Thanks for the good link. I agree with earlier poster that the full schematic is clearer than the snippet. There a lot of un-related (to this thread) going on and the full diagram makes it easier to see what should be ignored.
It might be clearer still if it were redrawn without the the diodes in the bridge configurations, but that can be done mentally. The main difference between this simple blocker earlier in the that is that the opposed caps are parallel rather than series an are protected by two rather than one diode. I've no idea which is superior, but a 35A bridge rectifier is a convenient way to get four hefty diodes.
It might be clearer still if it were redrawn without the the diodes in the bridge configurations, but that can be done mentally. The main difference between this simple blocker earlier in the that is that the opposed caps are parallel rather than series an are protected by two rather than one diode. I've no idea which is superior, but a 35A bridge rectifier is a convenient way to get four hefty diodes.
I'm not clear about something. What is the function of the diodes? If one diode drop is 0V6, then 2 diodes series, back to back, will limit the drop at the caps to +or -1V2. But at the time this happens, isn't that the diodes are actually passing the DC (that is blocked by the caps)?
Hi Sam,
The advantage of the serial connected caps is that on every half cycle one is protecting the other and so provided both do not degrade significantly there should be no failure. But, the series caps need to be 4times bigger to match the parallel cap set-up to present the same impedance to the operational current.
edit:
I hope this also answers Luman.
I guess the parallel caps are there so that if a long term and fairly stable DC is present then at least one of the caps is polarised in the correct direction and continues to operate as intended. The other may degrade over time if the DC polarity does not change.sam9 said:
The advantage of the serial connected caps is that on every half cycle one is protecting the other and so provided both do not degrade significantly there should be no failure. But, the series caps need to be 4times bigger to match the parallel cap set-up to present the same impedance to the operational current.
The peak voltage across a single diode is 600mV to 700mV. This limits the voltage across the blocking caps. This in turn limits the maximum current that can pass without the diode turning on and bypassing the DC blocking action. Two series diodes will require twice the voltage to turn on and that allows the load to be about double for a fixed value of blocking cap. i.e. cheaper caps for the same blocking action.sam9 said:
edit:
I hope this also answers Luman.
Indeed a cheap and convenient way of packaging this device, but may be overkill for most domestic sized transformers, say upto 1000VA. I use 1n5404 which have a sufficiently high peak current capability, provided they only turn on very infrequently i.e. at start up and during a fuse blowing fault event. (a 50r soft start on 240Vac will see an occasional current of 7.2Apk)sam9 said:
if using the "DaBlock" on a line conditioner (ala Jon Risch design), I assume the diodes must be able have to be as least 10A (assuming contacts are rated at 10A). how about the voltage rating? must be be 240V and higher?
I'm thinking of using 22mF. using Andrew's calculation below, the voltage drop across the caps is .68VAC. Though I am unsure how he arrived at the 27Apeak for a 1.2Vf.
Thank you
ps. is it a good idea to install "DaBlock" on the line conditioner or better if it's in the individual components? thanks again
I'm thinking of using 22mF. using Andrew's calculation below, the voltage drop across the caps is .68VAC. Though I am unsure how he arrived at the 27Apeak for a 1.2Vf.
Thank you
ps. is it a good idea to install "DaBlock" on the line conditioner or better if it's in the individual components? thanks again
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This is easy to see before you click on the link, as the truncated portion will be black.
All you needed to do was click on the link, and then copy and paste the truncated portion onto the end.
The server truncated " (Oct02).pdf "
This is easy to see before you click on the link, as the truncated portion will be black.
All you needed to do was click on the link, and then copy and paste the truncated portion onto the end.
FYI: Epcos (capacitor maker) has reliable information that (their) electrolytics can only withstand 0.8 Volts reverse bias for extended periods:
http://www.epcos.de/web/generator/W...__nn.pdf;/PDF_GeneralTechnicalInformation.pdf
(see section 3.1.6 on page 8. This doc also is a good primer on electrolytics)
Cornell Dubilier has quite similar information:
http://www.cde.com/tech/appguide.pdf
(pages 6&7)
Therefore the back-to-back series connection of caps is mandatory, as is a clamping of extensive voltage across this compound capacitor. Ripple current / dissipation is the main spec that must be paid attention to (use a good safety margin there).
Regards, Klaus
http://www.epcos.de/web/generator/W...__nn.pdf;/PDF_GeneralTechnicalInformation.pdf
(see section 3.1.6 on page 8. This doc also is a good primer on electrolytics)
Cornell Dubilier has quite similar information:
http://www.cde.com/tech/appguide.pdf
(pages 6&7)
Therefore the back-to-back series connection of caps is mandatory, as is a clamping of extensive voltage across this compound capacitor. Ripple current / dissipation is the main spec that must be paid attention to (use a good safety margin there).
Regards, Klaus
Hi,
this is a DC blocking circuit.
It achieves this by inserting capacitors into the AC line.
The capacitors allow the AC to pass.
While doing this they develop an AC voltage across them that is directly related to the impedance of the caps.
To make the DC blocker economic to build, we use low voltage capacitors.
These low voltage capacitors must be protected from overvoltage.
We protect the low voltage capacitors by adding a bypass circuit to pass peak currents that would otherwise drop too high a voltage across the caps.
We can use a single diode in each direction (sometimes referred to as inverse parallel), or we can string together series diodes to bypass at a higher voltage. Single pair diodes start to pass at about 500mV to 600mV, double pairs @ between 1V to 1.2V, and triple pairs between 1.5V to 1.8V. The higher the diode voltage the higher the cap voltage drop before the bypass starts to operate.
DC is a completely different problem.
I am not too sure of the physics here so more knowledgible views are welcome.
The DC impedance of the mains and transformer is very low.
The DC is blocked by the caps and most of the time (99.9%) all the current flow passes through the caps. If the voltage across the caps exceeds the passing voltage of the diodes the current starts to bypass. After the bypass opens the DC component is shorted to neutral and the AC component generates the voltage across the transformer primary.
Prior to the bypass opening all the current passes through the caps. The DC voltage is relatively small and usually never approaches the diode forward voltage. Keep in mind it is distortion of the AC waveform that produces the asymmetric half waves and it is this unbalance that approximates to the net DC. As said earlier the transformer has almost zero impedance at DC and so quite tiny DC components in the AC can cause a disproportionate current to flow generating the flux that overloads the transformer core.
It would be interesting to see the voltage across the blocking caps. Would anyone with a storage scope care to experiment for us?
this is a DC blocking circuit.
It achieves this by inserting capacitors into the AC line.
The capacitors allow the AC to pass.
While doing this they develop an AC voltage across them that is directly related to the impedance of the caps.
To make the DC blocker economic to build, we use low voltage capacitors.
These low voltage capacitors must be protected from overvoltage.
We protect the low voltage capacitors by adding a bypass circuit to pass peak currents that would otherwise drop too high a voltage across the caps.
We can use a single diode in each direction (sometimes referred to as inverse parallel), or we can string together series diodes to bypass at a higher voltage. Single pair diodes start to pass at about 500mV to 600mV, double pairs @ between 1V to 1.2V, and triple pairs between 1.5V to 1.8V. The higher the diode voltage the higher the cap voltage drop before the bypass starts to operate.
DC is a completely different problem.
I am not too sure of the physics here so more knowledgible views are welcome.
The DC impedance of the mains and transformer is very low.
The DC is blocked by the caps and most of the time (99.9%) all the current flow passes through the caps. If the voltage across the caps exceeds the passing voltage of the diodes the current starts to bypass. After the bypass opens the DC component is shorted to neutral and the AC component generates the voltage across the transformer primary.
Prior to the bypass opening all the current passes through the caps. The DC voltage is relatively small and usually never approaches the diode forward voltage. Keep in mind it is distortion of the AC waveform that produces the asymmetric half waves and it is this unbalance that approximates to the net DC. As said earlier the transformer has almost zero impedance at DC and so quite tiny DC components in the AC can cause a disproportionate current to flow generating the flux that overloads the transformer core.
It would be interesting to see the voltage across the blocking caps. Would anyone with a storage scope care to experiment for us?
DC blocker sim
Adding to Andrews very good explanation I have some curves from a simulation of a DC blocker.
Mains: 230V(rms)=325V(peak)
Offset: 0.5V
Blocking Cap: 10mF
Diodes: antiparallel 1N4001
Transformer, values taken from an actual smaller toroid:
- 15 Ohms DC resistance of primary, 0.6H inductance
- 0.5 Ohms DC resistance of secondary, 3mH inductance
- coupling factor 0.9 (deliberately set to get some leakage inductance)
Load circuit is very typical:
- 4x1N4001 bridge
- 10mF Cap, 0.1 ohms ESR
- 5 ohms load, drawing ~2.6A(rms)
Mains inductance was neglected as the 15 ohms DC-R and the leakage inductance of the Xformer dominiate this.
Yellow trace is the mains voltage.
Blue trace is the voltage across the blocking circuit (plotted inverted polarity, to see the phase realtionships a little better).
Red trace is the current through one of the diodes (the other one doesn't get forward biased).
Green trace is the current through the cap. It is non-sinusodial due to the load circuit.
Max. positive voltage across the cap is 0.83V , max. negative is -0.32V. The diode starts to conduct at the positive peak. This is not because of the mains DC but due to the too high impedance of the cap @50Hz (0.32 Ohms). Peak cap current is ~1.7A. This gives us +-0.55V momentary voltage across the cap (which in total happens to be the sum of 0.83V and 0.32V). But, this momentary voltage does not directly add to the DC offset of 0.5V because of the phase shifts and the non-sinusodial cap current.
Regards, Klaus
Adding to Andrews very good explanation I have some curves from a simulation of a DC blocker.
Mains: 230V(rms)=325V(peak)
Offset: 0.5V
Blocking Cap: 10mF
Diodes: antiparallel 1N4001
Transformer, values taken from an actual smaller toroid:
- 15 Ohms DC resistance of primary, 0.6H inductance
- 0.5 Ohms DC resistance of secondary, 3mH inductance
- coupling factor 0.9 (deliberately set to get some leakage inductance)
Load circuit is very typical:
- 4x1N4001 bridge
- 10mF Cap, 0.1 ohms ESR
- 5 ohms load, drawing ~2.6A(rms)
Mains inductance was neglected as the 15 ohms DC-R and the leakage inductance of the Xformer dominiate this.
Yellow trace is the mains voltage.
Blue trace is the voltage across the blocking circuit (plotted inverted polarity, to see the phase realtionships a little better).
Red trace is the current through one of the diodes (the other one doesn't get forward biased).
Green trace is the current through the cap. It is non-sinusodial due to the load circuit.
Max. positive voltage across the cap is 0.83V , max. negative is -0.32V. The diode starts to conduct at the positive peak. This is not because of the mains DC but due to the too high impedance of the cap @50Hz (0.32 Ohms). Peak cap current is ~1.7A. This gives us +-0.55V momentary voltage across the cap (which in total happens to be the sum of 0.83V and 0.32V). But, this momentary voltage does not directly add to the DC offset of 0.5V because of the phase shifts and the non-sinusodial cap current.
Regards, Klaus
Attachments
No, not actually the DC. Even with no DC, when there is too high a current through the caps, producing too much voltage drop, the diodes start to conduct. This distorts the waveforms, but symmetrically ==> no DC effect.
If we have DC one of the diodes start to conduct earlier than the other, which result in a part of the DC being effective again (due to the asymmetry).
If we have DC one of the diodes start to conduct earlier than the other, which result in a part of the DC being effective again (due to the asymmetry).
Hi,
the red trace showing the diode conducting on every cycle tells the user the caps are too small.
The diodes should not conduct in normal operation.
I think the diode bypass condition should ONLY operate during the start-up current peak and during a fuse blowing incident.
The diodes may have to be very sturdy to survive the fuse blowing incident. Even a T3.1A will pass many hundreds (and possibly thousands) of amps during the rupture period. I would like the diodes to survive this very short term transient. The risk of damage to the caps increases enormously if the diodes are unknowingly open circuit.
If the red trace is the bypass conducting, then I would expect to see a flattening of the blue voltage (across the blocker) trace. But it's not evident.
How much current needs to bypass for the cap measurment to show that bypass action is ocurring?
Is there a better way to measure when the diode conducts?
What really happens if one puts a scope across the blocker (oops it's earthed) when feeding a capacitor input filter?
How does one scope the blocker safely?
There's no point isolating the scope. That just brings the scope chassis up to mains voltage.
Is isolating the amplifier the only safe solution?
the red trace showing the diode conducting on every cycle tells the user the caps are too small.
The diodes should not conduct in normal operation.
I think the diode bypass condition should ONLY operate during the start-up current peak and during a fuse blowing incident.
The diodes may have to be very sturdy to survive the fuse blowing incident. Even a T3.1A will pass many hundreds (and possibly thousands) of amps during the rupture period. I would like the diodes to survive this very short term transient. The risk of damage to the caps increases enormously if the diodes are unknowingly open circuit.
If the red trace is the bypass conducting, then I would expect to see a flattening of the blue voltage (across the blocker) trace. But it's not evident.
How much current needs to bypass for the cap measurment to show that bypass action is ocurring?
Is there a better way to measure when the diode conducts?
What really happens if one puts a scope across the blocker (oops it's earthed) when feeding a capacitor input filter?
How does one scope the blocker safely?
There's no point isolating the scope. That just brings the scope chassis up to mains voltage.
Is isolating the amplifier the only safe solution?
Hi,
In a typical domestic situation mains impedance is unlikely to be below 0.3Ohms. With 230V mains this would give up to 1000A of surge current in case of a full short. Therefore bigger bridge rectifier modules seem to be adequate, and maybe even two of them in parallel. A KBPC50xxx can handle up to 400A single pulse per diode, two or four of them in parallel seem to handle the surge. Also, with their 50A(rms) continuous rating (when properly heatsunk) they even seem to survive a situation when the caps went completely ineffective in passing current.
With a completely unloaded secondary and with the 0.6H of primary inductance (plus the 15 ohms DC-R) the Z @50Hz of the primary is ~200Ohms, almost purely inductive. This is the highest possible primary impedance. Hence peak cap current is never below a hefty 1.6 amps! Of course average power draw from the mains is close to zero. With the load resistor shorted, cap current increases to 2.6A(peak), while power draw reaches 300 watts (with a perfect xformer not driven into saturation plus other "perfect" parts). Peak diode current is 120mA (open circuit) and 350mA (shorted load). These results are quite counter-intuitive. After all, who would expect a high cap current with an unloaded secondary?
If one takes a closer look on the blue curve, one can find a little distortion, it's not perfectly sinusodial. This is best seen when taking the 1st or 2nd time-derivate of the voltage.
To scope diode or cap currents I suggest the use of inductive current probes. This is very safe, no galvanic coupling.
FOR DIYer: A DC-Blocker is **NOT** a project that you just hookup from arbitrary parts from your tinker box in a "Let's just see how it works with what I have here" approach. Calculate, design, simulate and build it with utmost care, triple checking everything. If you have any doubts, don't build it. Use ground fault interruptors, wear safety glasses and take all precautions that are necessary when dealing with mains voltages.
WARNING: the parts I used in the simulation (post #51) are **NOT** adequate for a real project. The simulation is and was only for explanatory purposes.
Regards, Klaus
In a typical domestic situation mains impedance is unlikely to be below 0.3Ohms. With 230V mains this would give up to 1000A of surge current in case of a full short. Therefore bigger bridge rectifier modules seem to be adequate, and maybe even two of them in parallel. A KBPC50xxx can handle up to 400A single pulse per diode, two or four of them in parallel seem to handle the surge. Also, with their 50A(rms) continuous rating (when properly heatsunk) they even seem to survive a situation when the caps went completely ineffective in passing current.
With a completely unloaded secondary and with the 0.6H of primary inductance (plus the 15 ohms DC-R) the Z @50Hz of the primary is ~200Ohms, almost purely inductive. This is the highest possible primary impedance. Hence peak cap current is never below a hefty 1.6 amps! Of course average power draw from the mains is close to zero. With the load resistor shorted, cap current increases to 2.6A(peak), while power draw reaches 300 watts (with a perfect xformer not driven into saturation plus other "perfect" parts). Peak diode current is 120mA (open circuit) and 350mA (shorted load). These results are quite counter-intuitive. After all, who would expect a high cap current with an unloaded secondary?
If one takes a closer look on the blue curve, one can find a little distortion, it's not perfectly sinusodial. This is best seen when taking the 1st or 2nd time-derivate of the voltage.
To scope diode or cap currents I suggest the use of inductive current probes. This is very safe, no galvanic coupling.
FOR DIYer: A DC-Blocker is **NOT** a project that you just hookup from arbitrary parts from your tinker box in a "Let's just see how it works with what I have here" approach. Calculate, design, simulate and build it with utmost care, triple checking everything. If you have any doubts, don't build it. Use ground fault interruptors, wear safety glasses and take all precautions that are necessary when dealing with mains voltages.
WARNING: the parts I used in the simulation (post #51) are **NOT** adequate for a real project. The simulation is and was only for explanatory purposes.
Regards, Klaus
Is this "DC Blocker" is the same with commercially available "Line Conditioner"? If they are not the same, what is it that a "Line Conditioner" do?
My impression is that products called "line conditioners" do different things depending on who makes them. Some of the more expensive appear to include an isolation transformer. Some are high frequency filters that may or may not be effective. Some seem designed to take your money and little else.
In short, I don't think you can generalize, but have to dig into the manufacture's literature or even ask the manufacturer directly.
lumanauw said:
Hi lumanauw. They are not the same.
sam9 said:
sam's got it right 🙂 Line conditioners are supposed to clean up the AC voltage for the equipment downstream. Most include AC line filtering and surge suppression. Some provide isolation and balanced power (converting 0-120VAC to 60-060 VAC). And some are just out to get your money. As sam said, you really have to investigate the product to know what are you are really going to get. One easy way to tell if there's an isolation transformer inside is the weight. A 1:1 isolation transformer at any respectable VA rating is not going to be light!