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Hi Chris
I will get this done, but I will have to get the diodes 1n004 *3. I have had a look at the maplin website, but they don't seam to have them so I will not be able to get them tomorrow, it will take a few days.
Can you take a look at this link, will any of these do the job:
http://www.maplin.co.uk/search?criteria=diodes
Let me know as I can get these tomorrow!
Regards
Justin
I will get this done, but I will have to get the diodes 1n004 *3. I have had a look at the maplin website, but they don't seam to have them so I will not be able to get them tomorrow, it will take a few days.
Can you take a look at this link, will any of these do the job:
http://www.maplin.co.uk/search?criteria=diodes
Let me know as I can get these tomorrow!
Regards
Justin
Yes, the site is annoying tonight. I did a search and they have only odd numbers in stock,,,,,1 * 1n4006 and 2 * 1n4002s
Would it be ok to buy these two types and mix these diodes.
Why on earth they can't stock more than 1 or 2 diodes of a specific type is beyond me???
Regards
Justin
Justin, when you make an omelet you have to break a few eggs. Well, this is my metaphor for this circuit test. As you can see, I moved one of your LEDs to the negative (Green) lead of the uPuck's output. My guess is that it'll work fine there but since there is no data sheet of the internals of the puck I can't guarantee that it won't fry. I just wanted you to know this.
Since I know you're learning I'm sure you're curious why we're doing this test. Well, the reason is this. Switching out (shunting) one of the LEDs in its original configuration would require the use of PNP BJTs or PCHAN MOSFETs. Most of us like to avoid using either.
Chris
Edit: Those diodes are too inexpensive to buy just three. You might as well buy at least ten of them. I'd have you buy some transistors too but I want to be sure this works first. I hope it's not a long trip to the supplier.
Since I know you're learning I'm sure you're curious why we're doing this test. Well, the reason is this. Switching out (shunting) one of the LEDs in its original configuration would require the use of PNP BJTs or PCHAN MOSFETs. Most of us like to avoid using either.
Chris
Edit: Those diodes are too inexpensive to buy just three. You might as well buy at least ten of them. I'd have you buy some transistors too but I want to be sure this works first. I hope it's not a long trip to the supplier.
Last edited:
Chris
I'm sorry I haven't had a chance to get to the store today, I was the wrong end of the county and I have only just got home. They close at 8PM so I will have to wait until tomorrow.
Can you explain to me what shunting means, as I have often heard this phrase but as you didn`t say which LED is being shunted, I can't work it out.
Also,,,, OMG, her he goes again
How hard would it be to produce a circuit that ensures that a battery (6S Lipo) does not discharge below 18V. ie so once the voltage of the cell reaches 18V, the device switches off.
I know that I could do this with the axe and the circuit we produced earlier using the voltage divider to monitor the cell Voltage and then simply switch off once the voltage is reached but is there an easy way, ie just a simple small circuit that switches off once the cell get to 18V.
As always, with respect
Justin
I'm sorry I haven't had a chance to get to the store today, I was the wrong end of the county and I have only just got home. They close at 8PM so I will have to wait until tomorrow.
Can you explain to me what shunting means, as I have often heard this phrase but as you didn`t say which LED is being shunted, I can't work it out.
Also,,,,
OMG, her he goes againHow hard would it be to produce a circuit that ensures that a battery (6S Lipo) does not discharge below 18V. ie so once the voltage of the cell reaches 18V, the device switches off.
I know that I could do this with the axe and the circuit we produced earlier using the voltage divider to monitor the cell Voltage and then simply switch off once the voltage is reached but is there an easy way, ie just a simple small circuit that switches off once the cell get to 18V.
As always, with respect
Justin
- Joined
- Jan 21, 2010
- Messages
- 25,510
It's actually harder than you might expect.
One problem with circuits like this is that the battery falls below 18V under load, so some (unspecified magic) turns the circuit off. Immediately, the battery voltage rises again and the circuit is re-enabled. The current rises, the voltage falls, back off again. No load, higher voltage, on again. The pattern repeats.
There are 2 ways to handle this.
The first is that the circuit is switched off in a way that can't be automatically turned back on again. This allows the user to try again, at which point the circuit presumably switched off (either immediately, or after a short delay).
The second is to have the circuit switch off at some voltage, but require a substantially higher voltage to turn back on again. So it might turn off at 16V, but require 18V to turn on again (automatically, or manually).
The former is simpler, the latter is suitable for things that turn on automatically.
However, not wanting to discharge LiPo batteries below a certain point is a pretty common requirement. I would be looking for power management devices to do this first. My second option would be to think about designing one.
IMHO it's a bad idea to post that within this thread. It can only yuck up the continuity of it. Oh,... "shunt" is synonymous with bypass, divert & avert. The Diode string is wired to shunt current around LED1.
Chris
Chris
Thanks for the reply Steve,,Thats clears some things up for me..
Chris, your right, I should have started a new thread, but I think I have my answer in one!
Ill be picking up the diodes later and get back to you tonight with the figures you need.
Regards
Justin
Chris, your right, I should have started a new thread, but I think I have my answer in one!
Ill be picking up the diodes later and get back to you tonight with the figures you need.
Regards
Justin
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