Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] DC voltage isolation

Status
Not open for further replies.
Is R100 a dummy load or are you yet to indicate your load? What's the label LEAK*CURR? There should be a resistor (maybe 10k) from the base to the source terminals of the MOSFET.
 
Last edited:

Is R100 a dummy load or are you yet to indicate your load? What's the label LEAK*CURR? There should be a resistor (maybe 10k) from the base to the source terminals of the MOSFET.

R100 + connector J9 is to connect a panel mount LED. LEAK*CUUR is the output of the PCB and it will be connected to a high Z digital input switch,
 

The grounds are well separated though.
 

Hi,

YES!

Your DCDC converter can deliver ±33 mA. The relay coil resistance is 360 Ω, resulting in a current of I = 15 V / 360 Ω = 4.17 mA. Check the rest of the ±15 V circuitry if the DCDC converter delivers enough power/current.

greets


UPDATE: This is the corresponding answer to post #20
 

R100 + connector J9 is to connect a panel mount LED...
What current is required to flow through the panel-mount LED and what is the forward voltage?
...LEAK*CUUR is the output of the PCB and it will be connected to a high Z digital input switch,
The hi-Z switch works with 24V input?
 

The 0805 resistors are not appropriate for the task you're giving to them.

- - - Updated - - -

Some of them should not be in the circuit.
 

Hi

I am going to use the next panel LEDs:

https://uk.farnell.com/cml-innovative-technologies/192a0350/led-indicator-24vdc-satin-red/dp/1105495

The output of the relays are connected to the next optocoupler :

https://www.digikey.es/product-detail/es/broadcom-limited/ACPL-247-560E/516-2219-1-ND/2403185

Which 0805 resistors are not appropriate for this? The 0805 resistors of the right side of my pic are used to provide a positive or negative logic output according to the application. The current will flow through them will be very low...
 

Hi,

The current will flow through them will be very low...
And the current through the relay contacts, too?
If so ... and if this is a long time installation.... then consider not to use a "power relay" but a "signal relay" preferably with Au contacts.

Read about "wetting current".
In the past I had such problems where the contacts went high ohmic...

Klaus

Added:
Datasheet says: minimum load 100mA/5V.
 

Calculate for resistor power dissipation so as to select suitable sizes. Also for the relay, it seems it's absolute max power dissipation is exceeded due to the direct connection of the 15V (which is higher than 12V) across the coil. You may need to connect a series resistor to limit the coil current.

- - - Updated - - -

Calculate for resistor power dissipation so as to select suitable sizes. Also for the relay, it seems it's absolute max power dissipation is exceeded due to the direct connection of the 15V (which is higher than 12V) across the coil. You may need to connect a series resistor to limit the coil current.

For the LEDs you just posted link, you should consider selecting an appropriate value (and probably, size) for R100 depending on the number (and configuration) of LEDs that you will be connecting through the JP9 connector.
 
Hi,

Hi,


And the current through the relay contacts, too?
If so ... and if this is a long time installation.... then consider not to use a "power relay" but a "signal relay" preferably with Au contacts.

Read about "wetting current".
In the past I had such problems where the contacts went high ohmic...

Klaus

Added:
Datasheet says: minimum load 100mA/5V.

I agree with Klaus, a signal relay with gold contacts is more suitable for your application. I was only focused on the isolation topic, thus haven't had a look on the load side expect the load voltage. The linked power relay should actually only show the relay coil-voltage rating.

Also for the relay, it seems it's absolute max power dissipation is exceeded due to the direct connection of the 15V (which is higher than 12V) across the coil. You may need to connect a series resistor to limit the coil current.

The stated 400 mW is the power dissipation at 12 V (P = 12² V² / 360 Ω = 0.4 W). A coil-voltage of 15 V is within the relay spedification, where the maximum coil voltage is given by 12 V • 1.7 = 20.4 V.

greets
 

Hi,


And the current through the relay contacts, too?
If so ... and if this is a long time installation.... then consider not to use a "power relay" but a "signal relay" preferably with Au contacts.

Read about "wetting current".
In the past I had such problems where the contacts went high ohmic...

Klaus

Added:
Datasheet says: minimum load 100mA/5V.

Hi!

Yes the current flows through relay contacts are also very low (arround 10mA). Therefore, the option of the signal relays makes sense for this application. After checking commercial parts of signal realys on the market, I designed the next "final approach" (I hope :)).

Final.png

Thanks

- - - Updated - - -

Calculate for resistor power dissipation so as to select suitable sizes. Also for the relay, it seems it's absolute max power dissipation is exceeded due to the direct connection of the 15V (which is higher than 12V) across the coil. You may need to connect a series resistor to limit the coil current.

- - - Updated - - -

Calculate for resistor power dissipation so as to select suitable sizes. Also for the relay, it seems it's absolute max power dissipation is exceeded due to the direct connection of the 15V (which is higher than 12V) across the coil. You may need to connect a series resistor to limit the coil current.

For the LEDs you just posted link, you should consider selecting an appropriate value (and probably, size) for R100 depending on the number (and configuration) of LEDs that you will be connecting through the JP9 connector.

There will be just one LED connected to the JP9 terminal. Therefore, the maximum current flowing through the LED will be 17mA (according to the manufacturer)
 
Last edited:

Hi,

the relay fits good. The resistor R100 has still a too small case size as its consumed power is P_100 = V²/R_100 = 24² V²/2200 Ω = 0.26181 W. Use a 1210 case size, see [1].

There will be just one LED connected to the JP9 terminal. Therefore, the maximum current flowing through the LED will be 17mA (according to the manufacturer)

If you assume a current of 17 mA flowing through the LED, this current also flows through R99, which leads to a power dissipation of P_99 = I² • R_99 = 0.017² A² • 2200 Ω = 0.6358 W. This is too much for a 0805 SMD resistor, see [1]. Nevertheless, the curent of 17 mA will flow through your LED at 24 V without a series resistor, thus you may expect a lower current but probably still to high for a 0805 resistor with a power rating of 0.125 W.

[1] https://www.resistorguide.com/resistor-sizes-and-packages/

greets
 

Hi,

It's recommended that you use a suitable relay.
It's recommended that you calculate power dissipation of resistors and use suitable package.
Both I can not see this in your new design.

Klaus

Added: Ah yes, now the relay is good. My eyes are weak :-(
 

I think R95 is too large. Use 100 ohm.
 

Hi,

I think R95 is too large. Use 100 ohm.

yes, that one is too large compared to your coil restistance. But as your maximum voltage rating is +200 %, you may also omit R95 (0 Ω as in your previous circuitry).

greets
 

Hi!

Then I propose the next solution changing resistor packages by the ones you recommend :)

final2.png

Thanks a lot for your help :)
 

Hi,

1206 can handle 0.25 W. R_100 has to handle ~0.26 W, so as I already suggested you need a 1210 resitor.

greets

- - - Updated - - -

UPDATE: R99 is with an expected power disipation of 0.22 W also critical. I would replace it also by a 1210 resistor.
 

Hi,

you choose 1206 now. It may be rated for 0.4W. Your expected power dissipation is more than 200mW.

You need to know that the rated 0.4W is for chip temperature of 150°C.
With 220mW you still may expect more than 100°C chip tepmperature. It surely additionally depends on layout and ambient temperature..

Klaus
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top