Alright, lets do the math and look at the datasheet. We need a minimum load of 5mA for the LM7805 to function properly. This is according to the datasheet.
I=V/R
I=5/1000
I=.005
So we can see that while putting the 1k resistor in series is generally a fine idea there is no way that it is possible to present a larger load than 5mA to the LM7805 without putting a resistor in parallel before the 1k resistor. It has to be done to get a stable 5V from the regulator or we have to use a different regulator and dropping a resistor from OUT to GND of the reg circuit is pretty easy to do.
Thats all. Thats why you have a crazy voltage and thats why the LDRs are acting up.
Uriah
I=V/R
I=5/1000
I=.005
So we can see that while putting the 1k resistor in series is generally a fine idea there is no way that it is possible to present a larger load than 5mA to the LM7805 without putting a resistor in parallel before the 1k resistor. It has to be done to get a stable 5V from the regulator or we have to use a different regulator and dropping a resistor from OUT to GND of the reg circuit is pretty easy to do.
Thats all. Thats why you have a crazy voltage and thats why the LDRs are acting up.
Uriah
Fred, 1,65 from lateral pins to ground when the pot is completely turned up or down.
I agree with Uriah, i did the same math and read the datasheet.
It can be a possible solution, i will put a 470ohm 1/2W in parallel to the cap and see what happens. It will draw about 10mA. I can do it on tuesday afternoon, i'm not at home today.
Other ideas are welcome too.
Thanks for your help,
Gianni
I agree with Uriah, i did the same math and read the datasheet.
It can be a possible solution, i will put a 470ohm 1/2W in parallel to the cap and see what happens. It will draw about 10mA. I can do it on tuesday afternoon, i'm not at home today.
Other ideas are welcome too.
Thanks for your help,
Gianni
Hi Gianni,
I don't think the 7805 not turned on is the problem because the 10uF cap will draw more then 5ma, and once it is charged up, it can provide the voltage. The evidence is you had 1.65V from the pins to ground. If the 7805 is not turned on, there will be no voltage. I do suggest to use a bigger cap here.
I suspect the 1K is to aggressive to throttle the current (the reason why I asked you to take voltage measurement across it). 1.65V from pin to ground indicated the 1K was eating up 3.35V which implies 3.35ma from both sides, or approx. 1.67ma from each side. I suggest to // the 470R to the 1K (resultant R is approx 320). That should effectively raise the max operating voltage from pins to ground to close to 2V. Give this a try and us how it goes.
I don't think the 7805 not turned on is the problem because the 10uF cap will draw more then 5ma, and once it is charged up, it can provide the voltage. The evidence is you had 1.65V from the pins to ground. If the 7805 is not turned on, there will be no voltage. I do suggest to use a bigger cap here.
I suspect the 1K is to aggressive to throttle the current (the reason why I asked you to take voltage measurement across it). 1.65V from pin to ground indicated the 1K was eating up 3.35V which implies 3.35ma from both sides, or approx. 1.67ma from each side. I suggest to // the 470R to the 1K (resultant R is approx 320). That should effectively raise the max operating voltage from pins to ground to close to 2V. Give this a try and us how it goes.
I think it is good to be busy these days
Hopefully, we are moving in the right direction.
Ok, new tests, here's the result.
I think that 1k resistor is the problem, i try to explain.
Black pin of dmm fixed to ground
4,95V out of 78l05, any volume position, before 1k.
1.64V from center pin of pot, so AFTER 1k, at top and bottom positions.
4.92V from center pin of pot at -6dB.
1,40V each led at -6dB
Doing the math:
3,3V loss from 78l05 to center pin of the pot (3,3/1000=3,3mA total flowing/4=0,825 each ldr)
For me, 1k is definitely too much...
Regards,
Gianni
I think that 1k resistor is the problem, i try to explain.
Black pin of dmm fixed to ground
4,95V out of 78l05, any volume position, before 1k.
1.64V from center pin of pot, so AFTER 1k, at top and bottom positions.
4.92V from center pin of pot at -6dB.
1,40V each led at -6dB
Doing the math:
3,3V loss from 78l05 to center pin of the pot (3,3/1000=3,3mA total flowing/4=0,825 each ldr)
For me, 1k is definitely too much...
Regards,
Gianni
pchw, you're right, better work hard these days
Now i'm reading again your previous post and found that we are saying the same thing!
I don't think that, for example, 2 ldr full on with 250k in series will draw more than 5ma.
So, even without the resistor or with only 100ohm, i think a 1/2W pot will last for a long time. Obviously with this set of ldr, showing 10k with a 400k in series (udailey chart).
Regards,
Gianni
Now i'm reading again your previous post and found that we are saying the same thing!
I don't think that, for example, 2 ldr full on with 250k in series will draw more than 5ma.
So, even without the resistor or with only 100ohm, i think a 1/2W pot will last for a long time. Obviously with this set of ldr, showing 10k with a 400k in series (udailey chart).
Regards,
Gianni
Gianni,
Please post a pic.
250k will give you 20uA.
Is your 5V going straight to the wiper of the single gang pot? The middle pin?
Something is weird. Did your 1k resistor go in series from the output of the LM7805 to the wiper of the pot or did you put it in parallel from output to ground?
Did you ever put a resistor parallel to load the LM7805?
Uriah
Please post a pic.
250k will give you 20uA.
Is your 5V going straight to the wiper of the single gang pot? The middle pin?
Something is weird. Did your 1k resistor go in series from the output of the LM7805 to the wiper of the pot or did you put it in parallel from output to ground?
Did you ever put a resistor parallel to load the LM7805?
Uriah
Hi Gianni,
You absolutely right about that 100R is good enough to protect the LDR's because when the pot is fully attenuated or open, only one ldr of the series/shunt pair is drawing current (the other one is negligible). In these position, the value of the pot doesn't matter because one of the pins is shorted to the wiper. The only thing that is protecting the ldr's from the full 5V is the resistor in between. However, 1K is too high.
When the ldr's of both channel draw up to 12.5ma each (total 25ma), the 100R will drop 2.5V which is the max voltage spec of the LDR's. I was just a little conservative to suggest 300R just in case the ldr's don't enough curret to drop 2.5V FWIW, in my test, 1ma will bring the ldr's resistance to 100R - 200R. That's another reason that I suggest 300R. Anything between 100R and 300R will do, don't go below 100R. If you see improvement but not to the degree that you want, the next step is to use a lower pot, e.g 250K.
Cheers,
You absolutely right about that 100R is good enough to protect the LDR's because when the pot is fully attenuated or open, only one ldr of the series/shunt pair is drawing current (the other one is negligible). In these position, the value of the pot doesn't matter because one of the pins is shorted to the wiper. The only thing that is protecting the ldr's from the full 5V is the resistor in between. However, 1K is too high.
When the ldr's of both channel draw up to 12.5ma each (total 25ma), the 100R will drop 2.5V which is the max voltage spec of the LDR's. I was just a little conservative to suggest 300R just in case the ldr's don't enough curret to drop 2.5V
FWIW, in my test, 1ma will bring the ldr's resistance to 100R - 200R. That's another reason that I suggest 300R. Anything between 100R and 300R will do, don't go below 100R. If you see improvement but not to the degree that you want, the next step is to use a lower pot, e.g 250K.Cheers,
Uriah, i assure you that a pic will be the nearly identical to the schematic i posted before, i'll do tests, modifications and photo during the weekend.
Yesterday i only did measurements. Yes, 5V to the 1k in series with the middle pin of the pot, nothing in parallel, but i have 4,95V out from the 78l05, measured from the output pin of the regulator, in any position of the pot, as i wrote before, so the problem is not its stability.
As pchw said, the problem is the falling voltage on 1k resistor.
With 3ma it eats 3V, with 5ma 5V. So, what remains to the LDRs?
It is a right current limiter when you know exactly the voltage to drop and the current taken from the load. At this time i don't know the current but it's over your expectations, so the current required by the LDRs let that resistor produce an overdropping voltage across itself.
I have to try that 330, than down to 100 or a different value of the pot if it doesn't work.
Has anyone built the attenuator strictly following Uriah schematic with the single 500k pot using July/August shipped matched set?
Thanks Uriah and Fred, stay tuned...
Gianni
Yesterday i only did measurements. Yes, 5V to the 1k in series with the middle pin of the pot, nothing in parallel, but i have 4,95V out from the 78l05, measured from the output pin of the regulator, in any position of the pot, as i wrote before, so the problem is not its stability.
As pchw said, the problem is the falling voltage on 1k resistor.
With 3ma it eats 3V, with 5ma 5V. So, what remains to the LDRs?
It is a right current limiter when you know exactly the voltage to drop and the current taken from the load. At this time i don't know the current but it's over your expectations, so the current required by the LDRs let that resistor produce an overdropping voltage across itself.
I have to try that 330, than down to 100 or a different value of the pot if it doesn't work.
Has anyone built the attenuator strictly following Uriah schematic with the single 500k pot using July/August shipped matched set?
Thanks Uriah and Fred, stay tuned...
Gianni
Yes. Built using George's Lightspeed Mk2 circuit. (same as Uriah posted) and worked first time.
Only deviation was to use 1k resistors to each ldr instead of the 100R specified. (extra insurance
)Had to experiment with several cable types to get frequency balance correct but that is probably to do with the cd/amp combo.
Plenty of drive, good useable range from the 500k pot, 11 o'clock seems to be preferred level.
Had to tear myself away from listening to post this and heading back now.
I am rediscovering my music collection
Best improvement I have ever experienced.
It has breathed new life into my aging system. ( and ears !!)
The 700-1k resistor is more for the protection of the pot. The pot will degrade and slowly die with the 100R resistors and the 100R resistors put the current right at the max of the LDRs. The current necessary at almost any volume with this circuit to power the LDRs is less than 1mA so using 100R puts an unnecessary strain on the LEDs in the LDRs.
If you read Rod Elliots page on pots you will find that when we use them in this way we are pushing their current limits pretty hard.
Yes other people have built it this way and so have I. I used a battery, 9V and dropped it twice with LM317 then put 1k trimmer in series and turned it halfway up then put another in parallel and turned it about halfway up.
Uriah
If you read Rod Elliots page on pots you will find that when we use them in this way we are pushing their current limits pretty hard.
Yes other people have built it this way and so have I. I used a battery, 9V and dropped it twice with LM317 then put 1k trimmer in series and turned it halfway up then put another in parallel and turned it about halfway up.
Uriah
a 1k resistor will reduce the worst case pot current to <=3mA.
That is equivalent to 4.5W across the 500k pot.
The maximum current for a 500k 500mW pot is <=1mA.
The 750r to 1k0 resistor does not provide sufficient worst case current reduction to protect the 500k pot fed from a 5V source.
I would rather use the full range of safe and reliable LED current and face the possibility of replacing a cheap pot periodically.
The maximum current for a 100k 500mW pot is ~2.2mA
That is equivalent to 4.5W across the 500k pot.
The maximum current for a 500k 500mW pot is <=1mA.
The 750r to 1k0 resistor does not provide sufficient worst case current reduction to protect the 500k pot fed from a 5V source.
I would rather use the full range of safe and reliable LED current and face the possibility of replacing a cheap pot periodically.
The maximum current for a 100k 500mW pot is ~2.2mA
Andrew, i don't understand where 4,5W come out.
1k resistor at 3ma produce 3V drop, so the pot will have 2V in the middle pin
and 3*0,003= 6mW. Where i'm doing wrong?
Please, what do you mean?
Regards,
Gianni
So you have a 500 ohm in series and 500 in parallel, that is a split by two, so you have 2,5V in the middle pin, right?
Regards,
Gianni
So is there an updated/definitive schematic that we can reference ?.
When I built mine there was a discussion about the 500k pot and value of resistors to use (after pot and before ldrs) and 1k was mentioned.
Since then there have been comments about a resistor before the pot and now trimmers before and across it.
I'm confused.
When I built mine there was a discussion about the 500k pot and value of resistors to use (after pot and before ldrs) and 1k was mentioned.
Since then there have been comments about a resistor before the pot and now trimmers before and across it.
I'm confused.
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