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PICList Thread
'[EE] Getting 15-ish volts from 5V'
2009\02\28@174410 by Rikard Bosnjakovic

picon face
Having been absent from electronics and this list for about a year or
so due to lack of ideas and apathy, my father-in-law asked me a week
ago for help with a project that finally brought me to the surface and
I found some brain cells that made me like to play with electronics
again.

To cut a long story short, for this project I have a relay that (with
the help of a PIC) trigs on any voltage between 16-30VDC. The system
will be powered using a 9 or 12V wallwart. Perhaps 5, if I manage to
find one in my junkbox (where I found the relay). What I need to do is
to increase the voltage up to the level where the relay triggers. It
will be triggered about 4-5 times each day, and 10 minutes each time

What I don't want to do is to use a high(er) voltage wallwart, because
I expect that the energy consumed in standby will be higher, and this
project will be plugged in 24 hours a day.

So basically, what I'm looking for is a way to raise the voltage. I've
googled around and found things about "step up-converters" and
"chargepumps". Some people tell me that a step-up is way overkill for
this tiny project. Myself, I have no idea except that I need to keep
the energy usage dead low. And the budget as well, that's why I cannot
buy a 5V-relay.

Can anyone point me in a direction of what kind of construction I want to use?

PS: I'm sorry if this post contains too little information. It was a
long time ago since I sat in my lab being creative and analytic.
Please bear with me, four years of unemployment can seriously kill
your brain. And your wallet. Hence the low, non-existing, budget.

Thanks.


--
- Rikard - http://bos.hack.org/cv/

2009\02\28@180541 by olin piclist

face picon face
Rikard Bosnjakovic wrote:
> To cut a long story short, for this project I have a relay that (with
> the help of a PIC) trigs on any voltage between 16-30VDC. The system
> will be powered using a 9 or 12V wallwart. Perhaps 5, if I manage to
> find one in my junkbox (where I found the relay). What I need to do is
> to increase the voltage up to the level where the relay triggers. It
> will be triggered about 4-5 times each day, and 10 minutes each time

What voltage do you need accross the relay when on, and what current does it
then draw?

> What I don't want to do is to use a high(er) voltage wallwart, because
> I expect that the energy consumed in standby will be higher, and this
> project will be plugged in 24 hours a day.

It will probably be easier and cheaper to make a buck regulator to run the
PIC from a higher voltage than to use a boost regulator to run the relay
from a lower voltage.  This is because the PIC will likely draw a lot less
power, so you can get by with a smaller and cheaper inductor.

> So basically, what I'm looking for is a way to raise the voltage. I've
> googled around and found things about "step up-converters" and
> "chargepumps".

You will likely need too big and expensive capacitors to make a charge pump
produce the current the relay needs.  You need a basic boost switcher.
Fortunately these are easy to make.  You may be able to use the PIC to drive
it directly, depending on what else the PIC has to do and which PIC it is.
I use 10F204 a lot for running simple switching power supplies.

> And the budget as well, that's why I cannot
> buy a 5V-relay.

But 5V relays are cheap and available.  By the time you build a boost
switcher, you've probably paid for most of the relay in $$ and a lot more in
hassle.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\02\28@180542 by Harold Hallikainen

face
flavicon
face
I think the cheapest solution would be a relay at the desired voltage.
Looking at more expensive approaches are boost and charge pump converters.
I've used a charge pump converter from TI to get 3.3V from a single AAA
cell. The advantage of a charge pump converter is that no inductor is
required. However, these tend to be relatively low power converters. A
standard boost converter has an inductor and diode between the voltage
source and the load, and a capacitor across the load. The converter has a
switch that grounds the junction of the inductor and the diode, causing
the inductor current to ramp up. The switch is then opened and the
inductor current ramps down dumping energy into the load. Several
companies make chips for this. I've mostly used stuff from Linear
Technology (http://www.linear.com).

Looking at power consumption, it is true that you'd have dissipation in a
regulator dropping from a higher voltage (enough to run the relay) down to
the lower logic voltage, but this may be minimal compared with losses in
the power supply itself, especially if it's a linear wall wart type
supply.

Good luck!

Harold


--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2009\02\28@181624 by Bob Blick

face
flavicon
face
Hi Rikard,

I think most wall-warts waste about the same amount of power, so a lower
voltage one may not use less. Especially if your application uses less
than a watt at idle, sine wall-warts use that just plugged in.

But if you do decide to use some step-up, don't use a charge pump. They
are only for tiny currents and you have a relay to drive. If you can
generate a high duty cycle pulsetrain from your pic, all you need to add
is a transistor, coil, diode and capacitor to step up your voltage.
Under load your voltage multiplication will be proportional to the duty
cycle, plus the input voltage. So if your duty cycle is 75%, that is 3:1
plus 1 so your output voltage will be 4 times the input. Generally.
Certainly close enough for a relay.

Use data sheets from power supply chips to help you choose components
and operating frequency, just don't use their chips :)

National Semiconductor has lots of design data available for their
somewhat dated and mediocre Simple Switcher chips you can use when
designing your own. Or just grab a coil out of an old dead compact
fluorescent lamp and experiment. Remember to alyats test under load
since you will run this without feedback or regulation.

Cheerful regards,

Bob

Rikard Bosnjakovic wrote:
{Quote hidden}

2009\02\28@182131 by Rikard Bosnjakovic

picon face
On Sun, Mar 1, 2009 at 00:05, Olin Lathrop <spam_OUTolin_piclistTakeThisOuTspamembedinc.com> wrote:

> What voltage do you need accross the relay when on, and what current does it
> then draw?

16V minimum. The relay triggers on anything between 16-30V (I haven't
got any supply that goes above 30V), so there's no need for precision.
18V +/- some volt is fine. I will use any inductors or components
that's in my junkbox, and that's definately a guarantee that precision
is no-go.

The relay consumes about 50mA at 20V.


> You will likely need too big and expensive capacitors to make a charge pump
> produce the current the relay needs.  You need a basic boost switcher.
> Fortunately these are easy to make.  You may be able to use the PIC to drive
> it directly, depending on what else the PIC has to do and which PIC it is.
> I use 10F204 a lot for running simple switching power supplies.

I haven't yet decided which PIC to use, but it will probably be an
8-pin 12F. The outline for the PIC is this:

1. sleep until an external component makes the PIC wake up
2. drive the relay for about 10 minutes
3. repeat

The last 12F-PICs I used had 5 I/O-lines. One will be for the external
wake up and one for the relay. Will that be enough to drive a booster?


--
- Rikard - http://bos.hack.org/cv/

2009\02\28@185641 by olin piclist

face picon face
Rikard Bosnjakovic wrote:
> 16V minimum. The relay triggers on anything between 16-30V (I haven't
> got any supply that goes above 30V), so there's no need for precision.
> 18V +/- some volt is fine.

Not really.  This is probably a "24V" relay if it first pulls in at 16V.

> I will use any inductors or components
> that's in my junkbox, and that's definately a guarantee that precision
> is no-go.

Precision can come from feedback.  As long as the inductor can supply the
power, the PIC can do the rest.

> The relay consumes about 50mA at 20V.

That's 1W.  You definitely want a boost switcher, not a charge pump.

But again, why not use a 24V wall wart and use a buck regulator to run the
PIC?  The PIC will require a lot less than 1W.  If the wall wart is a dumb
transformer or transformer followed by full wave bridge, then it's probably
dissipating 1W just plugged in anyway.  Even a 7805 from 24V running the PIC
should be low enough power compared to the wall wart idle so that it doesn't
matter anymore.

If you start with 24V you can probably make the inductor for the buck
switcher yourself by winding some enamel coated wire around a chunk of
ferrite from a junked AM radio, or even a few small nails insulated from
each other.

> I haven't yet decided which PIC to use, but it will probably be an
> 8-pin 12F. The outline for the PIC is this:
>
> 1. sleep until an external component makes the PIC wake up
> 2. drive the relay for about 10 minutes
> 3. repeat

That sounds like a job for a 10F204.  You can use the built in comparator
and absolute voltage reference to regulate either the relay current or
volage when on.  One pin for your wakeup signal, another to drive the boost
swither switching element, and another for the 600mV voltage or current
feedback.  So what are you going to do with the fourth I/O pin?


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\02\28@193336 by Jinx

face picon face
An option that may work for you, together with boosting the 5V, is

http://www.piclist.com/techref/io/relay/LVwHVcoil.htm

Take note of the comments

The relay is energised with a higher voltage pulse (energy supplied by a
cap) and held at a lower voltage

I've used this with 10V relays on 5V and they hold reasonably well. 12V
relays are sometimes not so well held. If V+ was 10V they'd probably be
OK, but you have to experiment with the particular relay to make sure
what the hold limit is, to your specifications

2009\02\28@220140 by Rolf

face picon face
Jinx wrote:
> An option that may work for you, together with boosting the 5V, is
>
> http://www.piclist.com/techref/io/relay/LVwHVcoil.htm
>
> Take note of the comments
>
> The relay is energised with a higher voltage pulse (energy supplied by a
> cap) and held at a lower voltage
>
> I've used this with 10V relays on 5V and they hold reasonably well. 12V
> relays are sometimes not so well held. If V+ was 10V they'd probably be
> OK, but you have to experiment with the particular relay to make sure
> what the hold limit is, to your specifications
>
>  
Just when I think I am getting a good grasp on simple things, someone
throws a curve-ball like that circuit.

I just don't understand how the circuit puts -5V at the base of the coil
when the transistor initially conducts.

When the input is deactivated the cap will have a 5V potential through
it giving +5V on the resistor side, and 0V on the coil side.

Activating the input causes the transistor to conduct, and the only
thing I see happening is that the cap will discharge to ground without
any change on the coil side? Does the inductor-like nature of the coil
play a part?

Even if the 100R resistor was a diode, I just don't 'get it'.

Obviously there is something I am missing... anyone care to enlighten
me. Is there a reference in H&H's Art of electronics anyone can point me to?

Thanks

Rolf

2009\02\28@230636 by peter green

flavicon
face

> Activating the input causes the transistor to conduct, and the only
> thing I see happening is that the cap will discharge to ground without
> any change on the coil side? Does the inductor-like nature of the coil
> play a part?
>  
When the input is sitting low the transistor side of the cap will be at
5V (the cap seems to be in backwards BTW) and the coil side of the cap
will be at ground. The other end of the coil will also be at ground
(since the input is low) so the relay is off.

When the input goes from low to high the transistor switches on and the
transistor end of the cap is pulled to ground. Since a capacitor resists
change in voltage (remember I=C(dV/dt) ) the voltage accross the
capacitor will remain roughly the same dragging the other side of the
capacitor down to a negative voltage.

Gradually the capacitor will discharge through both the coil and the 100
ohm resistor but by this time the relay has already switched on so it
doesn't matter that the voltage accross it gradually decays.

2009\02\28@231221 by Jinx

face picon face
part 1 693 bytes content-type:text/plain; (decoded 7bit)

> I just don't understand how the circuit puts -5V at the base of the coil
> when the transistor initially conducts

Rolf,

Bearing in mind that the cap should be the other way around .....

....because with 5V supplied, it charges up through the 4k7 and 100R (or
diode with cathode to ground)

Now, when 5Vin is 5V, this turns the transistor on, connecting the cap's
+ve terminal to 0V. Therefore, relative to the rest of the circuit, the cap's
-ve terminal is -5V. The voltage across the relay coil is thus 5Vin (5V) to
cap -ve terminal (-5V), or 10V

The energy is enough to pull the relay in, and 5Vin holds it after the
capacitor has discharged


part 2 1651 bytes content-type:image/gif; (decode)


part 3 35 bytes content-type:text/plain; charset="us-ascii"
(decoded 7bit)


'[EE] Getting 15-ish volts from 5V'
2009\03\01@100418 by olin piclist
face picon face
Jinx wrote:
> Bearing in mind that the cap should be the other way around

No, it shouldn't.  Think of the quiescient case where the relay on signal
(confusingly labeled "5V in") is 0 to hold the relay off.  In that case,
both sides of the relay coil are at 0V and the transistor side of the cap
charges up to 5V with the polarity as shown.  When "5V in" goes high to
cause the relay to turn on, the voltage on the cap initially is still +5V
with it eventually discharging to the the drop on the bottom diode minus the
saturated transistor drop.  This should only be a few 100 mV.  Not great,
but the major voltage accross the cap is definitely as shown.

If I had to use this circuit, I would make both diodes Schottky.  The top
should be schottky for the speed and the bottom Schottky to minimize the
reverse voltage on the cap and to leave a little more badly needed voltage
for the relay.

Overall though, I don't like this circuit.  Yes it's kindof clever how it
puts twice the supply voltage accross the relay to initially pull it in.
However, look at most relay specs and 1/2 the pull in voltage is usually
below the guaranteed holding voltage.  While it probably does work with most
relays, it will be susceptible to mechanical vibrations and could be flaky
from batch to batch.

Here's a heretical thought: Get the right relay in the first place!


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\03\01@135216 by Brooke Clarke

flavicon
face
Hi Rikard:

Yes, wall warts do consume a lot of power and that's why the EPA has new Energy
Star recommendations that don't support the classical transformer type wall
warts.
www.energystar.gov/ia/partners/product_specs/program_reqs/EPS_FAQs_June_2008.pdf
http://www.energystar.gov/ia/partners/prod_development/downloads/power_supplies/powersupplyreport.pdf

The new generation wall warts contain a switch mode power supply that consumes
almost no power when there's no load.  This also allows for a single worldwide
wall wart (i.e. it works at any line frequency or voltage).  For example see:
http://www.prc68.com/I/KPC350.shtml#DC <- came with battery charger
http://www.prc68.com/I/MHC401FS.shtml <- came with TV camera

12 VDC is a common output.  This way you don't need to do a voltage conversion
and you get a very energy efficient setup.

Jameco has a number of the AC-to-DC Switching / Wall Transformers in 3.3, 5,
12, 15, 24 and other common DC voltages.

--
Have Fun,

Brooke Clarke
http://www.prc68.com

>What I don't want to do is to use a high(er) voltage wallwart, because
>I expect that the energy consumed in standby will be higher, and this
>project will be plugged in 24 hours a day.

2009\03\01@161238 by cdb

flavicon
face
As this may not need a high current, what about using a 555 as a DC-DC
inverter, just switch it on via a Pic pin when the latch voltage is
needed or even Roman Blacks discrete inverter?

Colin
--
cdb, .....colinKILLspamspam@spam@btech-online.co.uk on 2/03/2009

Web presence: http://www.btech-online.co.uk  

Hosted by:  http://www.1and1.co.uk/?k_id=7988359







2009\03\01@193818 by Rikard Bosnjakovic

picon face
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On Sun, Mar 1, 2009 at 00:16, Bob Blick <bobblickspamKILLspamftml.net> wrote:

> If you can
> generate a high duty cycle pulsetrain from your pic, all you need to add
> is a transistor, coil, diode and capacitor to step up your voltage.
> Under load your voltage multiplication will be proportional to the duty
> cycle, plus the input voltage. So if your duty cycle is 75%, that is 3:1
> plus 1 so your output voltage will be 4 times the input. Generally.
> Certainly close enough for a relay.

Hi Bob, thanks for the input.

Googling a basic schematic for a step-up, and using your tips above, I
came up with the circuit attached to this post. For the pulsetrain, I
made this simple code for a PIC12F683 using 4MHz INTOSC:

void delay(unsigned char ctr) {
 while (ctr-- > 0) {}
}

void main(void) {
 ANSEL = 0;
 CMCON0 = 0x7;
 TRISIO = 0;

 while(1) {
   GPIO2 = 1;
   delay(3);
   GPIO2 = 0;
   delay(1);
 }
}

75 and 25 for the delay are just the values which you suggested (75%
duty). I haven't got a scope so I cannot verify that it really shoots
a pulsetrain, but I expect it does so.

However, I'm not getting a higher voltage over the load (which I'm
measuring with a Fluke 179). Peeking the disassembler code and
counting the number of instructions executed, it looks like the PIC is
outputting at a frequency near 110kHz which could be too high.
Changing the delays to 75 and 25 respectively, the freq should be
around 10kHz which - according to wikipedia's text on step-up
converters - should be better.

Still, my DMM doesn't show me a higher voltage. I get about 4.3V when
disconnecting the transistor completely, and about 4.2V when using the
oscillating PIC. I don't know the value of the inductor, so this could
be a source of the error. I tried two different ones from my junkbox,
but the voltage doesn't change much on any of them.

The diode could possibly be another error source. Other peoples'
schematics tells me to use a fast recovery schottky-diode, but I have
only got standard silicon ones (0.7V drop, hence the 4.3V over the
load).

The transistor is a BJT, a general purpose NPN (BC547, or if it was
557). Perhaps this one can't handle the high frequency, so this might
be another error source as well.

Could you make a guess of which of the above parts (code and
components) could be the bottleneck? All of them?


--
- Rikard - http://bos.hack.org/cv/

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2009\03\01@195124 by Bob Blick

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Hi Rikard,

If you're using a bipolar transistor you should use a resistor in series
with the base, to limit current from the PIC and allow it to reach a
real logic "1" value. Also, your diode needs to be fast. It doesn't have
to be shottky but it can't be a typical 1 amp rectifier like the 1N4000
series. You can use a 1N4148 small signal diode, it is fast enough but
don't expect more than 30 milliamps through it.

Cheerful regards,

Bob



Rikard Bosnjakovic wrote:
{Quote hidden}

2009\03\01@203048 by Richard Prosser

picon face
2009/3/2 Bob Blick <EraseMEbobblickspam_OUTspamTakeThisOuTftml.net>:
{Quote hidden}

If it's a BC557, then it's a PNP.
As Bob said. you will need a fast diode &  you  need a resistor in the
base circuit.

RP

2009\03\02@105635 by Rikard Bosnjakovic

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On Mon, Mar 2, 2009 at 01:50, Bob Blick <@spam@bobblickKILLspamspamftml.net> wrote:

> If you're using a bipolar transistor you should use a resistor in series
> with the base, to limit current from the PIC and allow it to reach a
> real logic "1" value. Also, your diode needs to be fast. It doesn't have
> to be shottky but it can't be a typical 1 amp rectifier like the 1N4000
> series. You can use a 1N4148 small signal diode, it is fast enough but
> don't expect more than 30 milliamps through it.

Thanks Bob, a 1N4148 made it work. I had to tweak a lot with the
inductor, capacitor, transistor and the pulsetrain to get it to work,
but now it works like a charm. I'm aware that this trials and
error-approach is not "scientific", but again - it made me able to
think in analytical ways again.

I'm posting a schematic of the circuit. I know it's basic for just
about everyone on the list, but describing one's work is a way to
learn it even more.

R2 and T1 controls whether the relay should be on or not. If T1 wasn't
there, there would be leak currents through L1, D2 and the relay down
to ground. Since this project was supposed to be as power conserving
as possible, I added T1. But there's a second reason for adding it:
When the relay had triggered, it wouldn't break even if the voltage
dropped down to 5V (or below). I don't know why, but I believe that
there's something to do with electromagnetics which is an area I have
no knowledge of.

Before adding T1 I tried to make a voltage divider, letting the
booster trig itself via a transistor. I had unfortunately no luck in
making this work as I didn't find a balance. When the divider had
correct voltage on top of it (same net that feeds the relay), the
voltage over the transistor base was too low (0.2V) and it wouldn't
switch. When I changed resistors letting the base have 0.8V, the relay
net had too little voltage and couldn't trig. So I ditched the idea.
Too bad it didn't work, would have saved me an I/O on the PIC.

The step-up converter itself consists of L1, D2, T2 and C2. Like I
said, I had to do a LOT of tweaking with these components.
Particularily because the value of L1 was (and still is) unknown so I
was not able to do any kind of maths.

The PIC runs on 8MHz INTOSC. I found that if I run it on lower speed,
it would take longer for C2 to charge. This was true also if C2 was
bigger. Itself this is no problem, but the relay is supposed to
trigger "immediately" when the circuit powers up. Using an 1uF-cap
seems to be the fastest, but it wouldn't store enough charge to make
the relay trigger. 22uF was a nice value that balances both the charge
time and the charge amount, and it works well.

To make the circuit work, a pulsetrain is sent to T2 to make the
inductor build up some voltage. After about 250ms the voltage is
enough, and T1 is being closed - letting the relay trig. If T1 is
closed before T2, there is not enough voltage for the relay and it
won't trig.

I realize that the simplest solution would have been to get a
5V-relay, but this was a fun starter project for my one year absence
of electronics. Again, this approach was certainly not scientific in
any sort of matter, but it made me able to play and think.

Thanks Bob and everyone else for the input and feedback, I appreciate it.


--
- Rikard - http://bos.hack.org/cv/

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2009\03\02@111910 by olin piclist

face picon face
Rikard Bosnjakovic wrote:
> I'm posting a schematic of the circuit.

You left out the base resistor on T2.

You have MCLR pulled down to ground instead of up to Vdd.

1N4148 is a fast diode but has very limited current capability.  Remember
that the peak current thru D2 will be more than the average current.  I see
that the diode accross the relay is a plain old 1N4004.  If a 1N4148 is good
enough for D2, then it should certainly be good enough for D1.  However, I
suspect it may be underrated for D2.  I would use the same Schottky for
both.

10Kohms on the base of T1 seems like too much.  Assuming Vdd is 5V, that
means you get about 430uA of base current.  I don't know what a BC547 is
rated at, but assuming a guaranteed gain of 50 yields a collector current of
only 21.5mA.  Your relay probably needs more than that.  Without knowning
the relay current and looking up a BC547 I can't tell for sure, but R2 is
probably too large.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\03\02@115240 by Rikard Bosnjakovic

picon face
On Mon, Mar 2, 2009 at 17:19, Olin Lathrop <KILLspamolin_piclistKILLspamspamembedinc.com> wrote:

>> I'm posting a schematic of the circuit.
>
> You left out the base resistor on T2.
> You have MCLR pulled down to ground instead of up to Vdd.

Thanks, this is an error in the schematic. MCLR is tied to Vdd on my
breadboard, and T2 has a base resistor as well. I just forgot to draw
them.

> 10Kohms on the base of T1 seems like too much.  Assuming Vdd is 5V, that
> means you get about 430uA of base current.  I don't know what a BC547 is
> rated at, but assuming a guaranteed gain of 50 yields a collector current of
> only 21.5mA.  Your relay probably needs more than that.  Without knowning
> the relay current and looking up a BC547 I can't tell for sure, but R2 is
> probably too large.

The circuit works fine with 10k, but you are probably right; I should
lower R2 to 4k7 or similiar.

I don't have any Schottky-diodes at home, that's why I didn't use them
instead of a 1N4148. I will try a 1N4004 there instead to make sure it
can handle the current. Last time I tried a 1N4004 it didn't work, but
at that time I had another transistor (a power NPN) and I guess that
transistor couldn't handle the high frequency pulse train which this
BC547 does, so it's worth a shot.


--
- Rikard - http://bos.hack.org/cv/

2009\03\02@120731 by Isaac Marino Bavaresco

flavicon
face
Olin Lathrop escreveu:
> Rikard Bosnjakovic wrote:
>  
>> I'm posting a schematic of the circuit.
>>    
>
> You left out the base resistor on T2.
>
> You have MCLR pulled down to ground instead of up to Vdd.
>
> 1N4148 is a fast diode but has very limited current capability.  Remember
> that the peak current thru D2 will be more than the average current.  I see
> that the diode accross the relay is a plain old 1N4004.  If a 1N4148 is good
>  
The free-wheeling diode speed is not critical, it conducts only when the
relay turns off. If it takes more time to conduct, just the induced
voltage in the coil grows larger. while the total voltage seen by T1's
collector is under 35V, it's everything OK.

I usually use 1N4007 or 1N5408 as free-wheeling diodes, depending on the
load current, and never had a damaged transistor.

> enough for D2, then it should certainly be good enough for D1.  However, I
> suspect it may be underrated for D2.  I would use the same Schottky for
> both.
>
> 10Kohms on the base of T1 seems like too much.  Assuming Vdd is 5V, that
> means you get about 430uA of base current.  I don't know what a BC547 is
>  

Assume beta to be anything between 150 and 800, depending on the variant
(BC547A, B or C).

> rated at, but assuming a guaranteed gain of 50 yields a collector current of
> only 21.5mA.  Your relay probably needs more than that.  Without knowning
> the relay current and looking up a BC547 I can't tell for sure, but R2 is
> probably too large.
Regards,

Isaac


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2009\03\02@124059 by olin piclist

face picon face
Rikard Bosnjakovic wrote:
> Thanks, this is an error in the schematic. MCLR is tied to Vdd on my
> breadboard, and T2 has a base resistor as well. I just forgot to draw
> them.

OK, so I'll waste less time on your schematics and take them less seriously
in the future.  If you can't be bothered, I see no reason I should be.

> The circuit works fine with 10k, but you are probably right; I should
> lower R2 to 4k7 or similiar.

The point was you should do the math.  Look up the minimum gain of the
transistor and get the relay current, then guarantee there is enough base
current to support the collector current.

> I don't have any Schottky-diodes at home, that's why I didn't use them
> instead of a 1N4148. I will try a 1N4004 there instead to make sure it
> can handle the current.

Bad idea, assuming "there" means the switching power supply diode.  1N4004
are very slow as diodes go.  Unless you are somehow guaranteeing
discontinuous mode, you don't want to use a slow diode.  I believe you are
actually assuming continuous mode, so you want a fast recovery or Schottky
diode.  Schottky diodes are inhrently fast due because they don't have the
problem of minority carriers loitering about the depletion region to the
extent silicon diodes have.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\03\02@134248 by Rikard Bosnjakovic

picon face
On Mon, Mar 2, 2009 at 18:41, Olin Lathrop <RemoveMEolin_piclistTakeThisOuTspamembedinc.com> wrote:

> OK, so I'll waste less time on your schematics and take them less seriously
> in the future.  If you can't be bothered, I see no reason I should be.

There's no need to be harsh, it's not about I'm not being bothered. I
built the circuit on my breadboard, took a few photos of it and then,
on another machine, drew a schematic from the photos.

I simply missed two details during the transfer from the photos to the
schematics.


--
- Rikard - http://bos.hack.org/cv/

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