Understanding Schottky Diode

[koala]

Functions in a boost circuit.

So I am looking to buy some Schottky Diode to build a boost circuit. I am not sure about some specifications of the Schottky Diode.

I know that the lower the forward voltage is better. The higher the forward current(cont & peak) is better for high output. My choice of Schottky Diode is the Zetex ZHCS2000. Note this is from the datasheet, I have not tested lower current.

It has a forward voltage of
290-325mV at 500mA current
500mV at 2A current

So my aim is to look for something with lower forward voltage. There are a lot of very low forward voltage schottky but they can only sustain ~200mA current.

My idea is to parallel the super low forward voltage diodes to increase current. Any disadvantage to the efficiency to the boost circuit? Secondly, what does the Vr, reverse voltage means, in terms of a diode in a boost circuit. Thanks for reading.

Vince.

 

[sysadmn]

You can put diodes in parallel to increase current capacity, and in series to increase voltage capacity. In extreme situations (really high power or really small signal) you might go to the trouble to match diodes (or buy all diodes from the same manu. batch) so their characteristics are the same. For the most part, you don't need to worry about that.

Search for "parallel connected diodes" here: http://www.powerdesigners.com/InfoWeb/design_center/articles/Diodes/diodes.shtm

Reverse voltage is the highest voltage a diode will still reliably block current flow.

Reverse Voltage

When a reverse voltage is applied a perfect diode does not conduct, but all real diodes leak a very tiny current of a few µA or less. This can be ignored in most circuits because it will be very much smaller than the current flowing in the forward direction. However, all diodes have a maximum reverse voltage (usually 50V or more) and if this is exceeded the diode will fail and pass a large current in the reverse direction, this is called breakdown.

 

[koala]

Thanks for the link, will take sometime to digest. So it's ok as long as the reverse voltage is more than the voltage flowing across the LED? What about the peak voltage surge from the inductor?

 

[MrAl]

Hi there,

You do have to be a bit careful. If say you want to pass 4 amps of current
using two 2 amp diodes it probably wont work because you cant get the
two currents to be exactly half for every temperature. Using say three
diodes would probably work ok however.

Also, the reverse voltage rating is the max the diode should ever see,
in that there is a good chance it will be destroyed if a voltage higher
than the rating is applied even for a short period of time. In other words,
the diode does not recover from a reverse overvoltage. To find out
what the reverse will be for your diode, look at the different states
of the circuit (transistor on, transistor off) and calculate the worst case.
I think you already understand this however, but it does depend on the
extact topology of your particular boost circuit--some are different than
others.

Also, the reverse resistance is much lower than a 'normal' signal diode.
A normal diode could be 500 Megohms in reverse, where a Schottky might
read only 20k ohms. Putting them in parallel makes this worse, because two
in parallel might only be 10k. In signal conditioning circuits this could really
be a problem, but in power supply circuits this is not usually a problem
because 10k is still quite high. There is a tiny bit of loss of efficiency, but
the Schottkys lower forward voltage more than makes up for it.

 

[ScottB]

Hmph.

The lowest Schottky barrier on Earth is found in an unexpected place, the lowly Zener diode.

The Zener is operated in the reverse bias regime and breaks down in avalanche beyond a specified applied potiential.

But, if you forward bias a Zener you will find that it turns on at about 50mV.

I discovered this when searching for a maximally efficient diode to detect AM. Standard diodes turn on about 700mV, specialized "low-Schottky" devices turn on in the 350-500 mV region. A forward biased Zener turns on about 50 mV. I was able to build a passive AM tuner that could drive an 8-ohm speaker rather loudly with no external power or amplification.

There, now you know more about Schottky barriers than 99.99% of mankind. :sick2:

Scott

 

[jsr]

With a zener, you will lose reverse protection/blocking and the schottky in a boost circuit is used to block any reverse current, so it will not be useful for this purpose. There are various schottky types with different barrier metals. There is always a compromise between VF, VR, IR, and runaway. The lowest VF schottkys have the highest chance of thermal runaway.

 

[koala]

:bow::bow: Thanks for all your knowledge. I still haven't found a better spec schottky diode for parallel operations. The problem with spec sheet is that the manufacturers like to advertise the minimum forward voltage of the diode @ low current, then bury the ugly side of it somewhere in the datasheet. I have learned that going through devices takes some time.

I am thinking of building a variable boost output current of 5mA-200mA max from a 1AA NiMH battery. So max output power is about 0.66watts assuming 3.3v at the LED. Not exactly super high current, but I am aiming for longer running and high efficiency circuit. Cost is not a problem as it isn't mass production but it has to be small.

 

[MrAl]

Hi again,

koala, way back in 2002 i think when we did extensive testing on the
first Zetex circuit ever presented on CPF, the conclusion was that the
2 amp Zetex Schottky diode was the best. It's small and has a very
very low forward voltage as measured on a scope during actual circuit
operation. The forward current was about 1 amp though, not near 2 amps.
If i remember right, the forward voltage measured under this condition
was between 0.2 and 0.3 volts, which is low even for a Schottky.
Hopefully they still make them the same these days.

 

[koala]

It is very good indeed that is why I am having a hard time looking for a better one. I don't remember which was the one that was tested back then but I think the one mention in the ZXSC300 datasheet is the 1Amp ZHCS1000. The ZHCS2000 is rated 2Amp.

So you guys did some test and found out it was actually 1Amp I think it's probably the ZHCS2000. Otherwise the test would just match the specs and there's nothing to post. If the forward voltage is 0.2v to 0.3v then I think I will settle with it. Many Schottky diodes that I am looking at will vF at 0.2 to 0.3v but that's like moving 1mA of current.

 

[MrAl]

It is very good indeed that is why I am having a hard time looking for a better one. I don't remember which was the one that was tested back then but I think the one mention in the ZXSC300 datasheet is the 1Amp ZHCS1000. The ZHCS2000 is rated 2Amp.

So you guys did some test and found out it was actually 1Amp I think it's probably the ZHCS2000. Otherwise the test would just match the specs and there's nothing to post. If the forward voltage is 0.2v to 0.3v then I think I will settle with it. Many Schottky diodes that I am looking at will vF at 0.2 to 0.3v but that's like moving 1mA of current.

Hi again koala,

I am not sure if i made the point about the ZHCS2000 clear so i will do
that now...

This diode is rated at 2 amps, but in the circuit it was driven to about
1 amp peak (maybe a little more). During that time it showed a forward
voltage of about 0.2 to 0.3 volts on the scope. Thus, the diode was
driven to only half it's actual rating.

 

[koala]

Hi again,

I've studied datasheet and done some research for the pass few days I had another idea.

Take this for an example... target output current to the LED 100mA. So we source a Schottky that could handle a forward current of 300mA continuous. It has a forward voltage of 200mV@50mA and 400mV@100mA.

If two of these Schottky(voltage matched) are in parallel config, each of them will be doing 50mA each, right? So the forward voltage will drop to 200mV at the same time?

So if there's a Schottky that forward 100mV@50mA, gang up 2 of them we'll get a 50mV@50mA Schottky? Terrific or I am dreaming? Is there a minimum forward voltage on a Schottky? I measured the Zetex ZHCS2000 Schottky with my DMM diode mode, it showed 0.091V, not sure how much current my DMM push but the forward voltage sure looks low to me.

If you guys agree with my theory, I will do 5x parallel Zetex ZHCS2000 schottky to a test Zetex 300 circuit. I wonder how much more power and efficiency I can extract from it. It looks like it's doing between 200mV to 300mV at 500mA and the higher the operating temperature the better. I guess I can shave at most 150mV.

Regarding the reverse resistance, is it important in the Zetex 300 circuit? I measured and it is about 12k ohms, in 5x parallel config it will be about 2.4k ohms, some current will leak?

 

[MrAl]

Hi again,

Well, a diodes voltage current relationship isnt normally that linear,
in that it follows a more exponential curve. For your example
of 400mv at 100ma, a diode like that will more likely have 300mv
drop at 50ma, not half (200mv).

If you parallel 5 of them the resistance of 2.4k probably wont matter
with a 1 watt LED, but it probably a good idea to also calculate the
total capacitance and do a rough simulation to find out the effect
of the extra capacitance, which will be 5 times higher than a single diode.
Capacitance like this can eat up efficiency, but it's hard to say until
you point out a diode and do a simulation. If you find a diode perhaps
i can do the simulation for you if you prefer, but i would like it if you
would look up the specs on the diode first and tell me what the
capacitance is for a single diode.

To statically test a single diode, simply run a current through it and measure
the voltage drop, but do this for several currents like 10ma, 20ma, 40ma, and
80ma (it's convenient when testing diodes to step the current by doubling
the previous value because of the way the voltage changes with current).
To statically test 5 diodes in parallel, run the same test using the same
current levels, then compare with the previous test to see how much
voltage drop is being saved.

If i get a chance today i'll do a simulation with a common Schottky
i have a model for and report back in this thread what the results were.
This device is rated for 1 amp, so i'll do higher currents, but the results
will still be interesting, and this makes it easy to try paralleling 2 through
10 devices just to see what happens.

 

[koala]

Hi MrAl,

Thanks again for your guidance. It was a scenario I know it isn't linear by looking at the datasheet. The most suitable schottky diode I have on hand is the ZHCS2000, it has a capacitance of 50pF at 1Mhz. The reverse current is between 160uA and 300uA. If that helps.

I have no idea how capacitance in the schottky can have effect on efficiency in a boost circuit. Can you please enlighten me? :help:

I will do the 10ma, 20ma, 40ma, and 80ma voltage drop test when I get home from work tomorrow. It's getting late here got to :sleepy:.

 

[MrAl]

Hi again koala,

Oh ok great. 50pf doesnt sound too bad really, but i'll have to do a check
to see what happens when you parallel five such units. Roughly, we get
250pf with five in parallel.

The capacitance of the diode affects the efficiency of a boost circuit by
effectively shorting out the diode during the first instants when the
voltage across the diode reverses at the time the inductor charges starts
to charge up. The transistor has to dump all of the caps energy before
the diode starts to look like an open circuit (reverse bias). Because the
diode is in series with the output cap, the voltage across the transistor
just before turn on is equal to the LED voltage plus the diode voltage,
which can be around 4v, and so the transistor has to discharge the cap
by switching a 4v low impedance source to ground, which eats up energy.
The amount of energy wasted is dependant on the value of the internal
cap. When the transistor again opens up (switches off) the current
through the inductor has to charge this cap back up again, which takes
a little more energy. Because the transistor is turning on and off very
quickly, this takes place as fast as the switching frequency of the circuit,
and the higher the frequency the more that cap has to charge and discharge,
so as the frequency increases the efficiency will get lower and lower.
Since 50pf isnt that high, we might not see too much of a problem
even when connecting five in parallel, but it's worth a quick check.
The Zetex circuit runs around 300kHz i think, and that's not too high either.

With 300ua leakage and a nominal LED drive current of 100ma the
efficiency lost for five diodes in parallel would be about 1.5 percent,
which isnt too bad.
The efficiency gained driving a 3.3v LED after obtaining a 0.1v drop
across the diode would be about 3 percent, so the net overall effect
of connecting five in parallel would be a gain of 1.5 percent, not
counting the capacitance effects because i havent done that test yet.



Here's the data i obtained by a static simulation of a 1N5817 Schottky diode:

1 amp (rated current)
Single: 439.4mv
Two in parallel: 364.3mv
Four in parallel: 315.6mv

1/2 amp (half rated current)
One: 364.3mv
Two: 315.6mv
Four: 280.2mv

Next i'll do another diode, the 1 amp Zetex.
Unfortunately i dont have a model for the 2 amp Zetex diode.
Maybe they have one available now on the web site.

 

[MrAl]

Hello again,

Ok the test with the extra capacitance varied according to what
diode was used.

Using a diode with 2pf internal C and connecting four in parallel
the efficiency rose by 4 percent.
Using a diode with 250pf internal C and connecting four in parallel
the efficiency rose by 2 percent.

The tests were done using a square wave at 300kHz.

For a diode with 50pf internal C, the result would be somewhere between
2 and 4 percent. This means the decrease in voltage drop more than
makes up for the extra capacitance.

Apparently the 50pf diodes will be ok for this kind of connection,
although an eff gain of around 3 percent isnt that much really.

 

[koala]

Hi MrAl,

Many thanks for running the simulation and sharing the results. I appreciate it.

I had to read the 2nd paragraph in post #14 many times and finally understand.
Sorry I am not very good at this, I know how a boost and buck circuit works but
this is difficult to master. I came home from work late tonight, dead tired.

Anyways, it's all cool after a coffee and beaming the wall. So the capacitance
in the schottky acts as a bucket. We'll need to fill up the bucket before the
water can flow. And water has to be dumped and wasted to before the next flow.
All right so the lower the capacitance the better? Look for a diode with lower
capacitance or lower the switching frequency? In this case, the Zetex switching
frequency is determined by the Vout and Vin.

Assuming single AA cell 1.2v input and output of 3.4v, the onTime 3.116uS and
offTime 1.7uS(fixed). That's about 208kHz operating frequency. For 2xAA cell the
switching frequency is 320kHz. Your memory is all good MrAl

In the post #13, I mention the reverse current is between 160uA and 300uA. Unfortunately,
That's for 1 diode not 4. Which is a problem when parallel, worst case 300uA x5, that's
1.5mA? Is it included in the 1.5% efficiency gain in 3rd paragraph, post #14?

So to counter the internal capacitance problem, is to parallel less diode as possible,
otherwise the increase in capacitance will defeat the purpose of a lower forward voltage
diode. Use 2-3 parallel instead of 5 right? Suddenly, with all these variables, it
seems not so worthy to have parallel diodes in exchange for the 2-3% in efficiency.

I haven't got the chance to do the forward voltage test yet, will do it.
Getting late now, got to :sleepy:

 

[MrAl]

Hi MrAl,

Many thanks for running the simulation and sharing the results. I appreciate it.

I had to read the 2nd paragraph in post #14 many times and finally understand.
Sorry I am not very good at this, I know how a boost and buck circuit works but
this is difficult to master. I came home from work late tonight, dead tired.

It does take some time, but it's not impossible.

Anyways, it's all cool after a coffee and beaming the wall. So the capacitance
in the schottky acts as a bucket. We'll need to fill up the bucket before the
water can flow. And water has to be dumped and wasted to before the next flow.
All right so the lower the capacitance the better? Look for a diode with lower
capacitance or lower the switching frequency? In this case, the Zetex switching
frequency is determined by the Vout and Vin.

Yes lower capacitance would be better, but the Zetex 2000 diode looks ok with it's 50pf.
at the frequencies we intend to run.

Assuming single AA cell 1.2v input and output of 3.4v, the onTime 3.116uS and
offTime 1.7uS(fixed). That's about 208kHz operating frequency. For 2xAA cell the
switching frequency is 320kHz. Your memory is all good MrAl

Ha ha, i remembered something

In the post #13, I mention the reverse current is between 160uA and 300uA. Unfortunately,
That's for 1 diode not 4. Which is a problem when parallel, worst case 300uA x5, that's
1.5mA? Is it included in the 1.5% efficiency gain in 3rd paragraph, post #14?

Yes except it's a loss, not a gain. 1.5ma means 1.5 percent lost,
300ua means 0.3 percent lost, etc.
After reviewing this information however i realized that the transistor
wont be 'on' all the time, so the loss will be greater when the
transistor is on and less when the transistor is off, because the
voltage across the diode changes levels. With a 66 percent duty
cycle the loss would be around 1.1 percent rather than 1.5 percent,
but no need to calculate this accurately because this
is the loss caused only by the reverse leakage, and we are
assuming worst case. In an actual circuit we could end up seeing
only a 0.5 percent loss.
The simulations take this parameter into account, but again they
assume worst case, so perhaps we could see 3.5 percent overall.

So to counter the internal capacitance problem, is to parallel less diode as possible,
otherwise the increase in capacitance will defeat the purpose of a lower forward voltage
diode. Use 2-3 parallel instead of 5 right? Suddenly, with all these variables, it
seems not so worthy to have parallel diodes in exchange for the 2-3% in efficiency.

That's the long and the short of it, but it mainly depends on if
the decrease in forward voltage causes enough gain to *more* than
make up for the loss due to capacitance. In other words, one is
a gain and one is a loss, and the net sum tells us if this is
favorable or not to our circuit. With higher output voltages,
the decrease in drop will not make as much of a difference as
when working as low as 4v, so we might end up with a total net
*loss* if the output is higher. Lucky we are at 4v where we
see a little gain.

You lose some, you win some, and some get rained out.
In our case, we lose some, gain some, and the net gain is not as
large as we would like to see, but then again at least we got a gain
so we can say we accomplished something positive

I haven't got the chance to do the forward voltage test yet, will do it.
Getting late now, got to :sleepy:

Ok, have a good night.

Oh BTW, the diode models i used for the simulation to get an overall idea how
much gain or loss using four in parallel had all of the parameters that
would affect efficiency so yeah we can guess around 3 percent gain
with four in parallel. I ended up with four instead of five cause it was a
bit easier to manage, and four is probably enough anyway.

 

[koala]

Tonight I got home early and went straight for the bench.
Here's the result of a single Zetex ZHCS2000 diode, connected to a K vf bin Lux I. The circuit is fed with a constant current.
The ambient temperature is 18c/64.6F, I had to stop testing at 1600mA because the Lux I color shifted and I don't want it
to explode in front of my face. The vF is measured across the diode anode and cathode while powering the Lux I.
A Dale Power Strip 0.5% 0.1 ohm resistor used as a current sense resistor.

Please note: The ZHCS2000 forward voltage in the datasheet is result of pulsed current with a pulse width of 300uS, 2% duty cycle.
My below results are with constant current 8-10mV ripple. Hope you all will find something useful from the results.

Current(mA) Forward Voltage(mV) Diode Temp(C) Diode Temp(F)
0.1         59.5                18.6          65.48
0.2         74.6                18.5          65.3
0.4         91.2                18.3          64.94
0.8         109.2               18            64.6
1           114.1               18.5          65.3
2           131.3               18.7          65.66
4           148.6               19            66.2
8           166.4               19            66.2
10          172.7               18.9          66.02
20          192.5               18.5          65.3
40          209.5               18.8          65.84
80          227                 19.1          66.38
100         231.8               19.8          67.64
200         247                 22            71.6
400         263.5               26            78.8
800         287.6               32            89.6
1000        296.6               37.1          98.78
1200        307.4               41.1          105.98
1400        318.7               42.9          109.22
1600        335.5               52.1          125.78
1800        351                 61.7          143.06

 

[koala]

From the data below, the single diode have a forward voltage of 247mV at current of 200mA. Two diodes in parallel will cut the forward voltage to 231.8mV that's like 15.2mV only. Where as 4 diodes in parallel will drop it to about 21XmV. The ZHCS2000 doesn't seems to be a good candidate at this range. It should work a bit better in parallel for higher output.

Current(mA) Forward Voltage(mV) Diode Temp(C) Diode Temp(F)
0.1         59.5                18.6          65.48
0.2         74.6                18.5          65.3
0.4         91.2                18.3          64.94
0.8         109.2               18            64.6
1           114.1               18.5          65.3
2           131.3               18.7          65.66
4           148.6               19            66.2
8           166.4               19            66.2
10          172.7               18.9          66.02
20          192.5               18.5          65.3
40          209.5               18.8          65.84
80          227                 19.1          66.38
100         231.8               19.8          67.64
200         247                 22            71.6
400         263.5               26            78.8
800         287.6               32            89.6
1000        296.6               37.1          98.78
1200        307.4               41.1          105.98
1400        318.7               42.9          109.22
1600        335.5               52.1          125.78
1800        351                 61.7          143.06

 

[MrAl]

Hi again,

Thanks for the taking the data. I'll have to stick that in my notes because
real world data like that can be very valuable when you go to think about
how well your circuit might work.

The bottom line now is, comparing one diode to four in parallel seems to
provide an overall increase in efficiency of only 1.5 percent approximately.
Because the frequency is only around 200kHz to 300kHz the capacitance
doesnt seem to affect the efficiency enough to worry about for this
circuit.
This time i set up the circuit as an actual boost, with typical output cap
of 10uf and inductor of 22uH, but a good inductor with Rs=0.01 ohms.
The diode(s) i used was(were) the Zetex 1000 diodes (1 amp rating).
If you feel like checking the Zetex site for the 2000 diode see if they
have the spice model published yet and i'll do the circuit again with
that diode, but i dont think it will make that much difference now.

The Zetex 2000 diode should be ok for this circuit, but paralleling four
of them will probably only mean a gain of 1.5 percent or so. Not much
really.

One of the other facts that came out here is that the peak current is
what we need to consider, not the average output current. The peak
current is almost four times the average current (400ma) for 1.2v in
and 3.3v out. This means the diodes operate close to 400ma, not 100ma.
This also means paralleling 4 of them only brings the current down to 100ma
per diode. Since the voltage only goes down by 10 percent for this condition,
and the diode is (roughly) responsible for 10 percent of the efficiency,
taking 10 percent off of 10 percent leaves us with 9 percent, so we would
only increase the efficiency by roughly 1 percent. This agrees somewhat
with the simulation, which showed 1.5 percent with the Z1000 diode.