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Reduce PS voltage

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jonlaury

Joined: 06 Nov 2006
Posts: 336

Posted: Sat Oct 22, 2011 7:58 am Post subject: Reduce PS voltage

I'm trying to reform the caps in an old unused Whelen strobe PS by using an independent PS.

A Whelen tech suggested powering up the Whelen PS at 6vdc to see if it works. And if it does, then power up to 12v

Whelen's white paper says use 9v for 15 min, then 12v.



Other people have reformed in steps, 6v, 9v, 12v.



How do I step down the voltage of my independent 13.5 PS?



The Whelen PS says it uses 7a. I tried using 4 'C' cells in series but the voltage dropped to mV, so I'm deducing that I need an independent supply capable of more amperage. At 6v, will I be trying to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a big honkin' resistor to the 13.5 v supply. Don't know if there are resistors that big.  I think there must be a more sophisticated way to do this.



Thanks,

John

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klehman(at)albedo.net
Guest

Posted: Sat Oct 22, 2011 9:59 am Post subject: Reduce PS voltage

I think I used a 12 volt headlight bulb as a "resistor" when I did this

but too long ago to be sure. IIRC the average current is considerably 

less than 7 amps.  Suspect I would have paralleled a second bulb for the

intermediate voltage. I always keep a spare headlight bulb or two in 

stock. Other things that you can play with include resistance 

appliances. A 1500 watt kettle would be around 1500/120v = 12.5 ohms. 

Two in parallel would be 6.25 ohms.

Ken



On 22/10/2011 11:58 AM, jonlaury wrote:

 	  Quote: 	
		   

 "jonlaury"<jonlaury(at)impulse.net>



 I'm trying to reform the caps in an old unused Whelen strobe PS by

 using an independent PS. A Whelen tech suggested powering up the

 Whelen PS at 6vdc to see if it works. And if it does, then power up

 to 12v Whelen's white paper says use 9v for 15 min, then 12v.



 Other people have reformed in steps, 6v, 9v, 12v.



 How do I step down the voltage of my independent 13.5 PS?



 The Whelen PS says it uses 7a. I tried using 4 'C' cells in series

 but the voltage dropped to mV, so I'm deducing that I need an

 independent supply capable of more amperage. At 6v, will I be trying

 to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a

 big honkin' resistor to the 13.5 v supply. Don't know if there are

 resistors that big.  I think there must be a more sophisticated way

 to do this.



 Thanks, John

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retasker(at)optonline.net
Guest

Posted: Sat Oct 22, 2011 4:13 pm Post subject: Reduce PS voltage

Good ideas, but your calculations for a 1500 watt kettle are wrong.  You have calculated the current consumption (Current=Watts/Voltage).  To calculate ohms use: watts = (voltage * voltage) / 

resistance or more specifically resistance = (voltage * voltage) / watts.  Or in this case (120*120)/1500 = 9.6 ohms.  It is probably somewhat less than this since it won't heat to the normal 

operating temperature and the resistance goes up as it heats.

Dick Tasker

Ken wrote:

 	  Quote: 	

 I think I used a 12 volt headlight bulb as a "resistor" when I did this

 but too long ago to be sure. IIRC the average current is considerably less than 7 amps.  Suspect I would have paralleled a second bulb for the

 intermediate voltage. I always keep a spare headlight bulb or two in stock. Other things that you can play with include resistance appliances. A 1500 watt kettle would be around 1500/120v = 12.5 

 ohms. Two in parallel would be 6.25 ohms.

 Ken

 On 22/10/2011 11:58 AM, jonlaury wrote:

> 

> "jonlaury"<jonlaury(at)impulse.net>

>

> I'm trying to reform the caps in an old unused Whelen strobe PS by

> using an independent PS. A Whelen tech suggested powering up the

> Whelen PS at 6vdc to see if it works. And if it does, then power up

> to 12v Whelen's white paper says use 9v for 15 min, then 12v.

>

> Other people have reformed in steps, 6v, 9v, 12v.

>

> How do I step down the voltage of my independent 13.5 PS?

>

> The Whelen PS says it uses 7a. I tried using 4 'C' cells in series

> but the voltage dropped to mV, so I'm deducing that I need an

> independent supply capable of more amperage. At 6v, will I be trying

> to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a

> big honkin' resistor to the 13.5 v supply. Don't know if there are

> resistors that big.  I think there must be a more sophisticated way

> to do this.

>

> Thanks, John

-- 

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klehman(at)albedo.net
Guest

Posted: Sat Oct 22, 2011 6:20 pm Post subject: Reduce PS voltage

Yes thanks - I should read before sending.



1500 watts/120volts = 12.5 AMPS



120volts/12.5 amps=9.6 ohms



Ken



On 22/10/2011 8:08 PM, Richard E. Tasker wrote:

 	  Quote: 	
		   

 <retasker(at)optonline.net>



 Good ideas, but your calculations for a 1500 watt kettle are wrong. You

 have calculated the current consumption (Current=Watts/Voltage). To

 calculate ohms use: watts = (voltage * voltage) / resistance or more

 specifically resistance = (voltage * voltage) / watts. Or in this case

 (120*120)/1500 = 9.6 ohms. It is probably somewhat less than this since

 it won't heat to the normal operating temperature and the resistance

 goes up as it heats.



 Dick Tasker



 Ken wrote:

> 

>

> I think I used a 12 volt headlight bulb as a "resistor" when I did this

> but too long ago to be sure. IIRC the average current is considerably

> less than 7 amps. Suspect I would have paralleled a second bulb for the

> intermediate voltage. I always keep a spare headlight bulb or two in

> stock. Other things that you can play with include resistance

> appliances. A 1500 watt kettle would be around 1500/120v = 12.5 ohms.

> Two in parallel would be 6.25 ohms.

> Ken

>

> On 22/10/2011 11:58 AM, jonlaury wrote:

>> 

>> "jonlaury"<jonlaury(at)impulse.net>

>>

>> I'm trying to reform the caps in an old unused Whelen strobe PS by

>> using an independent PS. A Whelen tech suggested powering up the

>> Whelen PS at 6vdc to see if it works. And if it does, then power up

>> to 12v Whelen's white paper says use 9v for 15 min, then 12v.

>>

>> Other people have reformed in steps, 6v, 9v, 12v.

>>

>> How do I step down the voltage of my independent 13.5 PS?

>>

>> The Whelen PS says it uses 7a. I tried using 4 'C' cells in series

>> but the voltage dropped to mV, so I'm deducing that I need an

>> independent supply capable of more amperage. At 6v, will I be trying

>> to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a

>> big honkin' resistor to the 13.5 v supply. Don't know if there are

>> resistors that big. I think there must be a more sophisticated way

>> to do this.

>>

>> Thanks, John

>>

>>

>

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