Is there a device I can add to the end of a 16.9 V, 150 amp supply That will allow me to turn it down to 15.0 volts and keep the ability to supply 150 amps?
-
You can now help support WorldwideDX when you shop on Amazon at no additional cost to you! Simply follow this Shop on Amazon link first and a portion of any purchase is sent to WorldwideDX to help with site costs.
Lower voltage keep current
- Thread starter Larry A McClanahan
- Start date
There is a technique that can be used, but it's a transformer inserted into the 120-Volt end of the power supply.
Naturally if you plugged it into a Variac, variable AC transformer you could turn down the AC power-line voltage to get 15 Volts.
15 Volts is 88 percent of 16.9. If we feed 88 percent of 120 Volts into the primary, this would be roughly 106 Volts.
120 minus 106 is 14 Volts. Inserting a 14-Volt AC transformer secondary in series with your power supply's 120-Volt primary winding would have the same effect. But you have to wire that 14-Volt secondary winding with the correct phase. If you turn the leads of the 14-Volt secondary one way, you'll boost it 14 Volts higher. This is called a "series aiding" hookup. Turn it the other way and it will subtract 14 Volts from the 120, a trick called "buck" hookup.
The primary of this theoretical 14-Volt transformer gets hooked to the main 120-Volt source. The current rating of the 14-Volt winding should be more or less equal to the primary current you draw from the 120-Volt outlet.
Hmmm. Just thought to do the arithmetic for a 150Amp output. Would draw more than 22 Amps from a 120 outlet. Not legal.
But a 240 outlet would need only supply 11 Amps. Makes me suspect your power supply is wired this way. If so, just double the voltage numbers in the previous example, and cut the current in half. Makes it sound as if a 28-Volt 10 Amp or so transformer would buck your 240-Volt primary voltage down to where you want it.
73
Naturally if you plugged it into a Variac, variable AC transformer you could turn down the AC power-line voltage to get 15 Volts.
15 Volts is 88 percent of 16.9. If we feed 88 percent of 120 Volts into the primary, this would be roughly 106 Volts.
120 minus 106 is 14 Volts. Inserting a 14-Volt AC transformer secondary in series with your power supply's 120-Volt primary winding would have the same effect. But you have to wire that 14-Volt secondary winding with the correct phase. If you turn the leads of the 14-Volt secondary one way, you'll boost it 14 Volts higher. This is called a "series aiding" hookup. Turn it the other way and it will subtract 14 Volts from the 120, a trick called "buck" hookup.
The primary of this theoretical 14-Volt transformer gets hooked to the main 120-Volt source. The current rating of the 14-Volt winding should be more or less equal to the primary current you draw from the 120-Volt outlet.
Hmmm. Just thought to do the arithmetic for a 150Amp output. Would draw more than 22 Amps from a 120 outlet. Not legal.
But a 240 outlet would need only supply 11 Amps. Makes me suspect your power supply is wired this way. If so, just double the voltage numbers in the previous example, and cut the current in half. Makes it sound as if a 28-Volt 10 Amp or so transformer would buck your 240-Volt primary voltage down to where you want it.
73
I would rather adjust output voltage inside PS.
Does this PS have a name? Maybe adjusting will be easier than we think?
Mike
Does this PS have a name? Maybe adjusting will be easier than we think?
Mike
Last edited:
No, I have a 16.9 V automotive battery that I would like to be able to cut back to 15 volts and run my main amp in my vehicle on it.
Buy and use some Bridge Rectifies - Industrial style to handle the amps...
The exact outline is just for reference.
Why Bridge?
Well, each diode used, has an inherent voltage drop across it's junction - you can pretty much plan on about 1 volt drop at up to 150 amps or rated terminal capacity - these can be stacked (run n series) to attain any amount of drop you wish - each unit will provide a given voltage drop with little loss to heat because of the Semiconductor substrate - not a carbon composition - you're just passing power thru a PN junction.
It's a bit cost prohibitive, but where there is a will, there is a way...
The exact outline is just for reference.
Why Bridge?
Well, each diode used, has an inherent voltage drop across it's junction - you can pretty much plan on about 1 volt drop at up to 150 amps or rated terminal capacity - these can be stacked (run n series) to attain any amount of drop you wish - each unit will provide a given voltage drop with little loss to heat because of the Semiconductor substrate - not a carbon composition - you're just passing power thru a PN junction.
It's a bit cost prohibitive, but where there is a will, there is a way...
Have you checked the actual voltage the battery delivers with a 150-Amp load?
Probably falls very close to your 15-Volt target just from normal voltage drop under load.
73
Probably falls very close to your 15-Volt target just from normal voltage drop under load.
73
Andy has the right idea, using forward-biased rectifier diodes to lose a couple of Volts. But remember the current rating on a bridge-rectifier diode is for using it as a rectifier. In that service, each diode is turned on and dumping heat only half the time. The rest of the time it's shut off, reverse biased with no real current passing through it.
We learned to cut the current rating in half when using individual diodes in a DC circuit where it throws heat 100% of the time.
A bridge has two series pairs of diodes in parallel. If it says 25 Amps you can probably get away with 25 Amps of DC current through it. The two current paths will hopefully split the load evenly 50-50. But a set of diodes to take 150 Amps DC is a different nut to crack. 300-Amp rectifiers are out there. Used some in the last unregulated power supply we built here.
Oddly enough, you connect the positive side of your 16.9 Volt source to the terminal on a bridge package that's marked minus (-). The plus (+) terminal becomes the output side.
73
We learned to cut the current rating in half when using individual diodes in a DC circuit where it throws heat 100% of the time.
A bridge has two series pairs of diodes in parallel. If it says 25 Amps you can probably get away with 25 Amps of DC current through it. The two current paths will hopefully split the load evenly 50-50. But a set of diodes to take 150 Amps DC is a different nut to crack. 300-Amp rectifiers are out there. Used some in the last unregulated power supply we built here.
Oddly enough, you connect the positive side of your 16.9 Volt source to the terminal on a bridge package that's marked minus (-). The plus (+) terminal becomes the output side.
73
This has absolutely nothing to do with the subject at hand but my grandmother's maiden name was McClanahan.