Author Topic: Need a boost....  (Read 6905 times)

Offline WillieL

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Need a boost....
« on: September 16, 2012, 05:49:26 PM »
Can anyone give me some hints on boosting the output on this circuit?
I am severely mathematically challenged.   :doh:

I've put one together, but it's only putting out 68 mA and I'm trying to get 250mA out of it @ 300mV.

HELP?   :scratch:

WillieL

Midwestern USA

Offline AussieJimG

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Re: Need a boost....
« Reply #1 on: September 16, 2012, 07:21:26 PM »
What is the load when you get the 68mA at 300mV?

Jim

Offline WillieL

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Re: Need a boost....
« Reply #2 on: September 16, 2012, 09:06:42 PM »
Around 1.2 or 1.3 ohms I believe.  Basically whatever the resistance is running through my meter, measuring between the output leads.

I can use a soldering iron, but don't the first thing about circuit design I'm afraid.   :Doh:
WillieL

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Offline kwackers

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Re: Need a boost....
« Reply #3 on: September 17, 2012, 02:30:02 AM »
R3 limits the current.
Ohms law says I = V/R, so ignoring Q3 that's 3/47 = 63mA.

Simply reducing R3 may work, although obviously Q3 needs to be able to handle the current.

The other point is that to get 250mA through Q3 it needs to have enough gain.
Q1 and Q2 form a fixed voltage dropping aprox 0.7v each, so R2 has 0.7 volts across it  (Range: 0.7 to 1.4). That leaves 3-1.4 = 1.6v across R1 so the max current that can flow through R1 = 1.6 / 1k = 1.6mA.

However if you pull too much current through it you'll unbalance the whole divider chain so it's important the gain of Q3 is has high as possible.

Finally pulling 250mA from a D cell is a fair amount so you'll get some voltage drop on the rails which may compound your problems (and reduce battery life).

« Last Edit: September 17, 2012, 05:41:27 AM by kwackers »

Offline WillieL

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Re: Need a boost....
« Reply #4 on: September 17, 2012, 09:18:35 AM »
kwackers,

Thank you for that. I've made some changes and gotten some improvement, but still not where I need to be.  :(
I've tried reducing the R values to increase the current. This is where I'm at now:

R1 750 ohm, 1.0 W
R2 500 ohm, .75 W
R6 6.8 ohm, .5W
R3 6.8 ohm, .5W ------>  3/13.6 = 220mA correct? (R3 + R6)

Voltage drop across R1 is 1.63V (1.76V output shorted)
Voltage drop across R2 is .625V (.616V output shorted) Edit: End to end. Or did you mean between an end and the wiper? (Q3 base)  :scratch:

Oh, also I'm not using D cells. Power supply is a 3V 700mA wall wart.....

Still, I'm only reading 87mA at the output now. It will go much higher if I adjust R2, but then the output voltage goes sky high as well and I can't allow it to exceed the .3V limit.   :(

So is Q3 the bottleneck then?  :scratch:
If so, is there a different transistor that would work?

Again, I don't know what I'm doing so it's all guess work on my part.   :zap:

 :hammer:
WillieL

Midwestern USA

Offline kwackers

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Re: Need a boost....
« Reply #5 on: September 17, 2012, 12:10:05 PM »
First thing to note is Q1 & Q2 form two 0.7v constant voltages, therefore the voltage across R2 as part of the divider with R1 should be greater than 0.7v otherwise Q1 isn't doing it's job.
So 3 - 0.7 = 2.3v across r1 + r2. I = V/R = 2.3 / 1250 = 1.8mA, voltage across r2 = IR = 1.8mA * 500 = 0.92v.
That should be ok - except we really should take into account the current used by Q3.

So Q3 is just a constant voltage source. The output voltage will be the input - 0.7v (voltage drop across emitter/base), so R2 should be set to about at Q3's base.
The maximum collector current is equal to the base current x the transistor gain. DC current gain is listed as 100-300, so if we take the bottom number then to get 300mA out you'd need 3mA into the base...
However we can see that the max current flowing through the R1+R2+Q1+Q2 chain is 1.8mA and that doesn't include anything we need for the base current.

You can tell if the transistor is fully switched on by measuring the voltage across it's collector and emitter, therefore if you set it up for your max current and measure the voltage across Q2 C/E then it should be close to zero, if not the transistor base current isn't enough to switch it fully on.

So basically you need to find 3mA for Q3's base and this needs to be taken in such a way that the voltage across R2 doesn't fall below 0.7v (say 0.8-0.9 to be on the safe side).
If we double the current flowing through R1 to give us 6mA then there's 1.4 dropped across Q1+Q2 so R1 = V/I = 1.6 / 6mA = 266 ohms, call it 220...
Calulate R2 such that half the current flows through it, so 3mA across 0.7v is 233 ohms. (Could use a 100 ohm pot here with a couple of resistors either side to make it up to say 200 or so, just make sure the range is in the right place).

So with R1 = 220, R2 = 200 then current into Q3 base should be good for 3mA or so, with a transistor gain of 100 that should allow the transistor to switch fully on, providing then that R3 (and R6 if you're using it) will allow 300mA to flow then it should work...

R3+R6 are mainly current limiters, they should be chosen to give slightly more current than you need when across the supply (3v).

Hope that makes sense, I've typed it in as I go along...

Offline WillieL

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Re: Need a boost....
« Reply #6 on: September 17, 2012, 01:48:48 PM »


You can tell if the transistor is fully switched on by measuring the voltage across it's collector and emitter, therefore if you set it up for your max current and measure the voltage across Q2 C/E then it should be close to zero, if not the transistor base current isn't enough to switch it fully on.

It looks like you hit the nail on the head. I just checked the voltage drop across Q2 C/E and found .637V

So basically you need to find 3mA for Q3's base and this needs to be taken in such a way that the voltage across R2 doesn't fall below 0.7v (say 0.8-0.9 to be on the safe side).
If we double the current flowing through R1 to give us 6mA then there's 1.4 dropped across Q1+Q2 so R1 = V/I = 1.6 / 6mA = 266 ohms, call it 220...
Calulate R2 such that half the current flows through it, so 3mA across 0.7v is 233 ohms. (Could use a 100 ohm pot here with a couple of resistors either side to make it up to say 200 or so, just make sure the range is in the right place).

So with R1 = 220, R2 = 200 then current into Q3 base should be good for 3mA or so, with a transistor gain of 100 that should allow the transistor to switch fully on, providing then that R3 (and R6 if you're using it) will allow 300mA to flow then it should work...

OK, it looks like I'm going back to do some more resistor shopping. I went from 1K/1K down to 750/500 on R1/R2 (wild guess) and I suppose that is what gave me the partial increase in current. I just didn't go far enough.  :palm:

Hope that makes sense, I've typed it in as I go along...

From the parts that I could understand - yes!  I was just blindly replacing components.   :wack:

One other question. Is this going to affect the short indicator circuit? I found out that going below 5 ohms on R6 disabled the short LED. So I assume I upset the balance.   :doh:

Thank you SO much for helping me on this kwackers! I lost track of how many days I have spent trying to figure this out. 


 :nrocks:
WillieL

Midwestern USA

Offline kwackers

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Re: Need a boost....
« Reply #7 on: September 17, 2012, 02:35:13 PM »
For the short circuit LED to work the voltage across R6 needs to be 0.7v under a short circuit, so if you're pulling 300mA normally then the short circuit would need to deliver more than that - say 400mA. Which means your components need to be set up to provide 400mA.
Then R6 = 0.7/0.4 = 1.75 ohms.

Might be more trouble than it's worth, with the correct components the circuit is self limiting, it can't exceed the base current into Q3 x the transistor gain. In fact R3+R6 could in theory be removed and it would work fine providing that value was around your max current.
If you did this though all the power would be dissipated in the transistor rather than currently in Q3, R3 & R6.

Offline WillieL

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Re: Need a boost....
« Reply #8 on: September 29, 2012, 09:31:03 PM »
Well a little bit more progress, but still short of my goal. I've reduced the resistors yet again, but got no change in the output.  :scratch:

R1 100 ohms
R2 200 ohms
R3 6.8 ohms
R6 6.8 ohms

No change in the voltage across Q3.  :scratch: :scratch: :scratch:
Something isn't right here.

So I tried replacing Q3 with another 2N4401 from a different manufacturer.
BINGO!  Instantly jumped to 140 mA at the output.   :thumbup:
So then I did the same with Q2. No change.  I'm still about 100 mA short....   :(
WillieL

Midwestern USA

Offline John Swift

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Re: Need a boost....
« Reply #9 on: September 30, 2012, 08:33:46 AM »
I'd replace Q3 with a higher gain transistor like the MPSA14 darlingtor transistor
(a DIY version using two 2N4401 transistors  is shown as  Q3 + Q6 on the diagram)
when using a darlington transistor add Q5
 the extra diode connected transistor in series with  Q2

I'd  change R 6 to 2.2Ohms or 3.3 Ohms and R3 to 5 Ohms

If you assume Q3 when switched hard on has no resistance

when R6 = 3.3 ohms , the short circuit current will be 3V/(R3+R5+R6)     =  3v/9.3 = 322mA
by adding a 100 Ohm preset potentiometer in parallel with R6 you will be able to adjust the current that switches on the over current indicator LED


    John

Offline WillieL

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Re: Need a boost....
« Reply #10 on: September 30, 2012, 10:40:35 AM »
John,

Wow. Thank you for that! The diagram really helps.
So... I'll need to change R1 back to a 1K resistor then, is that correct?

I have plenty of the 2N4401 transistors so I guess I will use them instead of the darlington.
That means that Q5 is NOT required - correct? Also I don't see a diode in your drawing?   :scratch:

And thank you again for the idea about the adjustable over current LED. I would really like having that feature!   :clap:
I will also have to order a 100 ohm trimpot since I only have a 200 ohm on hand, but it will be well worth it to get this working the way I want it to.

Off to make another parts order.....   :palm:

Willie
WillieL

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Offline kwackers

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Re: Need a boost....
« Reply #11 on: September 30, 2012, 12:10:23 PM »
Looking at Johns diagram the only thing that bothers me is R1 & R2 are the same value.

Roughly speaking Q1, Q2 & Q5 each drop 0.7v for a total of 2.1v leaving 0.9v across R1 and 0.7v across R2.
Whilst not the same they're pretty close and make the voltage drops across the transistors and the supply voltage overly critical.
I'd be inclined to reduce R1 or increase R2.

You still need Q5. Because of the pair of transistors used as a darlington the BE voltage will be 1.4v instead of 0.7, Q5 ensures the voltage at the base is high enough to switch on the darlington.
The main use of the darlington is the gain of it is equal to the gain of both transistors multiplied, this means the current you need into the base can be tiny in order to fully switch it on.

Offline John Swift

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Re: Need a boost....
« Reply #12 on: September 30, 2012, 12:19:42 PM »
yes ,
 you will need to add Q5 and change R1  to  680 ohms to pass about 1.8mA through
Q1, Q2 and Q5   
 
assuming the transisors Q3 & Q6 each have a gain of 100
the darlington circuit they make will have a gain of 10000
and will only need 25 uA base current to output a 250 mA current

in the circuit diagram Q1 , Q2 and Q5 are used as diodes
and providing they are at the same temperature as Q3 and Q6 that form
the darlington transistor
the volt drop across the base emitter junctions of Q2 & Q5 (about 1.3V)
will be the same as the base emitter junctions of Q3 & Q6

the volt drop across R2 is fixed by Q2 to 0.65V (0.6V to 7V)

the output voltage across R4 will be the same as the voltage between the negative end of R2 (the connection to Q2) and the base of Q6
for a 0.3V output  the preset should be set about mid way



John

PS - WillieL
 you can use a 200 Ohm preset for the over current indicator

PPS  -Kwackers ,
I've had another look and  with the exra voltdrop across Q5 , it will be better with R1 being reduced to maintain the current through Q1 , Q2 and Q5
« Last Edit: September 30, 2012, 01:12:56 PM by John Swift »

Offline WillieL

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Re: Need a boost....
« Reply #13 on: September 30, 2012, 10:57:40 PM »
OK then, so.......

R1 680 ohms
R2 1K ohms
Add Q5
R9 100 or 200 ohm trimpot

I already ordered R9 (100 ohm) and a couple of darlingtons earlier today, but I'll have to get some 680 ohm resistors now.
I'm going to have quite a pile of extra parts laying around here....   :lol:

WillieL

Midwestern USA