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Power supply +VCC -VEE turnoff using n ch mosfets?
Power supply +VCC -VEE turnoff using n ch mosfets?
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Old 8th February 2018, 12:47 PM   #1
rhythmsandy is offline rhythmsandy
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Default Power supply +VCC -VEE turnoff using n ch mosfets?

want to turn on and turn off the power supply using mosfets. I will be using a microcontroller to turn off the amp. Is there any way to do it?
the following circuit doesnt work I have tried to simulate it.
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Last edited by rhythmsandy; 8th February 2018 at 01:24 PM.
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Old 8th February 2018, 02:49 PM   #2
Ketje is offline Ketje  Belgium
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Originally Posted by rhythmsandy View Post
want to turn on and turn off the power supply using mosfets. I will be using a microcontroller to turn off the amp. Is there any way to do it?
the following circuit doesnt work I have tried to simulate it.
Your drawing shows depletion mosfets,the have to be turned off with sufficient negative gate voltage.The other (more common) type is normaly turned off and need a positive voltage on the gate to turn on.
Mona
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Old 8th February 2018, 09:22 PM   #3
Monte McGuire is offline Monte McGuire
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Power supply +VCC -VEE turnoff using n ch mosfets?
It should work - I've rigged up a few and they do work. You need to pay attention to the polarity of the optocoupler output, and also make sure that the optocoupler can produce enough voltage to fully enhance the MOSFETs, and not provide too much to break down the gate relative to the source. Also make sure you're driving the LED properly - check the data sheet, but they generally want 10mA of drive, so get the resistor value right re. the LED's on voltage and your driving voltage. For example: (Vdrive - Vled)/0.01A = series resistor in ohms.
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Old 8th February 2018, 09:36 PM   #4
scottjoplin is offline scottjoplin  Wales
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Why not use a couple of mosfets before the rectifier? If one power rail is disconnected it could damage the amplifier and/or speaker
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Last edited by scottjoplin; 8th February 2018 at 09:41 PM.
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Old 9th February 2018, 01:18 AM   #5
rhythmsandy is offline rhythmsandy
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Quote:
Originally Posted by scottjoplin View Post
Why not use a couple of mosfets before the rectifier? If one power rail is disconnected it could damage the amplifier and/or speaker
im using almost 0.1F per rail so there will be considerable charge left to which leads into the load and hence damages more.
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Old 9th February 2018, 01:19 AM   #6
rhythmsandy is offline rhythmsandy
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Quote:
Originally Posted by Ketje View Post
Your drawing shows depletion mosfets,the have to be turned off with sufficient negative gate voltage.The other (more common) type is normaly turned off and need a positive voltage on the gate to turn on.
Mona
I believe the second type is better to use as there wont be micro controller noise entering into it. Do you have any schematic which worked for you?
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Old 9th February 2018, 01:45 AM   #7
Monte McGuire is offline Monte McGuire
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Power supply +VCC -VEE turnoff using n ch mosfets?
The photocouplers prevent microcontroller noise from having anything to do with anything - you get several thousand volts of galvanic isolation and only a tiny bit of capacitive coupling. Plus, if you get the right MOSFETs so that they are deeply enhanced when the photocoupler is on, any fluctuations in the LED current will not affect the conductivity of the MOSFET channel - t's already so enhanced that it can't be any further enhanced. The 0.1F of rail capacitance should also prevent problems.

Just get the basics of the circuit right and you'll be fine. The drawing you posted shows no series resistor into the photocoupler. You need one - see my first post on what value to make it.
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Old 9th February 2018, 02:12 PM   #8
rhythmsandy is offline rhythmsandy
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Is there be a continuous heat dissipation in the mosfet when its turned on?
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Old 9th February 2018, 02:57 PM   #9
Monte McGuire is offline Monte McGuire
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Power supply +VCC -VEE turnoff using n ch mosfets?
Yes, equal to the product of the current through it and its "on resistance". You can get MOSFETs with very low on resistances now, 30mΩ for pretty small units and well under 1mΩ for larger ones. A typical 2A rated MOSFET (a tiny one) will probably have an on resistance of around 50mΩ. So, that'll be 50mW at 1A on current.

Look at the graph of Rds vs. Vgs in the datasheet and see how it varies with increasing Vgs. At a certain point, additional Vgs will decrease Rds very little, since the channel is almost completely enhanced. This is the region you want to drive a MOSFET to, since it will be maximally efficient.

The photocoupler will generate a voltage, usually around 7-8V, and initially, most of the charge generated is spent on charging the gate capacitance. After that's done, the voltage rises further, and gets to its ultimate value. So, look at the Rds versus Vgs graph and see what the value is at the open circuit output voltage rating of the photocoupler, and you can estimate these losses pretty accurately.

BTW, that photocoupler mentioned in the schematic you posted, the PV1050N, is long obsolete. Vishay and Toshiba make modern ones that you can actually buy. Digikey stocks them. I'd post a list, but my list is only of ultra-thin packaged versions for SMT that are truly annoying to use outside of a custom designed PCB. I dug around and here's a product selector from Toshiba: Photovoltaic-Output Photocouplers | TOSHIBA Semiconductor & Storage Products | Europe(EMEA) Cross reference that to Digikey's stock and you can find something that can actually be ordered.

Last edited by Monte McGuire; 9th February 2018 at 03:25 PM.
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Old 9th February 2018, 06:34 PM   #10
rhythmsandy is offline rhythmsandy
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but depletion mosfets which are available have little current ratings can be ok for 1 pair of output transistors rather than 6 or 7 pairs of output transistors.

I have checked the following case: Can anyone answer the following fault condition scenario.

1. One of the Positive and negative transistor is blow up and there is DC going into the load. Consider in that case using a Microcontroller we can cutoff the speaker load but we see the positive voltage is applied to the emitter of the bottom transistor (PNP). So my question is will it destroy the bottom transistor also? I am worried about the latched state of the mechanical relay which is used for the power supply cuttoff for both positive and negative.

2. I can use two relays one for positive voltage rail after the PSU caps and one mech relay for the negative. Mosfet based protection for the speaker. Now if one of the transistor blows off then the mosfet at the speaker will trigger off and the speaker is saved. Now when the positive transistor is blownup then the current will flow heavily into the negative side about all the current gets to the negative transistor also and hence I believe that will also burn out. So Im worried about the relay latch I will be using a 20Amp rated relay but will the relay latches up and burnup the whole amp PCB?
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