mode supply without loss the average input current at 25V will be 1A.
But it´s low pass filtered by the current. The current in the swith is 5A but with 20% duty cycle.

At the coil and the output you have 5A at 5V with 100% duty cycle. in the diode you have 5A 80% duty cycle.

For sure there will be ripple at the coil current. This depends on duty cycle, frequency and inductance.
And the coil current is allways in the same direction, but not the voltage.
If you measure V(coil-in) - V(coil-out), then you will have a positive voltage at charging the coil (switch is ON)
and a negative voltage during discharge (switch is off)

And when you calculate the current, then don´t use RMS, but average current.
RMS is good when U and I depend on each other, like a resistive, capacitive or inductive load.
But with a SMPS you assume that V(In) and V(out) to be constant.

Klaus
********************************
Look at: www.megausb.de (German)
********************************

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Thank you very much for the help and the further information, especially for the simulations which clarify the theory. Sometimes in the practical experience you see a lot of side effects, so that it is not always clear what the 'basic' or theoretical behaviour is.

Ossi: good explanation, what happens between coil and diode!

Maybe the misunderstanding of the increased current is because of the DC motor:
If a rotating DC motor is switched off (e.g. by opening a conducting MOSFET driver or a relay) it also has an additional rotational energy. If I only have a clamp diode in reversed direction over the load instead of an additional switch with a brake resistor, the motor will act as a generator while decreasing its speed.
This can be a huge amount of energy which must be transformed in another form (naturally warmth).

W = 0.5 J (2 pi f)² (J = moment of inertia)
By the way:
W = 0.5 L I²
(current is quadratical not the inductance)

A 'simple' inductive load like a relay, a loudspeaker (with low oscillating weight), a single coil of a stepping-motor etc. will not have such a behaviour. In my case I have the coil resistance still in the 'current circle', so the current through the diode will be limited.

Jayjay, I did not know about the hotspots. So I hope, that if I am beneath the peak limit of 600mA this should be OK, though I am not using a single 600mA-pulse of 1sec, but very short pulses (about 400mA) but repetitively.

I included the two gate resistors into the layout (everything is now close-fitting), I hope the pick and place machine will manage it.

Regards,
Michael

P.S.: It's interesting, how much I / you can learn, when trying to answer a simple question...

In the beginning was the Word, and the Word was with God, and the Word was God.