5
   

reducing voltage

 
 
jameslm
 
Reply Mon 29 Dec, 2014 01:42 pm
How would I get .65 volts out of 1.5 volt battery.
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Type: Question • Score: 5 • Views: 959 • Replies: 19

 
contrex
 
  5  
Reply Mon 29 Dec, 2014 01:59 pm
Why do you need to do this? What current is required?
0 Replies
 
dalehileman
 
  0  
Reply Mon 29 Dec, 2014 02:09 pm
@jameslm,
You could drop it using a diode or rectifier, most such of the silicon persuasion providing a drop of approximately the required magnitude. However it would waste half your power. Hence Con's q's above

https://www.google.com/?gws_rd=ssl#q=forward+drop+of+typical+silicon+rectifier

Still if you could find a 0.4-v battery somewhere of approximately the same current capacity you could connect in opposition to your 1.5'er
BillRM
 
  0  
Reply Mon 29 Dec, 2014 03:02 pm
@dalehileman,
Quote:
Still if you could find a 0.4-v battery somewhere of approximately the same current capacity you could connect in opposition to your 1.5'er


Off hand I do not know of any commerce battery with anywhere near a cell value of .4 volts.

An Edison battery Ni-Fe if memory serve me correctly have a value of around 1.2 volts per cell.

We need details as a DC/DC switching power supply is possible for large loads or perhaps the load is so small that throwing half the power always would not be a problem.
timur
 
  3  
Reply Mon 29 Dec, 2014 03:12 pm
In order to pragmatically deal with the OP's question, an answer to Contrex is needed..
BillRM
 
  0  
Reply Mon 29 Dec, 2014 03:15 pm
Footnote I remember dealing with the worst design ever where a 400 volts supply was used with a 20 watts wirewound resister and a 35 volt zener in order to get a 35 voltage reference source!!!!!!!!

Talk about throwing away power for not good reason as there was any number of lower voltages tap offs from the transformer that could had been used.
contrex
 
  1  
Reply Mon 29 Dec, 2014 03:32 pm
@timur,
timur wrote:
In order to pragmatically deal with the OP's question, an answer to Contrex is needed..

So why do you think my question got voted down?
0 Replies
 
timur
 
  2  
Reply Mon 29 Dec, 2014 03:36 pm
For reasons unknown.

I can only guess, maybe some personal animosity, which abound here...
0 Replies
 
dalehileman
 
  0  
Reply Mon 29 Dec, 2014 05:44 pm
@BillRM,
Quote:
Off hand I do not know of any commerce battery with anywhere near a cell value of .4 volts.
Nor do I Bill. Otherwise though isn't it a great idea

Quote:
perhaps the load is so small that throwing half the power always would not be a problem.
That's why we have to hear from Slim Jim
contrex
 
  1  
Reply Tue 30 Dec, 2014 02:07 am
@dalehileman,
dalehileman wrote:
That's why we have to hear from Slim Jim

Maybe he went away and read about Ohm's Law, and/or voltage regulators and he now understands what he needs to do. Seeing the 0.65 volt requirement makes me think he wants to drive an LED or maybe some kind of low power chip.

Chumly
 
  2  
Reply Tue 30 Dec, 2014 04:59 am
@jameslm,
You can use a simple two resistor voltage divider as long as the source voltage and the load resistance remain relatively fixed and you take into account the inevitable requirements of impedance matching, or (much better) you can use a DC to DC switching regulator like these here http://www.analog.com/static/imported-files/product_highlights/177228276DCtoDC_Regulator_E.pdf
Chumly
 
  2  
Reply Tue 30 Dec, 2014 05:05 am
@contrex,
contrex wrote:
dalehileman wrote:
That's why we have to hear from Slim Jim
Maybe he went away and read about Ohm's Law, and/or voltage regulators and he now understands what he needs to do. Seeing the 0.65 volt requirement makes me think he wants to drive an LED or maybe some kind of low power chip.
Nope the forward voltage drop of the LED may not only too high for a source voltage of only .65 volts but most importantly you would use a current limiting resistor (dropping resistor) in series with the LED so that the current rating of the LED is not exceeded .
Chumly
 
  2  
Reply Tue 30 Dec, 2014 05:11 am
@BillRM,
BillRM wrote:
Footnote I remember dealing with the worst design ever where a 400 volts supply was used with a 20 watts wirewound resister and a 35 volt zener in order to get a 35 voltage reference source!!!!!!!!

Talk about throwing away power for not good reason as there was any number of lower voltages tap offs from the transformer that could had been used.
Firstly simple zener regulator circuits always waste power and secondly the difference between the source voltage and the load voltage does not in and of itself dictate the power loss due to voltage regulation.
BillRM
 
  0  
Reply Tue 30 Dec, 2014 06:03 am
@Chumly,
Quote:
the source voltage and the load voltage does not in and of itself dictate the power loss due to voltage regulation.


You kidding me correct as power equal voltage time current and the power being wasted in the case I referred to is 400-35 volts time Current compare to the useful power of 35 volts time current.

So to sum up that design is wasting 365/400 or 91 percents of the total power being used.

Now if instead if a DC to DC step down design had been used then yes indeed the situation would had been far better as far as power being thrown away is concern and the voltage in compared to the voltage out would not have such a linear relationship to the power being wasted.
dalehileman
 
  0  
Reply Tue 30 Dec, 2014 12:29 pm
@Chumly,
Quote:
most importantly you would use a current limiting resistor (dropping resistor) in series with the LED so that the current rating of the LED is not exceeded
Yes Chum, I hope duly noted. However this recommendation makes the assumption that the 0.65-v load is also a semi of some sort

Alliteration unintended
0 Replies
 
dalehileman
 
  0  
Reply Tue 30 Dec, 2014 12:35 pm
@Chumly,
Quote:
you can use a DC to DC switching regulator
Of course Chum we don't know whether the load would tolerate it, but one way to power the 0.65-v gadget without significant loss of power is by a train of 1.5-v pulses

..readily conceding that the gadget one chooses to get that pulse train might reduce efficiency somewhat
0 Replies
 
Chumly
 
  2  
Reply Wed 31 Dec, 2014 01:50 pm
@BillRM,
BillRM wrote:
Quote:
the source voltage and the load voltage does not in and of itself dictate the power loss due to voltage regulation.


You kidding me correct as power equal voltage time current and the power being wasted in the case I referred to is 400-35 volts time Current compare to the useful power of 35 volts time current.

So to sum up that design is wasting 365/400 or 91 percents of the total power being used.

Now if instead if a DC to DC step down design had been used then yes indeed the situation would had been far better as far as power being thrown away is concern and the voltage in compared to the voltage out would not have such a linear relationship to the power being wasted.
I am not kidding you in any way whatsoever. I stand 100% by what I said. The fact that you wish to reference current calculations is pure straw man. Moreover I expect that you quote me in full should you wish me to explain further (not that I overly mind explaining).
BillRM
 
  0  
Reply Wed 31 Dec, 2014 04:12 pm
@Chumly,
Quote:
I stand 100% by what I said. The fact that you wish to reference current calculations is pure straw man. Moreover I expect that you quote me in full should you wish me to explain further (not that I overly mind explaining).


Somehow I did not think that I would need to explain ohm law but here is a link if you feel you need to bush up on the subject.

http://www.csgnetwork.com/ohmslaw2.html
Chumly
 
  2  
Reply Sat 3 Jan, 2015 12:20 pm
@BillRM,
At the risk of argumentum ad nauseum, and bearing in mind your lack of comprehension as to my simple stated facts:

I stand 100% by what I said. The fact that you wish to reference current calculations is pure straw man.

Firstly simple zener regulator circuits always waste power and secondly the difference between the source voltage and the load voltage does not in and of itself dictate the power loss due to voltage regulation.
BillRM
 
  0  
Reply Sat 3 Jan, 2015 02:02 pm
@Chumly,
Quote:
difference between the source voltage and the load voltage does not in and of itself dictate the power loss due to voltage regulation.


In linear power supplies the difference between the source and load voltage sure the hell dictate the power losses for any given load.
0 Replies
 
 

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