Symfony Profiler

Security

Token

There is no security token.

Firewall

main Name

Security enabled

Stateless

Configuration

Key	Value
provider	security.user.provider.concrete.app_user_provider
context	main
entry_point	App\Security\KbinAuthenticator
user_checker	App\Security\UserChecker
access_denied_handler	(none)
access_denied_url	(none)
authenticators	[`"two_factor" "remember_me" "App\Security\KbinAuthenticator" "App\Security\FacebookAuthenticator" "App\Security\GoogleAuthenticator" "App\Security\GithubAuthenticator" "App\Security\KeycloakAuthenticator"`]

Listeners

Listener	Duration	Response
Symfony\Component\Security\Http\Firewall\ChannelListener {#723`-map: Symfony\Component\Security\Http\AccessMap {#722 …} -logger: Monolog\Logger {#783 …} -httpPort: 80 -httpsPort: 443`}	0.00 ms	(none)
Symfony\Component\Security\Http\Firewall\ContextListener {#706-tokenStorage: Symfony\Component\Security\Core\Authentication\Token\Storage\TokenStorage {#1017 …} -sessionKey: "_security_main" -logger: Monolog\Logger {#783 …} -userProviders: Symfony\Component\DependencyInjection\Argument\RewindableGenerator {#705 …} -dispatcher: Symfony\Component\EventDispatcher\Debug\TraceableEventDispatcher {#747 …} -registered: false -trustResolver: Scheb\TwoFactorBundle\Security\Authentication\AuthenticationTrustResolver {#780 …} -sessionTrackerEnabler: Symfony\Component\Security\Core\Authentication\Token\Storage\UsageTrackingTokenStorage::enableUsageTracking(): void {#703 …}}	4.88 ms	(none)
Symfony\Component\Security\Http\Firewall\AuthenticatorManagerListener {#584`-authenticatorManager: Symfony\Component\Security\Http\Authentication\AuthenticatorManager {#595 …}`}	0.00 ms	(none)
Scheb\TwoFactorBundle\Security\Http\Firewall\TwoFactorAccessListener {#582`-twoFactorFirewallConfig: Scheb\TwoFactorBundle\Security\TwoFactor\TwoFactorFirewallConfig {#842 …} -tokenStorage: Symfony\Component\Security\Core\Authentication\Token\Storage\UsageTrackingTokenStorage {#1018 …} -twoFactorAccessDecider: Scheb\TwoFactorBundle\Security\Authorization\TwoFactorAccessDecider {#581 …}`}	0.05 ms	(none)
Symfony\Component\Security\Http\Firewall\AccessListener {#579`-tokenStorage: Symfony\Component\Security\Core\Authentication\Token\Storage\UsageTrackingTokenStorage {#1018 …} -accessDecisionManager: Symfony\Component\Security\Core\Authorization\TraceableAccessDecisionManager {#937 …} -map: Symfony\Component\Security\Http\AccessMap {#722 …}`}	0.00 ms	(none)
Symfony\Component\Security\Http\Firewall\LogoutListener {#786`-tokenStorage: Symfony\Component\Security\Core\Authentication\Token\Storage\UsageTrackingTokenStorage {#1018 …} -options: [ "csrf_parameter" => "_csrf_token" "csrf_token_id" => "logout" "logout_path" => "app_logout" ] -httpUtils: Symfony\Component\Security\Http\HttpUtils {#841 …} -csrfTokenManager: Symfony\Component\Security\Csrf\CsrfTokenManager {#1015 …} -eventDispatcher: Symfony\Component\EventDispatcher\Debug\TraceableEventDispatcher {#747 …}`}	0.00 ms	(none)

Authenticators

No authenticators have been recorded. Check previous profiles on your authentication endpoint.

Access Decision

affirmative Strategy


"Symfony\Component\Security\Core\Authorization\Voter\AuthenticatedVoter"
"Scheb\TwoFactorBundle\Security\Authorization\Voter\TwoFactorInProgressVoter"
"Symfony\Component\Security\Core\Authorization\Voter\RoleHierarchyVoter"
"Symfony\Component\Security\Core\Authorization\Voter\ExpressionVoter"
"App\Security\Voter\EntryCommentVoter"
"App\Security\Voter\EntryVoter"
"App\Security\Voter\MagazineVoter"
"App\Security\Voter\MessageThreadVoter"
"App\Security\Voter\MessageVoter"
"App\Security\Voter\NotificationVoter"
"App\Security\Voter\OAuth2UserConsentVoter"
"App\Security\Voter\PostCommentVoter"
"App\Security\Voter\PostVoter"
"App\Security\Voter\UserVoter"

#	Voter class
1	"Symfony\Component\Security\Core\Authorization\Voter\AuthenticatedVoter"
2	"Scheb\TwoFactorBundle\Security\Authorization\Voter\TwoFactorInProgressVoter"
3	"Symfony\Component\Security\Core\Authorization\Voter\RoleHierarchyVoter"
4	"Symfony\Component\Security\Core\Authorization\Voter\ExpressionVoter"
5	"App\Security\Voter\EntryCommentVoter"
6	"App\Security\Voter\EntryVoter"
7	"App\Security\Voter\MagazineVoter"
8	"App\Security\Voter\MessageThreadVoter"
9	"App\Security\Voter\MessageVoter"
10	"App\Security\Voter\NotificationVoter"
11	"App\Security\Voter\OAuth2UserConsentVoter"
12	"App\Security\Voter\PostCommentVoter"
13	"App\Security\Voter\PostVoter"
14	"App\Security\Voter\UserVoter"

Access decision log


null
"App\Security\Voter\UserVoter"
App\Entity\Entry {#2419
  +user: Proxies\__CG__\App\Entity\User {#1970 …}
  +magazine: App\Entity\Magazine {#265
    +icon: Proxies\__CG__\App\Entity\Image {#246 …}
    +name: "askelectronics@discuss.tchncs.de"
    +title: "askelectronics"
    +description: """
      **For questions about component-level electronic circuits, tools and equipment.**\n
      \n
      Rules\n
      -----\n
      \n
      1: Be nice.\n
      \n
      2: Be on-topic (eg: *Electronic*, not *electrical*).\n
      \n
      3: No commercial stuff, buying, selling or valuations.\n
      \n
      4: Be safe.\n
      \n
      ---
      """
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      date: 2024-10-20 10:40:08.0 +02:00
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    -id: 11
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      date: 2024-09-25 09:01:54.0 +02:00
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    +apDeletedAt: null
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    +visibility: "visible             "
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      date: 2023-06-19 05:13:29.0 +02:00
    }
  }
  +image: Proxies\__CG__\App\Entity\Image {#1943 …}
  +domain: Proxies\__CG__\App\Entity\Domain {#1912 …}
  +slug: "How-could-a-DC-oscillator-produce-AC-after-a-capacitor"
  +title: "How could a DC oscillator produce AC after a capacitor?"
  +url: "https://discuss.tchncs.de/pictrs/image/d3d039c3-1c43-453a-863b-e1638fc5c7ab.jpeg"
  +body: """
    Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
    \n
    We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
    """
  +type: "image"
  +lang: "en"
  +isOc: false
  +hasEmbed: true
  +commentCount: 6
  +favouriteCount: 11
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  +lastActive: DateTime @1688908216 {#2414
    date: 2023-07-09 15:10:16.0 +02:00
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  +ip: null
  +adaAmount: 0
  +tags: null
  +mentions: null
  +comments: Doctrine\ORM\PersistentCollection {#1866 …}
  +votes: Doctrine\ORM\PersistentCollection {#1966 …}
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  +cross: false
  +upVotes: 0
  +downVotes: 0
  +ranking: 1688823757
  +visibility: "visible             "
  +apId: "https://discuss.tchncs.de/post/558897"
  +editedAt: null
  +createdAt: DateTimeImmutable @1688762257 {#1793
    date: 2023-07-07 22:37:37.0 +02:00
  }
}
"App\Security\Voter\UserVoter"
App\Entity\Entry {#2419
  +user: Proxies\__CG__\App\Entity\User {#1970 …}
  +magazine: App\Entity\Magazine {#265
    +icon: Proxies\__CG__\App\Entity\Image {#246 …}
    +name: "askelectronics@discuss.tchncs.de"
    +title: "askelectronics"
    +description: """
      **For questions about component-level electronic circuits, tools and equipment.**\n
      \n
      Rules\n
      -----\n
      \n
      1: Be nice.\n
      \n
      2: Be on-topic (eg: *Electronic*, not *electrical*).\n
      \n
      3: No commercial stuff, buying, selling or valuations.\n
      \n
      4: Be safe.\n
      \n
      ---
      """
    +rules: null
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    +postCount: 1
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      date: 2024-10-20 10:40:08.0 +02:00
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    +apDiscoverable: true
    +apManuallyApprovesFollowers: null
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    +apFetchedAt: DateTime @1727247714 {#269
      date: 2024-09-25 09:01:54.0 +02:00
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    +visibility: "visible             "
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      date: 2023-06-19 05:13:29.0 +02:00
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  }
  +image: Proxies\__CG__\App\Entity\Image {#1943 …}
  +domain: Proxies\__CG__\App\Entity\Domain {#1912 …}
  +slug: "How-could-a-DC-oscillator-produce-AC-after-a-capacitor"
  +title: "How could a DC oscillator produce AC after a capacitor?"
  +url: "https://discuss.tchncs.de/pictrs/image/d3d039c3-1c43-453a-863b-e1638fc5c7ab.jpeg"
  +body: """
    Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
    \n
    We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
    """
  +type: "image"
  +lang: "en"
  +isOc: false
  +hasEmbed: true
  +commentCount: 6
  +favouriteCount: 11
  +score: 0
  +isAdult: false
  +sticky: false
  +lastActive: DateTime @1688908216 {#2414
    date: 2023-07-09 15:10:16.0 +02:00
  }
  +ip: null
  +adaAmount: 0
  +tags: null
  +mentions: null
  +comments: Doctrine\ORM\PersistentCollection {#1866 …}
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  +notifications: Doctrine\ORM\PersistentCollection {#2426 …}
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  +cross: false
  +upVotes: 0
  +downVotes: 0
  +ranking: 1688823757
  +visibility: "visible             "
  +apId: "https://discuss.tchncs.de/post/558897"
  +editedAt: null
  +createdAt: DateTimeImmutable @1688762257 {#1793
    date: 2023-07-07 22:37:37.0 +02:00
  }
}
"App\Security\Voter\UserVoter"
App\Entity\Entry {#2419
  +user: Proxies\__CG__\App\Entity\User {#1970 …}
  +magazine: App\Entity\Magazine {#265
    +icon: Proxies\__CG__\App\Entity\Image {#246 …}
    +name: "askelectronics@discuss.tchncs.de"
    +title: "askelectronics"
    +description: """
      **For questions about component-level electronic circuits, tools and equipment.**\n
      \n
      Rules\n
      -----\n
      \n
      1: Be nice.\n
      \n
      2: Be on-topic (eg: *Electronic*, not *electrical*).\n
      \n
      3: No commercial stuff, buying, selling or valuations.\n
      \n
      4: Be safe.\n
      \n
      ---
      """
    +rules: null
    +subscriptionsCount: 1
    +entryCount: 154
    +entryCommentCount: 1230
    +postCount: 1
    +postCommentCount: 4
    +isAdult: false
    +customCss: null
    +lastActive: DateTime @1729413608 {#275
      date: 2024-10-20 10:40:08.0 +02:00
    }
    +markedForDeletionAt: null
    +tags: null
    +moderators: Doctrine\ORM\PersistentCollection {#237 …}
    +ownershipRequests: Doctrine\ORM\PersistentCollection {#233 …}
    +moderatorRequests: Doctrine\ORM\PersistentCollection {#222 …}
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      date: 2023-06-19 05:13:29.0 +02:00
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  +image: Proxies\__CG__\App\Entity\Image {#1943 …}
  +domain: Proxies\__CG__\App\Entity\Domain {#1912 …}
  +slug: "How-could-a-DC-oscillator-produce-AC-after-a-capacitor"
  +title: "How could a DC oscillator produce AC after a capacitor?"
  +url: "https://discuss.tchncs.de/pictrs/image/d3d039c3-1c43-453a-863b-e1638fc5c7ab.jpeg"
  +body: """
    Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
    \n
    We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
    """
  +type: "image"
  +lang: "en"
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  +hasEmbed: true
  +commentCount: 6
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    date: 2023-07-09 15:10:16.0 +02:00
  }
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    date: 2023-07-07 22:37:37.0 +02:00
  }
}
"App\Security\Voter\UserVoter"
null
"App\Security\Voter\UserVoter"
App\Entity\EntryComment {#4026
  +user: App\Entity\User {#3974 …}
  +entry: App\Entity\Entry {#2419
    +user: Proxies\__CG__\App\Entity\User {#1970 …}
    +magazine: App\Entity\Magazine {#265
      +icon: Proxies\__CG__\App\Entity\Image {#246 …}
      +name: "askelectronics@discuss.tchncs.de"
      +title: "askelectronics"
      +description: """
        **For questions about component-level electronic circuits, tools and equipment.**\n
        \n
        Rules\n
        -----\n
        \n
        1: Be nice.\n
        \n
        2: Be on-topic (eg: *Electronic*, not *electrical*).\n
        \n
        3: No commercial stuff, buying, selling or valuations.\n
        \n
        4: Be safe.\n
        \n
        ---
        """
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        date: 2024-10-20 10:40:08.0 +02:00
      }
      +markedForDeletionAt: null
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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    The standard way of looking at this is to consider a capacitor-resistor series combination going to ground. Connect a 10v (wrt ground) supply to the capacitor and the voltage across the resistor rises to +10v, then decays. Now connect that capacitor to ground and that same resistor gets -10v across it, which then decays. Whatever is connected to the capacitor “top” terminal has to be able to sink current as well as source it.\n
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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    The standard way of looking at this is to consider a capacitor-resistor series combination going to ground. Connect a 10v (wrt ground) supply to the capacitor and the voltage across the resistor rises to +10v, then decays. Now connect that capacitor to ground and that same resistor gets -10v across it, which then decays. Whatever is connected to the capacitor “top” terminal has to be able to sink current as well as source it.\n
    \n
    That’s what generators in simulators do - they have zero internal impedance (usually). They sink currents as well as source them.
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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    The standard way of looking at this is to consider a capacitor-resistor series combination going to ground. Connect a 10v (wrt ground) supply to the capacitor and the voltage across the resistor rises to +10v, then decays. Now connect that capacitor to ground and that same resistor gets -10v across it, which then decays. Whatever is connected to the capacitor “top” terminal has to be able to sink current as well as source it.\n
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        \n
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
      \n
      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
      """
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    True AC is sort of “balanced” in that it has just as much positive as negative. The positive area of the waveform is the same size as the negative area. For waveforms that are sort of symmetric across the 0V with a time offset, such as a sine or square wave, this means that it is centered along the 0V line. A DC source, on the other hand, never changes voltage.\n
    \n
    The 0V to +10V source you have is actually a -5V to +5V square AC plus a +5V DC. The capacitor is getting rid of the DC component leaving just the AC, which happens to be the -5V to +5V AC that you are getting.
    """
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
      \n
      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
      """
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    True AC is sort of “balanced” in that it has just as much positive as negative. The positive area of the waveform is the same size as the negative area. For waveforms that are sort of symmetric across the 0V with a time offset, such as a sine or square wave, this means that it is centered along the 0V line. A DC source, on the other hand, never changes voltage.\n
    \n
    The 0V to +10V source you have is actually a -5V to +5V square AC plus a +5V DC. The capacitor is getting rid of the DC component leaving just the AC, which happens to be the -5V to +5V AC that you are getting.
    """
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
      \n
      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
      """
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    True AC is sort of “balanced” in that it has just as much positive as negative. The positive area of the waveform is the same size as the negative area. For waveforms that are sort of symmetric across the 0V with a time offset, such as a sine or square wave, this means that it is centered along the 0V line. A DC source, on the other hand, never changes voltage.\n
    \n
    The 0V to +10V source you have is actually a -5V to +5V square AC plus a +5V DC. The capacitor is getting rid of the DC component leaving just the AC, which happens to be the -5V to +5V AC that you are getting.
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
      \n
      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
      """
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    In my opinion an oscillator always produces an AC sine wave. There is usually no need for a DC overlapped oscillator signal. The DC supply of an oscillator produces a AC sine wave relative to GND.\n
    \n
    Where exactly did you measure a DC sine wave, relative to what, and what do you mean by “AC removes a DC component”?
    """
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
      \n
      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
      """
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    In my opinion an oscillator always produces an AC sine wave. There is usually no need for a DC overlapped oscillator signal. The DC supply of an oscillator produces a AC sine wave relative to GND.\n
    \n
    Where exactly did you measure a DC sine wave, relative to what, and what do you mean by “AC removes a DC component”?
    """
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      Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.\n
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      We’ve all heard that AC removes DC component and let’s AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!
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