Beginners question about using a switch with a mosfet

Thread Starter

daeiron

Joined Feb 28, 2022
11
Hi,
I'm trying to figure out the best way to use a switch to power on an amplifier board. But I don't want to run all the current trough the switch.
I thought maybe something like this:

switch.png

Would this work? Or can this be done with a simple transistor (my thought was to go for a mosfet because of the 5A current)? And how can I estimate the heat the mosfet will generate?

Any help or tips would be appreciated!
 

ronsimpson

Joined Oct 7, 2019
2,058
Why are is there a ground connected to the bluetooth (5) and another ground the Q1 is connected to? Generally we don't interrupt ground(s).
MrChip wants you to use a P-MOSFET on the power supply pin.
 

AnalogKid

Joined Aug 1, 2013
10,053
In your drawing, you show a ground connection at pin 5 and a switched ground connection through the FET. What pin is the FET switching? Please post information about the amplifier; link to product/vendor page, user manual, part number, *photo*, etc.

Also, a MOSFET *is* a transistor; the T stands for transistor. If you are asking about using a bipolar transistor instead of a MOSFET, both can work. But for a 5 A load, the FET will draw much less current than a BJT through the switch.

ak
 

Thread Starter

daeiron

Joined Feb 28, 2022
11
Thanks for the quick replies. So it should be either of these:
Where the left one is high side and the right side low side switch. The extra ground on the amplifier was a mistake.
What would be the better option? And would the mosfet be ok without a heatsink or how do I estimate that?

@AnalogKid Yes, I meant a bipolar transistor.
This would be the Amplifier board

switch1.png
 

Thread Starter

daeiron

Joined Feb 28, 2022
11
Okay thanks everyone I think I get it know. I assume the connection in the top left (white square) is gone (from vcc to switch), in @ronsimpson design. Is the 1k resistor still necessary though?

The Rds(on) is about 0.3ohm at 5ampere, so that should be ok I guess.
 

AnalogKid

Joined Aug 1, 2013
10,053
I assume the connection in the top left (white square) is gone (from vcc to switch), in @ronsimpson design. Is the 1k resistor still necessary though?
Nope. Each FET needs a relatively high value resistor (10K or so) from its gate to its source. this is what garanteed that the FET turns off when the switch is open. Without this resistor, the gate is floating and the gate charge (a small capacitance) could hold the FET on for a surprisingly long time.

The other resistor, the one in series between the FET gate and either VCC or GND, is not needed. Most power MOSFETs are rated to withstand 20 V on the gate continuously, and there is no current to limit. Check your datasheet.

If you keep the resistor in there, it forms a voltage divider with the gate pull-down resistor, you want to make sure the voltage at the common node is at least 10 V greater than (n-ch) or less than (p-ch) the source voltage, again, a common "full on" value for many power MOSFETs.

ak
 

sagor

Joined Mar 10, 2019
695
AT 0.3 ohms and 5A, you are generating 1.5W of heat, which will likely require a heatsink of some sort. Not a real big one, but a heatsink anyway...
 

MrChips

Joined Oct 2, 2009
26,077
In the schematic shown the P-channel MOSFET is upside down, i.e. you want the protection diode to be reversed biased.
 

Thread Starter

daeiron

Joined Feb 28, 2022
11
If you keep the resistor in there, it forms a voltage divider with the gate pull-down resistor, you want to make sure the voltage at the common node is at least 10 V greater than (n-ch) or less than (p-ch) the source voltage, again, a common "full on" value for many power MOSFETs.
So if I go with the p-channel design, the gate voltage should be 5v (when vcc is 15v), so that would mean a 10k (which is already there) and a 5k resistor right?
 

ronsimpson

Joined Oct 7, 2019
2,058
Look at the data sheet for your mosfet. Most can handle 20V from G to S. (some only 15V)
Most MOSFETs turn on at a voltage in the 3 to 4V range. (depends on temperature and what part you have) If you put a voltage G-S that is close to the turn on voltage the part will have higher resistance that you want. (will not turn on well at high current)
Bottom line, stay less than 20V and defiantly more than 'gate turn on threshold'.
 

sagor

Joined Mar 10, 2019
695
So if I go with the p-channel design, the gate voltage should be 5v (when vcc is 15v), so that would mean a 10k (which is already there) and a 5k resistor right?
Maximum gate-source voltage is +/- 25V, so feeding close to 15V to gate should be ok. At 5V, it may not be fully on, you need higher gate voltage. Your RDSon of 0.32 ohms is with 10V on gate.
 

shortbus

Joined Sep 30, 2009
9,372
Generally we don't interrupt ground(s).
I'm not understanding that statement. Isn't that what a low side switch does? In the original post schematic if he eliminated the amplifier extra ground(like he did in post #5 drawing using the Nmos) why wouldn't that work?
 

ronsimpson

Joined Oct 7, 2019
2,058
Reason for not interrupting ground. It is common for an amplifier to have a input that is connected to ground and a output connected to ground. When you cut open "power ground" the input and output are not happy. We try to connect all grounds together verry well. Switching power is normal.

As mentioned several ways, MOSFETs are turned on/off by the voltage from Gate to Source. (it does not matter what the voltage is from Gate to Ground)
 

crutschow

Joined Mar 14, 2008
29,773
The power dissipation is the current multiplied by the on resistance of the MOSFET you choose.
AT 0.3 ohms and 5A, you are generating 1.5W of heat
What happened to Ohms' law?

The dissipation is the current squared times the MOSFET on-resistance, thus the MOSFET ON power would be 5² * 0.32 = 8W, which definitely requires a heat-sink.

If you use a MOSFET with <40mΩ on-resistance to keep the dissipation below 1W, than you shouldn't need a heat-sink.
 
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