The Shop > Electronics & IC Programing
Need a boost....
kwackers:
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...
WillieL:
--- Quote from: kwackers on September 17, 2012, 12:10:05 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...
--- End quote ---
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:
kwackers:
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.
WillieL:
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.... :(
John Swift:
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
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