I am having trouble understanding why a shunt, or bleeder, resistor in parallel to a load can appear to keep the voltage from dropping when I make a power supply model in Duncan PSUD II.
For example; I modeled a unregulated DC tube heater design in Duncan PSUD II using the load I expect the heater to draw and I had to raise the voltage spec on the power transformer quite a bit to get the voltage I wanted at the load point. Then it occurred to me that I could add resistance and decrease the load and implement the extra resistance as a bleeder resistor in parallel to the load to lower the apparent load in the model. This let me specify a much lower voltage on the transformer in the model and maintain the voltage at the load.
I know I must be missing some implication. It seems to me that adding a parallel path to ground can't actually lighten the load.
Is it just a matter where the modeling software can not show me the localized details. It seems like the voltage is going to drop when it sees the actual load at the apogee of the circuit.
I worry about seeming annoying when I ask these basic questions. Thank You for your patience.
For example; I modeled a unregulated DC tube heater design in Duncan PSUD II using the load I expect the heater to draw and I had to raise the voltage spec on the power transformer quite a bit to get the voltage I wanted at the load point. Then it occurred to me that I could add resistance and decrease the load and implement the extra resistance as a bleeder resistor in parallel to the load to lower the apparent load in the model. This let me specify a much lower voltage on the transformer in the model and maintain the voltage at the load.
I know I must be missing some implication. It seems to me that adding a parallel path to ground can't actually lighten the load.
Is it just a matter where the modeling software can not show me the localized details. It seems like the voltage is going to drop when it sees the actual load at the apogee of the circuit.
I worry about seeming annoying when I ask these basic questions. Thank You for your patience.
When you keep the voltage across the load the same, anything placed in parallel does not decrease load current. It only adds load current as seen from the supply, so you actually go the wrong way.
The load current depends on the load impedance (current demand) and the load voltage. If neither changes, the load demand also doesn't change.
jan.
The load current depends on the load impedance (current demand) and the load voltage. If neither changes, the load demand also doesn't change.
jan.
I don't think I am asking the question very well.
Here is what I am looking at:
I am using a 42ohm resistor to represent the load of 0.15A at 6.3vDC on a triode tube heater and I am having to use a 40vAC secondary to keep the 6.3vDC level.
When I lower the secondary to something like 12vAC I found that the 42ohm load dropped the voltage down very low. It occurred to me that I could add a 458ohm resistance in parallel to the 42ohm load and model the power supply with a 250ohm resistor. That let me use a 12vAC secondary and get a 6.3vDC value at R5.
It doesn't seem like this will actually work in "real life" as the R5 in the second model will really be two loads in parallel and so the heater element will still draw it's 0.15 amps. I suspect that in real life the voltage will drop just as it did in the first example.
I am having a hard time accepting that I need 40vAC to get a filtered and smooth 6vDC and so I think I am either misunderstanding something badly, or I am just not accepting the fact that the voltage drop is really that large.
I feel like I might need to wire up a bench test rig and see what happens in real life. :-S
Thank you for considering the question.
Here is what I am looking at:
I am using a 42ohm resistor to represent the load of 0.15A at 6.3vDC on a triode tube heater and I am having to use a 40vAC secondary to keep the 6.3vDC level.
When I lower the secondary to something like 12vAC I found that the 42ohm load dropped the voltage down very low. It occurred to me that I could add a 458ohm resistance in parallel to the 42ohm load and model the power supply with a 250ohm resistor. That let me use a 12vAC secondary and get a 6.3vDC value at R5.
It doesn't seem like this will actually work in "real life" as the R5 in the second model will really be two loads in parallel and so the heater element will still draw it's 0.15 amps. I suspect that in real life the voltage will drop just as it did in the first example.
I am having a hard time accepting that I need 40vAC to get a filtered and smooth 6vDC and so I think I am either misunderstanding something badly, or I am just not accepting the fact that the voltage drop is really that large.
I feel like I might need to wire up a bench test rig and see what happens in real life. :-S
Thank you for considering the question.
Your heater resistance is 42 ohms, you have established that by Ohm's law. Adding 458 ohms in parallel to 42 ohms results in 38.5 ohms, not sure what you were thinking about there.
250 ohms might well give you 6.3V when you have a 12V transformer, but your heater resistance is 42 ohms. So the voltage drops as you found out.
With all those cascaded filters the voltage drop really is that large. You would be much better off using a single filter and/or voltage regulator with a 12V transformer. Performance will be much better and things will run much cooler (less power wasted).
250 ohms might well give you 6.3V when you have a 12V transformer, but your heater resistance is 42 ohms. So the voltage drops as you found out.
With all those cascaded filters the voltage drop really is that large. You would be much better off using a single filter and/or voltage regulator with a 12V transformer. Performance will be much better and things will run much cooler (less power wasted).
You don't need 40V AC to get smooth 6V DC, unless you use a very poor circuit. You have a very poor circuit, with lots of series resistance. The voltage drop (in your circuit) really is that large. Ohm's law always works (for ohmic resistors).whyvoltage said:
As SY told you in the other thread, and richie00boy has told you here, the proper solution is much less smoothing but add a regulator.
🙂 In a thread I started yesterday where I was asking about figuring out the current requirements for a power transformer used with a LM317 regulator and the 0.15A load my example was criticized for, well... having a regulator. 🙂
I am just trying to learn some stuff by comparing the different designs.
Thanks.
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Thank you for pointing out my mistake with the math.
I was thinking (458 + 42)/2 = 250. I obviously need to figure out what the equation is really supposed to be.
Thank You for your patience.
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I just looked at your other thread. What sreten was saying is that he thinks a basic rectified and smoothed arrangement is good enough, not that you need to try cascading filters.
The formula for resistors in parallel is
1/R1 + 1/R2 = 1/R
therefore invert the result to get R
The formula for resistors in parallel is
1/R1 + 1/R2 = 1/R
therefore invert the result to get R
Thank you! I was just coming back to say that I looked up the equation and realized what a gross misunderstanding I had with this.
I tried models like that in Duncan and it seems like there is still a lot of ripple being shown.
My primary interest has been to have a chance to assess the various choices and I am about to start trying to test rigs on the bench.
I would like to get to the point where I can listen to the results and get a feel for how the opinions about how much or what kind of filtering coincide with what I think I can actually hear.
I have used AC powered heaters in small guitar amps and never had any noise concerns with that, but this project is for a tube based condenser microphone and it seems like people that have experience with them either use regulated DC or heavily filtered and smooth DC. I'd like to learn for myself and the first step seems to be considering all the options and then perhaps experience some of those options with test builds.
I very much appreciate all the help and advice that is being offered here.
Thank You!
Or for just 2 resistors in parallel, the equation simplifies to:
Requal = (R1*R2)/(R1+R2)
which comes very handy since it is very easy to remember and calculate. I always use it.
Personally, I find that it is dumber to use a power supply capable of delivering (and wasting!) 3-5 times the required power just not to include a regulator and achieve the same smoothing. I would use a 3-terminal LM317.
Besides, I think that for such constant loads, a fixed regulator design can't be beaten, since you don't have to seek performace improvements (like with regulators used in preamps etc) - performance is already superb and requires ridiculously few parts. 🙄
Requal = (R1*R2)/(R1+R2)
which comes very handy since it is very easy to remember and calculate. I always use it.
Personally, I find that it is dumber to use a power supply capable of delivering (and wasting!) 3-5 times the required power just not to include a regulator and achieve the same smoothing. I would use a 3-terminal LM317.
Besides, I think that for such constant loads, a fixed regulator design can't be beaten, since you don't have to seek performace improvements (like with regulators used in preamps etc) - performance is already superb and requires ridiculously few parts. 🙄
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Valve microphone preamps are difficult to get right. I have never attempted one, despite my 45 years experience of electronics. There are threads on here started by others who have spent weeks or months trying to eliminate noise/buzz/hum. Good luck!
The only think you need to consider is selecting a reservoir capacitor capable of withstanding the required ripple current with ease, to ensure long-term reliability.
For example, a Panasonic FM 470uF 35V capacitor is rated at about 1.5A rms ripple current at 120Hz. Since I doubt you will go above 400mA rms secondary current, such a cap would be perfect for your application.
Position the LM317 close to the reservoir to avoid need for a 0.1uF input capacitor, add a 10uF electolytic (not low-esr) capacitor at the output, and provide a small heatsink for the LM317. Use 120R + 470R for the divider, add a 22uF 16V bypass capacitor and there you have it. 🙂
For example, a Panasonic FM 470uF 35V capacitor is rated at about 1.5A rms ripple current at 120Hz. Since I doubt you will go above 400mA rms secondary current, such a cap would be perfect for your application.
Position the LM317 close to the reservoir to avoid need for a 0.1uF input capacitor, add a 10uF electolytic (not low-esr) capacitor at the output, and provide a small heatsink for the LM317. Use 120R + 470R for the divider, add a 22uF 16V bypass capacitor and there you have it. 🙂
Looking at your earlier model, you have all those 50 Ohm resistors. Assuming your proposed current of 150mA is correct, that's a voltage drop of 7.5V across each of those 4 resistors. Put another way, your heater current is being dissipated at the rate of 4.5Watts across those 4 resistors, heating the enclosure instead of the tubes.
Try your model again but knock those resistors down to no more than 10 Ohms (or even a lot less). I think you'll see plenty of smoothing, and more voltage at the load, and a lot less wasted power.
Try your model again but knock those resistors down to no more than 10 Ohms (or even a lot less). I think you'll see plenty of smoothing, and more voltage at the load, and a lot less wasted power.
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