Multicamera automation project: step 1 - is partial Protective Earth disconnection safe

Thread Starter

niman

Joined Jun 7, 2018
3
My present photography system comprises:

8 by Canon Powershot Cameras
1 by PSU (5V X 20A with 10% voltage (+/-) variation) - to power cameras whose maximum current draw is 2A and, nominal, operating voltage is 3.2V
1 by PSU (5V X 20A with 10% voltage (+/-) variation) - to operate a camera shutter mechanism by way of a manual switch
1 by 10 port USB 2 powered hub - for ptp based communication with the cameras. The hub psu is mains 2 pin with no connection to mains Protective Earth (PE hereafter).
1 by Win 7 laptop - for ptp based communication (via the usb hub) with the cameras. It's psu has it's 3rd pin connected to mains PE.

In the first place I wish to replace the manual shutter switching mechanism with a microcontrolled version. I also wish to create a design so that shutter switching and ptp control of extra camera units can be accomadated as simply as possible.

Anyway, with limited electronic knowledge, I would appreciate any initial comment to:
1. support/refute my description of how the system presently operates [it does work, i'm just not necessarily sure how :)]
and
2. how I might make the first small change, safely, to facilitate the automation / expansion referred to.

My understanding (attached)
Please find attached my interpretation of the system as it is - simplified to include a single camera only.

I hope the diagram is mostly self explanatory but here are, what I think, the most important functional points:

The manual switch, when open, is connected to Camera mini usb port 1. When closed, mini usb port 1 is conned to the switch psu (DC V+ (3)). Within the camera mini usb port 1 is connected to the negative battery terminal (V Batt -) via an effective resistance of 12.1 KΩ (that value was determined by experiment).

DC V+ (1) of the Camera psu (PSU (1)) is connected to the Camera positive battery terminal (V Batt +) by way of a power diode and a 3 metre length of copper cable (of 0.5Ω). Within the camera, V Batt + is connected to the negative battery terminal (V Batt -) by way of a 1.75Ω load (calculated as measured voltage drop between V Batt + & V Batt divided by maximum fuse value (2A) of PSU (1)..

Within the camera, V Batt - is short-circuit to mini usb port 5.

Camera V Batt - is connected to DC V- (1) by way of a 3 metre copper cable (of 0.5Ω).
Camera mini usb port 5 is connected to Protective Earth (PE) by way of 0.4Ω associated with the Win 7 laptop PSU (that physical path is indicated by the dashed yellow line in the diagram.

DC V- (1) terminal is connected (by short wire) to PE (1) terminal
DC V- (3) terminal is connected (by short wire) to PE (3) terminal

DC V- (1) terminal is connected by thick gauge wire to DC V- (3) terminal

Switch operation:
when the switch is closed the rising edge of the 5.15V applied to camera mini usb port 1 causes the camera to perform autofocus and autoexposure and stop in a tight loop just before where the actual exposure would start. When the switch is opened, the negative edge of this 5.15V step is used by the camera to proceed with the exposure - very tight synchronisation can be achieved across a high camera count.

Switch current flow description:
When the switch is closed, current flow into the camera at mini usb port 1 is 0.43mA. This current divides in two paths at the common potential point V Batt- / mini usb port 5 as 4/9th's (0.19mA) to DC V- (1) and 5/9th's (0.24mA) to laptop psu PE. The fraction of current at DC V- (1) then then returns to the switch psu (DC V- (3)) via short wire and the fraction at laptop psu PE via the mains earth conductor.

If all of that is based in reality, the first change I would like to contemplate is:

A/ remove the short wire connections between DC V- (1) & PE (1) and DC V- (3) & PE (3)
B/ disconnect Camera mini usb port 1 from laptop psu PE

Laptop psu PE is then the only system point connected to mains PE and the switch current can only return by a single path. My understanding is that the usb differential D+/D- ptp data comms should be unaffected by my proposal.

Initial Query:

with PSU (1) & PSU (3) no longer connected to PE they are, of course, "floating" with respect to it. Is there possibility that dangerous conditions could result in the system as a result of my proposal? where/how?

PS: the metal cases of PSU (1) & PSU (3) are connected to PE whether or not the short wire links between DC V- & PE are present.
 

Attachments

Hymie

Joined Mar 30, 2018
1,082
My present photography system comprises:

8 by Canon Powershot Cameras
1 by PSU (5V X 20A with 10% voltage (+/-) variation) - to power cameras whose maximum current draw is 2A and, nominal, operating voltage is 3.2V
1 by PSU (5V X 20A with 10% voltage (+/-) variation) - to operate a camera shutter mechanism by way of a manual switch
1 by 10 port USB 2 powered hub - for ptp based communication with the cameras. The hub psu is mains 2 pin with no connection to mains Protective Earth (PE hereafter).
1 by Win 7 laptop - for ptp based communication (via the usb hub) with the cameras. It's psu has it's 3rd pin connected to mains PE.

In the first place I wish to replace the manual shutter switching mechanism with a microcontrolled version. I also wish to create a design so that shutter switching and ptp control of extra camera units can be accomadated as simply as possible.

Anyway, with limited electronic knowledge, I would appreciate any initial comment to:
1. support/refute my description of how the system presently operates [it does work, i'm just not necessarily sure how :)]
and
2. how I might make the first small change, safely, to facilitate the automation / expansion referred to.

My understanding (attached)
Please find attached my interpretation of the system as it is - simplified to include a single camera only.

I hope the diagram is mostly self explanatory but here are, what I think, the most important functional points:

The manual switch, when open, is connected to Camera mini usb port 1. When closed, mini usb port 1 is conned to the switch psu (DC V+ (3)). Within the camera mini usb port 1 is connected to the negative battery terminal (V Batt -) via an effective resistance of 12.1 KΩ (that value was determined by experiment).

DC V+ (1) of the Camera psu (PSU (1)) is connected to the Camera positive battery terminal (V Batt +) by way of a power diode and a 3 metre length of copper cable (of 0.5Ω). Within the camera, V Batt + is connected to the negative battery terminal (V Batt -) by way of a 1.75Ω load (calculated as measured voltage drop between V Batt + & V Batt divided by maximum fuse value (2A) of PSU (1)..

Within the camera, V Batt - is short-circuit to mini usb port 5.

Camera V Batt - is connected to DC V- (1) by way of a 3 metre copper cable (of 0.5Ω).
Camera mini usb port 5 is connected to Protective Earth (PE) by way of 0.4Ω associated with the Win 7 laptop PSU (that physical path is indicated by the dashed yellow line in the diagram.

DC V- (1) terminal is connected (by short wire) to PE (1) terminal
DC V- (3) terminal is connected (by short wire) to PE (3) terminal

DC V- (1) terminal is connected by thick gauge wire to DC V- (3) terminal

Switch operation:
when the switch is closed the rising edge of the 5.15V applied to camera mini usb port 1 causes the camera to perform autofocus and autoexposure and stop in a tight loop just before where the actual exposure would start. When the switch is opened, the negative edge of this 5.15V step is used by the camera to proceed with the exposure - very tight synchronisation can be achieved across a high camera count.

Switch current flow description:
When the switch is closed, current flow into the camera at mini usb port 1 is 0.43mA. This current divides in two paths at the common potential point V Batt- / mini usb port 5 as 4/9th's (0.19mA) to DC V- (1) and 5/9th's (0.24mA) to laptop psu PE. The fraction of current at DC V- (1) then then returns to the switch psu (DC V- (3)) via short wire and the fraction at laptop psu PE via the mains earth conductor.

If all of that is based in reality, the first change I would like to contemplate is:

A/ remove the short wire connections between DC V- (1) & PE (1) and DC V- (3) & PE (3)
B/ disconnect Camera mini usb port 1 from laptop psu PE

Laptop psu PE is then the only system point connected to mains PE and the switch current can only return by a single path. My understanding is that the usb differential D+/D- ptp data comms should be unaffected by my proposal.

Initial Query:

with PSU (1) & PSU (3) no longer connected to PE they are, of course, "floating" with respect to it. Is there possibility that dangerous conditions could result in the system as a result of my proposal? where/how?

PS: the metal cases of PSU (1) & PSU (3) are connected to PE whether or not the short wire links between DC V- & PE are present.

I would strongly advise against disconnecting an earth connection which should be present on any electrical equipment.

Where equipment uses a protective earth connection, there need only be basic insulation between live parts and earthed parts. Should that basic insulation fail, the system fusing will prevent the earthed parts becoming live, but without the earth connection they would become live.

My advice is to investigate replacing the power supplies that you want floating – with class II power supplies which do not require an earth connection (as the 2 pin hub PSU in your attachment).
 

Thread Starter

niman

Joined Jun 7, 2018
3
@Hymie, thanks for your comment - it proved useful.

I have made some progress with my desire to automate the camera shutter mechanism using an optical transmitter, optical link, phototransistor and a logic level high side, small signal p channel mosfet.

I think (but in no way sure) the attached is a fair representation of the Gate charge/discharge circuit. I calculated the gate capacitance using the "typical" Total gate charge of 1.4nC divided by Vgs condition of -4.5V to give 0.31nF from the mosfet data sheet https://docs-emea.rs-online.com/webdocs/1539/0900766b815399c8.pdf

The phototransistor Ice is approx 150uA somewhere around 200 lux.

Vth for the mosfet is between -.03 & -1V.

I'd like to be able to approximate the time at which Vgs reaches -4.5V but have, so far (assuming my gate charge/discharge circuits are based in reality) been unable to identify the Thevenin equivalent circuits for the "With Light" condition in order to specify that time constant.

I've read this in connection with Thevenin equivalents https://www.allaboutcircuits.com/textbook/direct-current/chpt-16/complex-circuits/
but am unable to see how to apply that with the series gate resistance/capacitance and parallel pullup resistor.

Any comment to help appreciate the charge/discharge transient response in respect of this would be appreciated.

(The camera appears to act as an approximate 13KΩ load in the circuit).
 

Attachments

MaxHeadRoom

Joined Jul 18, 2013
25,469
by Win 7 laptop - for ptp based communication (via the usb hub) with the cameras. It's psu has it's 3rd pin connected to mains PE.
Are you sure the laptop P.S. is referenced to earth ground on the L.V. side?
Unlike a tower or desk top PC, laptops are generally not referenced to earth ground.
When you use a USB connection, on a desk top PC the USB power is referenced to earth GND as opposed to the lap top.
Max.
 

Thread Starter

niman

Joined Jun 7, 2018
3
Are you sure the laptop P.S. is referenced to earth ground on the L.V. side?
Unlike a tower or desk top PC, laptops are generally not referenced to earth ground.
When you use a USB connection, on a desk top PC the USB power is referenced to earth GND as opposed to the lap top.
Max.
Thanks for comment.

The present time constant investigations are without any laptop connection or any psu mains connection.

The optical transmitter is connected to a 5V battery (USB Battery Pack - 2200 mAh Capacity - 5V 1A Output) and a phyical switch. At the other end of the TOslink the phototransistor and mosfet are connected to a similar unit. The high side mosfet is connected to Camera USB Pin1 & the camera is powered by 2XAA batteries.

The physical switch connected to the optical transmitter causes the camera shutter to operate as expected.

I'm trying to gauge how quickly I can switch 5V (with practically no triode region voltage loss across the high side mosfet) to the Camera USB Pin1 in order to control its shutter in an appropriate timeframe - to do that I believe I need the time constant of the Thevenin equivalent circuit attached to my previous post. I'm not sure how to do that. .
 
Top