Power Supply
Power Supply
I don't understand why a larger resistor, allows a greater voltage, in this schematic?


Rickey- Member
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Re: Power Supply
I'm sorry, but, that schematic is way too small to read. Can you enlarge it so we cans see it, and, read values of components?
Thanks.
Bill Cahill
Thanks.
Bill Cahill
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Bill Cahill- Admin
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Re: Power Supply
Click on the photo Bill and it will enlarge.
Tony V- Moderator
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Re: Power Supply
Ah. Thanks, Tony.
Simple.......
R1 Is a suge protecting resistor.
R2 is for lowering Voltage to second filter.
R3 is a by pass resistor used in power supply. When you use a higher resistance there, you get a higher voltage.
Use a lower resistance, and, you get the lower voltage.
Since that resistor bypasses the second filter, that is, it's accross it, the lower resistance means less voltage is able to get to that section, as it puts a load on the power supply.
Hope this helps.
Bill Cahill
Simple.......
R1 Is a suge protecting resistor.
R2 is for lowering Voltage to second filter.
R3 is a by pass resistor used in power supply. When you use a higher resistance there, you get a higher voltage.
Use a lower resistance, and, you get the lower voltage.
Since that resistor bypasses the second filter, that is, it's accross it, the lower resistance means less voltage is able to get to that section, as it puts a load on the power supply.
Hope this helps.
Bill Cahill
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Bill Cahill- Admin
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Re: Power Supply
Thanks Bill, that makes sense!
Rickey- Member
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Re: Power Supply
Always happy to help........
Bill Cahill
Bill Cahill
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Bill Cahill- Admin
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Re: Power Supply
What would be the voltage with no resistor there?
Rickey- Member
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Re: Power Supply
Not sure, but, it would be high.
Bill Cahill
Bill Cahill
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Bill Cahill- Admin
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Re: Power Supply
Read up on Voltage Dividers and you'll learn something new.
repairtech- Member
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Re: Power Supply
Larger resistors (not physical size) mean, The higher value of resistance the less current will flow, since the last resistor is across the line, it diverts some current (like a short circuit to put it in simple terms), so a smaller resistor (less resistance) will act more like a bigger short (more current).
Just think of it in theses terms, Run a toaster, do the lights dim? That is a Higher resistance that drops some voltage, Now turn on an electric Furnace it has lower resistance and for a fraction of a second you will notice the Lights dim, that is because more current is being shunted and the voltage drops more at that second. That is why in your question why larger value resistances draw less current and that allows the voltage to remain Higher because of less of a load on the circuit. And note that the resistor is across (parallel to)the line acting as a load.
The resistors that are in series and connected to the other components limit the current through the circuit. This protects the diode from burning out, and also helps smooth out the filter capacitors and prevent voltage sags while the capacitors are charging and discharging. The capacitors also will boost the voltage some what. The first capacitor by the input is a small value and is used to filter out line noise and RF.
To get a better concept it was suggested to research voltage dividers which is a very good Idea. But also under stand that resistors cause voltage drops and / or restrict current flow depending on the circuit design.
It goes back to ohms law. If you take two or more resistors in series, the voltage is limited in one by the current flow. So if you have 1 volt and 2 resistors, with the value of each resistor being 1 ohm. Using ohms law you need to find the voltage drop across each resistor.
you frst need to find the total current through the circuit.
E=1 r1 = 1 r2 = 1
so add the resistances since its a series circuit.
r1 + r2 or 1+1 = 2
That is the total series resistance
Now apply ohms law to find the total current
we need to find It
It = Et/Rt = 1v/2ohms = .5amps
That is the total flow through a series circuit
now we take that current of .5amps
and again use ohms law to find the individual voltage drops of both resistors in series.
so I1 X R1 = .5amps X 1ohm = .5volts
since both resistors pass the same current and are the same resistance value you have the same voltage drop across each resistor. So if you connected a wire between the two resistors you would measure 1/2 volt. But across both resistors you would measure 1 volt. This is the principle of voltage dividers. Try ohm's law with just changing the value of only one resistor and do the calculation again using that set of formulas. You will find out that the voltage drop will be different across both resistors.
The two last resistors act as a voltage divider in the same way the resistor on top limits the current and the B+ connection in the diagram is like the wire between the two resistors in my example. Changing the value of the bottom resistor changes the supply voltage which is in essence the supply voltage by voltage drop
Whew!!!
Hope this helps when you read about voltage divider circuits.
Just think of it in theses terms, Run a toaster, do the lights dim? That is a Higher resistance that drops some voltage, Now turn on an electric Furnace it has lower resistance and for a fraction of a second you will notice the Lights dim, that is because more current is being shunted and the voltage drops more at that second. That is why in your question why larger value resistances draw less current and that allows the voltage to remain Higher because of less of a load on the circuit. And note that the resistor is across (parallel to)the line acting as a load.
The resistors that are in series and connected to the other components limit the current through the circuit. This protects the diode from burning out, and also helps smooth out the filter capacitors and prevent voltage sags while the capacitors are charging and discharging. The capacitors also will boost the voltage some what. The first capacitor by the input is a small value and is used to filter out line noise and RF.
To get a better concept it was suggested to research voltage dividers which is a very good Idea. But also under stand that resistors cause voltage drops and / or restrict current flow depending on the circuit design.
It goes back to ohms law. If you take two or more resistors in series, the voltage is limited in one by the current flow. So if you have 1 volt and 2 resistors, with the value of each resistor being 1 ohm. Using ohms law you need to find the voltage drop across each resistor.
you frst need to find the total current through the circuit.
E=1 r1 = 1 r2 = 1
so add the resistances since its a series circuit.
r1 + r2 or 1+1 = 2
That is the total series resistance
Now apply ohms law to find the total current
we need to find It
It = Et/Rt = 1v/2ohms = .5amps
That is the total flow through a series circuit
now we take that current of .5amps
and again use ohms law to find the individual voltage drops of both resistors in series.
so I1 X R1 = .5amps X 1ohm = .5volts
since both resistors pass the same current and are the same resistance value you have the same voltage drop across each resistor. So if you connected a wire between the two resistors you would measure 1/2 volt. But across both resistors you would measure 1 volt. This is the principle of voltage dividers. Try ohm's law with just changing the value of only one resistor and do the calculation again using that set of formulas. You will find out that the voltage drop will be different across both resistors.

The two last resistors act as a voltage divider in the same way the resistor on top limits the current and the B+ connection in the diagram is like the wire between the two resistors in my example. Changing the value of the bottom resistor changes the supply voltage which is in essence the supply voltage by voltage drop
Whew!!!
Hope this helps when you read about voltage divider circuits.
Resistance is Futile- Member
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Re: Power Supply
Thanks Cliff, I appreciate your time and patience!
Rickey
Rickey
Rickey- Member
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Re: Power Supply
Thanks!repairtech wrote:Read up on Voltage Dividers and you'll learn something new.
Rickey- Member
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Age : 66
Registration date : 2011-09-06
Re: Power Supply
Rickey,
the voltage there without the R3? is that what you are asking? if that is the question its about 165 to 169 volts. I built a few of these and experimented with them with different resistance on the R3
~jeff~
the voltage there without the R3? is that what you are asking? if that is the question its about 165 to 169 volts. I built a few of these and experimented with them with different resistance on the R3
~jeff~
Vintage Tunes- New Member
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Re: Power Supply
R2 and R3 form a voltage divider. The smaller the value of the lower resistor, the lower the voltage across it.
- Leigh
- Leigh
Leigh- Member
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Re: Power Supply
One thing to watch for based on the schematic presented--in the very least you should use a polarized AC plug as one side of the line is connected directly to the output making this a "hot" circuit and you could run the risk of a shock just like when dealing with ac/dc radios.
Dr. Radio- Member
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Re: Power Supply
Thanks, it's a little scary!Dr. Radio wrote:One thing to watch for based on the schematic presented--in the very least you should use a polarized AC plug as one side of the line is connected directly to the output making this a "hot" circuit and you could run the risk of a shock just like when dealing with ac/dc radios.
Rickey- Member
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Re: Power Supply
Dr. Radio wrote:One thing to watch for based on the schematic presented--in the very least you should use a polarized AC plug as one side of the line is connected directly to the output making this a "hot" circuit and you could run the risk of a shock just like when dealing with ac/dc radios.
Ahhhhhh....
the schematic pictured is a DC output for battery radios. If you plug it in you get 90 volts DC (or 67 vdc, depending on the resistor used) and if you turn the plug around....you still get the same output. Its no different. There is no more chance for a "hot chassis" with or without a polarized plug. The only thing that will change is that the negative and the positive will change poles and be either a positive 90vdc output or a negative vdc output. The polarized plug will have nothing to do with it.
The radio will not work if you hook it up incorrectly but it has no greater chance of a "hot chassis" either way.
Its still DC 90 (or 67) no matter how you plug it in
That is not to say that 90vdc won't hurt and you should probably avoid it if you can but a polarized plug will make no difference what so ever.
I just a non-polarized plug and just paint a little red dot on the prong that goes to the hot side of the plug (the small hole, not the big hole) but then again, I am the only one that uses them. If I were to sell them I would use a polarized plug so that the new owner would not have to be so careful in plugging it in as to make sure the correct polarization output is correct to the radio, but the polarization of the plug will NOT prevent a "hot chassis" on a battery (DC) radio.
Jeff
Vintage Tunes- New Member
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Re: Power Supply
Jeff,
I don't want to sound rude, but I think you need to review your electronics theory...
First off, this is a "hot" chassis risk--I'm referring to the circuit itself, when connected to the radio, that may be an entire additional risk to analyze.
Look carefully at the line cord connection. One side of the line IS DIRECTLY connected to the B- connection of this circuit. This means from start (at the wall outlet) to finish (the B- connection to connect into the radio) you are putting yourself at risk for a nasty electrical shock if the "hot" side of the 120 vac ac outlet is the B- line, either by chance or an incorrectly wired outlet. This circuit needs a polarized plug in the very least, that way you don't have a DIRECT connection to the hot side of the outlet. Really a transformer circuit is best (safer). And turning the plug around will not reverse polarity of the output. You have will not have "negative dc output".
Not the greatest design. C3 should not be a "ceramic" as indicated. This really needs to be a X2 safety capacitor since it is directly across the line input.
I don't want to sound rude, but I think you need to review your electronics theory...
First off, this is a "hot" chassis risk--I'm referring to the circuit itself, when connected to the radio, that may be an entire additional risk to analyze.
Look carefully at the line cord connection. One side of the line IS DIRECTLY connected to the B- connection of this circuit. This means from start (at the wall outlet) to finish (the B- connection to connect into the radio) you are putting yourself at risk for a nasty electrical shock if the "hot" side of the 120 vac ac outlet is the B- line, either by chance or an incorrectly wired outlet. This circuit needs a polarized plug in the very least, that way you don't have a DIRECT connection to the hot side of the outlet. Really a transformer circuit is best (safer). And turning the plug around will not reverse polarity of the output. You have will not have "negative dc output".
Not the greatest design. C3 should not be a "ceramic" as indicated. This really needs to be a X2 safety capacitor since it is directly across the line input.
Dr. Radio- Member
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Re: Power Supply
I've seen this schematic floating around the internet for a while.
Presumably, this is to serve as a line-in replacement for the now obsolete batteries that powered sets requiring 90 and 67 Volts, etc.
Antique Electronic Supply has a kit that provides these voltages (K-101A) ... and ... uses a transformer rather than a 'cutesy' way to step down the voltage straight from the wall.
Presumably, this is to serve as a line-in replacement for the now obsolete batteries that powered sets requiring 90 and 67 Volts, etc.
Antique Electronic Supply has a kit that provides these voltages (K-101A) ... and ... uses a transformer rather than a 'cutesy' way to step down the voltage straight from the wall.
NashvilleRad- Member
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