Hi,
I've got a circuit which needs to operate from either a 9V battery or 5V USB supply.
The twist is that the USB supply should be preferred, so I can't use a simple diode ORing scheme.
Total current demand is perhaps 200mA peak, with RMS well below 10mA.
My first thought looks like this, the intention being that the presence of the 5V supply should turn off the P-channel MOSFET, allowing SD2 to turn on and supply 5V to the load:
But of course that doesn't work, because the difference between the two supplies is higher than the threshold voltage of any commonly available p-channel MOSFET. If my battery supply was ~7V it would work perfectly, but as it is the 'FET is always switched on.
My second attempt, then, looks like this:
But the N-channel 'FET gives the opposite characteristic to that desired - now when 5V is present T2 is on, which grounds the gate of T1 and turns it on. So this version uses two 'FETs to achieve the same thing as no FET's - not terribly helpful.
So, adding a CMOS inverter in front of T2 brings us to this:
This circuit works correctly, but it uses 4 MOSFETs to do what I'd hoped to achieve with 1 or 2.
So, having applied everything I know about digital design to this problem, is there some easy solution I've overlooked?
I've got a circuit which needs to operate from either a 9V battery or 5V USB supply.
The twist is that the USB supply should be preferred, so I can't use a simple diode ORing scheme.
Total current demand is perhaps 200mA peak, with RMS well below 10mA.
My first thought looks like this, the intention being that the presence of the 5V supply should turn off the P-channel MOSFET, allowing SD2 to turn on and supply 5V to the load:
But of course that doesn't work, because the difference between the two supplies is higher than the threshold voltage of any commonly available p-channel MOSFET. If my battery supply was ~7V it would work perfectly, but as it is the 'FET is always switched on.
My second attempt, then, looks like this:
But the N-channel 'FET gives the opposite characteristic to that desired - now when 5V is present T2 is on, which grounds the gate of T1 and turns it on. So this version uses two 'FETs to achieve the same thing as no FET's - not terribly helpful.
So, adding a CMOS inverter in front of T2 brings us to this:
This circuit works correctly, but it uses 4 MOSFETs to do what I'd hoped to achieve with 1 or 2.
So, having applied everything I know about digital design to this problem, is there some easy solution I've overlooked?
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