The relay would be used to remove the power supply from the volt/amp meter that is mounted to the front of the enclosure. It currently stays on for about 30 seconds while the larger capacitor discharges.
Koford (Slot Car) Bench Power Supply Project
- Thread starter TheChad
- Start date
Okay, let's side track on the relay thing for a minute...
I connected the old 1k rheostat between the terminals of the capacitor, and turned the rheostat up to the 1k setting... It drained the capacitor in about 20 seconds (I didn't actually time this one, just a guess). It was much quicker...
I don't know what kind of wattage the rheostat is? 1/2 watt, 1 watt? or how much different that makes vs the ohm rating?
So what I'm thinking is what if I put say a 1w 750ohm resistor? What effect will this have on the capacitor itself? Will it even make a noticeable/measurable difference when powered on?
The capacitor bleed's off from ~19v down to under ~10v pretty quickly, but after it gets under ~7v it bleeds off much slower... When I had the rheostat (1k ohm) attached, It bleed off the sub 7v much quicker, more consistent with the +10v bleed off.
What difference does the higher or lower wattage of the resistor make vs the ohms? The other thing going threw my mind, is obviously this resistor will be burning off power the entire time the power supply is powered up... so how hot will this resistor get? Is that where a higher wattage resistor would come in?
Thanks!
-TheChad
I connected the old 1k rheostat between the terminals of the capacitor, and turned the rheostat up to the 1k setting... It drained the capacitor in about 20 seconds (I didn't actually time this one, just a guess). It was much quicker...
I don't know what kind of wattage the rheostat is? 1/2 watt, 1 watt? or how much different that makes vs the ohm rating?
So what I'm thinking is what if I put say a 1w 750ohm resistor? What effect will this have on the capacitor itself? Will it even make a noticeable/measurable difference when powered on?
The capacitor bleed's off from ~19v down to under ~10v pretty quickly, but after it gets under ~7v it bleeds off much slower... When I had the rheostat (1k ohm) attached, It bleed off the sub 7v much quicker, more consistent with the +10v bleed off.
What difference does the higher or lower wattage of the resistor make vs the ohms? The other thing going threw my mind, is obviously this resistor will be burning off power the entire time the power supply is powered up... so how hot will this resistor get? Is that where a higher wattage resistor would come in?
Thanks!
-TheChad
A Capacitor discharge rate is non-linear, so it will discharge quickly at first and taper off.
You can see that for every t amount of time, the charge will halve. Which eventually results in it only loosing 25%, then 12.5% then 6%... and
To mess with numbers, if you use a 750Ω resistor across the capacitor, you will be providing ~15V to it directly from the transformer.
So to do some basic math, we can determine that there will be 15V / 750Ω = 20mA of current passing through the resistor as a result of the voltage across it's pins.
We can use that to determine the power required for the resistor now. At 20mA, and 15V we end up with 15V * 0.2A = 0.3Watts
So a quarter watt resistor is not rated high enough, and will burn out, but a half watt should hold up. There is no downside that I am aware of for using a resistor that's rated for a higher wattage than required other than price.
This proposed solution will constantly draw an extra 0.3Watts that will no longer be available for your motors. The smaller the resistor Ω value, the more current it will draw, which will also result in it getting hotter.
Power = Voltage * Current
Current = Voltage / Resistance
Therefore Power = Voltage^2 / Resistance
Be very careful with using potentiometers like this... because there will be a point where you lower the resistance too much and the current that flows through will damage it... I've seen one of my peers at school release the smoke from one bad enough to watch a flicker of a flame shoot out...
Also, as you decrease the resistance, the affect of the capacitor will diminish. You have selected quite a large capacitor though due to the current draw requirements which is ideal... but to have it drain quickly you would require a smaller one, or wasting more energy through a capacitor... (I would leave the large one in there and explore other options like a resistor value, or a relay...)
You can see that for every t amount of time, the charge will halve. Which eventually results in it only loosing 25%, then 12.5% then 6%... and
To mess with numbers, if you use a 750Ω resistor across the capacitor, you will be providing ~15V to it directly from the transformer.
So to do some basic math, we can determine that there will be 15V / 750Ω = 20mA of current passing through the resistor as a result of the voltage across it's pins.
We can use that to determine the power required for the resistor now. At 20mA, and 15V we end up with 15V * 0.2A = 0.3Watts
So a quarter watt resistor is not rated high enough, and will burn out, but a half watt should hold up. There is no downside that I am aware of for using a resistor that's rated for a higher wattage than required other than price.
This proposed solution will constantly draw an extra 0.3Watts that will no longer be available for your motors. The smaller the resistor Ω value, the more current it will draw, which will also result in it getting hotter.
Power = Voltage * Current
Current = Voltage / Resistance
Therefore Power = Voltage^2 / Resistance
Be very careful with using potentiometers like this... because there will be a point where you lower the resistance too much and the current that flows through will damage it... I've seen one of my peers at school release the smoke from one bad enough to watch a flicker of a flame shoot out...
Also, as you decrease the resistance, the affect of the capacitor will diminish. You have selected quite a large capacitor though due to the current draw requirements which is ideal... but to have it drain quickly you would require a smaller one, or wasting more energy through a capacitor... (I would leave the large one in there and explore other options like a resistor value, or a relay...)
As we know at this point, I have much more power in this supply than I need or use.. I only need 12v and will never use more than 6a (Which I probably will never even use that much)... So it sounds like the correct size resistor on the Capacitor is the best solution (no?)..
I would get either a 1/2 watt or a 1 watt resistor, which should be large enough per your last post..
I used the potentiometer just to see what would happen, it got barely warm, but I don't care if it releases the smoke, because it's broke anyway and it's the old potentiometer.. But thanks for the heads up!
How much of the "affect of the capacitor" is diminishing? Since the capacitor is oversized anyway, I'm thinking it shouldn't even be a noticeable difference, but I could be wrong?
I am thinking if the top of the waveform is 100%, with the capacitor I have now, the drop between wave's is let's just say 2% (made up number)
with the resistor installed on the capacitor, would we be talking a drop between wave's to 3% or 5% or 20%? would it be even less like 0.25%?
With the rheostat set to 100 Ohm, the capacitor bleed's off in about 5-6 Seconds, which is acceptable to me. The voltage of the capacitor with the power on (No load) drops from ~19.36v to ~19.18v.
I'm considering ordering 25, 50, 75 & 100 Ohm 3w resistors.. I figure 3w will give plenty of safety window.
P.S. The magic number that released the smoke was 50 Ohm's.. LOL
-TheChad
I would get either a 1/2 watt or a 1 watt resistor, which should be large enough per your last post..
I used the potentiometer just to see what would happen, it got barely warm, but I don't care if it releases the smoke, because it's broke anyway and it's the old potentiometer.. But thanks for the heads up!
How much of the "affect of the capacitor" is diminishing? Since the capacitor is oversized anyway, I'm thinking it shouldn't even be a noticeable difference, but I could be wrong?
I am thinking if the top of the waveform is 100%, with the capacitor I have now, the drop between wave's is let's just say 2% (made up number)
with the resistor installed on the capacitor, would we be talking a drop between wave's to 3% or 5% or 20%? would it be even less like 0.25%?With the rheostat set to 100 Ohm, the capacitor bleed's off in about 5-6 Seconds, which is acceptable to me. The voltage of the capacitor with the power on (No load) drops from ~19.36v to ~19.18v.
I'm considering ordering 25, 50, 75 & 100 Ohm 3w resistors.. I figure 3w will give plenty of safety window.
P.S. The magic number that released the smoke was 50 Ohm's.. LOL
-TheChad
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Okay,
I ran by Radio Shack today and picked up a 100ohm 1-watt resistor, with the resistor installed, I get ~19.00v (+/- 0.12v) output from the capacitor. The capacitor bleeds off in ~3 seconds, which is perfectly acceptable.
So my only consideration left is: 1. What wattage should I order a 2, 3, 5-watt? (The 1w resistor was pretty hot, no load) 2. How much of the "smoothing effect of the capacitor" am I loosing by having a 100-ohm resistor installed across the capacitor?
It had a 33,000uf Capacitor from the factory running 6.3v 6A, I now have a 15000uF capacitor running 12.6v 8A, but we had previously determined that the 33,000uf was WAY overkill and the 15,000uF was still pretty big for the application.. But what's the realistic loss of the resistor across the capacitor? Am I really hurting it's smoothing affect? or is it a negligible amount?
If I am appling your above math correctly, I am taking 19v / 100Ω = 1.9A? so 19v * 1.9 = 36.1, which I'm guessing is suppose to be 3.61 Watts? If that is right, then I need a 5 watt resistor? and I am effectively wasting 1.9A of my 8A available? Giving me 6.1 available amps to my test leads? Is that even remotely correct? I am using 19v because that's what the voltage measures at the capacitor @ no load.
Thanks!
-TheChad
I ran by Radio Shack today and picked up a 100ohm 1-watt resistor, with the resistor installed, I get ~19.00v (+/- 0.12v) output from the capacitor. The capacitor bleeds off in ~3 seconds, which is perfectly acceptable.
So my only consideration left is: 1. What wattage should I order a 2, 3, 5-watt? (The 1w resistor was pretty hot, no load) 2. How much of the "smoothing effect of the capacitor" am I loosing by having a 100-ohm resistor installed across the capacitor?
It had a 33,000uf Capacitor from the factory running 6.3v 6A, I now have a 15000uF capacitor running 12.6v 8A, but we had previously determined that the 33,000uf was WAY overkill and the 15,000uF was still pretty big for the application.. But what's the realistic loss of the resistor across the capacitor? Am I really hurting it's smoothing affect? or is it a negligible amount?
If I am appling your above math correctly, I am taking 19v / 100Ω = 1.9A? so 19v * 1.9 = 36.1, which I'm guessing is suppose to be 3.61 Watts? If that is right, then I need a 5 watt resistor? and I am effectively wasting 1.9A of my 8A available? Giving me 6.1 available amps to my test leads? Is that even remotely correct? I am using 19v because that's what the voltage measures at the capacitor @ no load.
Thanks!
-TheChad
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First off, 19V / 100Ω is actually 0.19A 
I can't remember what the original capacitor was sized for, but I seem to recall that it was a little larger than required for the 8A the transformer is capable of. So it would not negatively impact the use of the capacitor as long as the current draw for the resistor + the current draw of the motor stays well below what the capacitor is sized for.
Because of this, I am sure you can easily pull half an amp and be fine. Considering you are pulling less than half of this now, I'm willing to bet this is a reasonable choice.
I should also mention that the power formula I gave you of Power = Voltage^2 / Resistance means that the power dissipated will exponentially grow with voltage... so 15V under load, and 19V would be a pig difference. Measuring the capacitor with nothing attached will give you a much closer reading to the peak voltage it sees. Once you have some kind of resistance on it, measure again and you will end up getting more of an average voltage which is what the resistor will be dissipating. Always oversize the wattage rating, and you can always use more than one resistor in series or parallel to divide the power dissipation between the two allowing you to use smaller rated resistors.
I can't remember what the original capacitor was sized for, but I seem to recall that it was a little larger than required for the 8A the transformer is capable of. So it would not negatively impact the use of the capacitor as long as the current draw for the resistor + the current draw of the motor stays well below what the capacitor is sized for.
Because of this, I am sure you can easily pull half an amp and be fine. Considering you are pulling less than half of this now, I'm willing to bet this is a reasonable choice.
I should also mention that the power formula I gave you of Power = Voltage^2 / Resistance means that the power dissipated will exponentially grow with voltage... so 15V under load, and 19V would be a pig difference. Measuring the capacitor with nothing attached will give you a much closer reading to the peak voltage it sees. Once you have some kind of resistance on it, measure again and you will end up getting more of an average voltage which is what the resistor will be dissipating. Always oversize the wattage rating, and you can always use more than one resistor in series or parallel to divide the power dissipation between the two allowing you to use smaller rated resistors.
Okay, so the capacitor charged (power on) no load measures ~19.6v +/- 0.2v... The capacitor with the 1 watt 100-ohm resistor installed, measures ~19, +/- 0.12v..
So, the wattage I got is still right? 3.61 watts? So a 5w resistor would be the proper size?
Thanks!
-TheChad
So, the wattage I got is still right? 3.61 watts? So a 5w resistor would be the proper size?
Thanks!
-TheChad
Yes, your wattage is correct. So you could use a single 5W 100Ω, or I'm sure using two 2W 200Ω in parallel will work fine for you.
I know we took off on a bit of a tangent with the relay. Everyone on here has different standards and views on the absolute best practice, and while much of it is gold and should be followed, sometimes the easier methods work just as well for specific applications.
- Joined
- Jan 21, 2010
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I would use wirewound resistors in this application as they are cap, easily available, and rugged.
Incidentally, if you were still using a relay, you could leave the resistor disconnected while power was applied. This would reduce the power wasted and the heart generated.
Incidentally, if you were still using a relay, you could leave the resistor disconnected while power was applied. This would reduce the power wasted and the heart generated.
http://www.mouser.com/ProductDetail...=sGAEpiMZZMtbXrIkmrvidCzA/KXd8dIybAo0bkBKWM8=
This is what I am looking at ordering, is that what you recommend?
-TheChad
Okay, I ordered the 100Ω 5watt resistor, but while I'm waiting for it to come, I thought i would so some testing....
I ran one of my fast motors, it pulled ~4amps @3v..
I ran it for about 10 minutes, the power supply was room temp!
Nothing even warm!
-TheChad
I ran one of my fast motors, it pulled ~4amps @3v..
I ran it for about 10 minutes, the power supply was room temp!
Nothing even warm!
-TheChad
Haha, Switch mod supplies are great arent they?
They are harder to build, but can run soo much cooler.
Because you bought a pre-assembled module
Because you bought a pre-assembled module
For the price, I can't imagine it being any cheaper to make the module my self! Just the components on the module would have cost me more than the whole module did!
How can I tell if the output voltage is Fluctuating or if it's the motor causing the voltage/amperage to fluctuate?
When I had the motor hooked up, it would run steady, then you would hear the rpm's increase and the amperage change, the voltage showed pretty consistant, but it would float a little...
But I don't know for sure that it's not the motor causing this...
Thanks!
-TheChad
The relationship between voltage current and resistance makes it impossible for a power supply to control the amount of amperage and voltage at the same time.
If the voltage is constant, the amperage will completely rely on the resistance of the motor.
Likewise, if the amperage is constant, the voltage will rely on the resistance of the motor.
The + output on the DC/DC supply, should be directly connected to the + output of your powersupply and the + input on the meter.
The - was confirmed a few posts back to be common on the input and output of the DC/DC supply, so that side should be no concern, but you can always connect them both (I can't remember if you did or not) Just make sure that the - portion after the bridge is connected to the - for the DC/DC, the meter, and the output for your power supply.
It may possibly be something funny with the meter... as the current must flow through it in order to display the amperage. Any components that the current must flow through to get to/from the motor could be suspect, but before you go digging too deep I'll draw up a picture.
I forgot to put in the resistor...
It would simply be in parallel with the capacitor.
Now, looking at the diagram, assuming we're both on the same page.
The current will travel from the DC/DC converter through the AmpMeter portion of your meter prior to getting to you motor.
The Meter will also be measuring voltage at one of two points... Either directly from the output of the DC/DC converter which should be very stable, or from the output of the AmpMeter circuit in the meter.
(The diagram shows directly from the DC/DC converter)
If the circuit is built as drawn, and the motor changes speeds with a constant voltage, the AmpMeter circuit may be suspect if you cannot guarantee it's not the motor. The meter will simply show you the voltage on the inside of the case behind the AmpMeter... and not what may be on the output pins on your power supply. The AmpMeter circuit should not alter the output by anything noticeable... but it can't be guaranteed.
If you connect the VoltMeter positive lead directly to the + terminal on your power supply (which gets directly connected to the motor) then you can determine that it is in fact the motor... as you have a voltage directly on the motor to confirm. Any changes in voltage shown on the display could be a result of the DC/DC converter or the AmpMeter circuit in your meter.
Hopefully this give you more info and not more questions. Let me know if this all makes sense for you. I sometimes do not explain myself properly.
It would simply be in parallel with the capacitor.
Now, looking at the diagram, assuming we're both on the same page.
The current will travel from the DC/DC converter through the AmpMeter portion of your meter prior to getting to you motor.
The Meter will also be measuring voltage at one of two points... Either directly from the output of the DC/DC converter which should be very stable, or from the output of the AmpMeter circuit in the meter.
(The diagram shows directly from the DC/DC converter)
If the circuit is built as drawn, and the motor changes speeds with a constant voltage, the AmpMeter circuit may be suspect if you cannot guarantee it's not the motor. The meter will simply show you the voltage on the inside of the case behind the AmpMeter... and not what may be on the output pins on your power supply. The AmpMeter circuit should not alter the output by anything noticeable... but it can't be guaranteed.
If you connect the VoltMeter positive lead directly to the + terminal on your power supply (which gets directly connected to the motor) then you can determine that it is in fact the motor... as you have a voltage directly on the motor to confirm. Any changes in voltage shown on the display could be a result of the DC/DC converter or the AmpMeter circuit in your meter.
Hopefully this give you more info and not more questions. Let me know if this all makes sense for you. I sometimes do not explain myself properly.
Attachments
Okay..
Resistors came in... It's works great, but having the resistor on all the time, it gets HOT! I tried connecting 2 to see if I could share the load, but them they both just got HOT!
So then I thought I would check to see what voltage the center tap on the transformer was... It is 7.2v... But does the center tap still give 8A? Or only 1/2 (4A)?
-TheChad
Resistors came in... It's works great, but having the resistor on all the time, it gets HOT! I tried connecting 2 to see if I could share the load, but them they both just got HOT!
So then I thought I would check to see what voltage the center tap on the transformer was... It is 7.2v... But does the center tap still give 8A? Or only 1/2 (4A)?
-TheChad
