earch those to find which ones will work down to what voltage with minimum drop out. This circuit limits the output voltage to about 12.4 volts and h as a minimum drop out of 0.15 volts down to a Vin of around 6 volts. This is powering a motor, so the current probably won't change as fast as I've s imulated.
cost is minimal even if the parts count is higher than with a regulator.
LTspice XVII. I wonder why it went from IV to XVII...?
17 volts max with minimum drop out voltage as the input drops to 10 volts? It needs to have a shutdown input as well and draw very low idle current,staying
lking about an output dropping once the input drops below the output plus s ome drop out. That's my point, regulators are spec'd for a minimum drop ou t where they meet full specs. I'm interested in the drop out once they are out of regulation and the pass transistor is turned on fully.
ombining the input DC @ 15V with the battery voltage through diodes to feed the main power rail. The motor is the only heavy draw. I'm wondering if it makes sense to use two FETs back to back to prevent reverse current flow between the two circuits. One pair for the main power source and a second pair for the battery resulting in a lower voltage drop so the battery can be used more effectively.
t flow to the main power input. So two diodes and a regulator or four FETs ?
but much lower power levels. Not sure which way I want to go.
on the raspberry pi to enable using a psu or USB power and not backfeed th e USB
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voltage when passing current in one direction. But it can't be used to lim it the output voltage so well because the lowest drop you can get is 0.7 vo lts or so through the parasitic diode. I need to scrub about 2.5 volt from the main power input or a varying amount from the battery, down to as clos e to zero as possible.
ombine the two supplies. Or maybe only one is needed on the battery. When the power source is unplugged, I'm not sure it would hurt anything to reve rse power the input. I only feeds the battery charger. I guess it would c reate some additional power drain. There is an OFF state where only the mi nimal circuitry is powered with the intent to use less power than the batte ry self drain current of ~100 uA. Heck, the FET diode circuit as drawn wou ld draw >>>
8uA aThe diagram you are likely looking at was posted by someone else to illust rate the properties of the circuit, not a suggestion for what I should use. I don't think the beta is a critical parameter. The circuit seem to prod uce a voltage drop of 80 mV used with an AOD4185 in my circuit.
You must have gotten out of bed on the wrong side these days. What is goin g on with you??? No one "posed" a circuit, "eXplicitly" or otherwise. It was an example of usage showing the concept... as I just explained. Do you not understand the difference?
I don't want 48 uA of leakage. Better to have lower leakage and the speed is not important here. Optimization is about improving the circuit for the particular application. The need here is to keep the vampire drain as low as practical so the self drain rate is not significantly increased. There are other parts of the circuit that are drawing current when off, so no re ason to increase this drain without purpose.
the response to the line power dropping resulting in a half volt drop in vo ltage for half a millisecond in the simulation. I consider that acceptable . The requirements on this circuit are not severe. It's powering a motor, not digital logic. There are also some whopping big caps on this rail to deal with the flyback current from the PWM turning off. Cutting power at t his FET is a different matter, so I added a flyback diode and small resisto r.
s continuous.
rs.
Sorry, I don't know what Zetek transistors you are talking about. The simu lation of this circuit is showing less than 100 mV drop out from the AOD418
This circuit is not directly driving the motor. There is an H-bridge in th e path and 2,000 uF of capacitor on the power to it. I don't think there i s going to be too much sass from the motor, but if there is a diode takes c are of that. No?
The bipolar transistors don't carry the motor current at all. Why do they need to be high current? Unless you are saying something very different. A picture is worth a thousand words.
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rch those to find which ones will work down to what voltage with minimum dr op out. This circuit limits the output voltage to about 12.4 volts and has a minimum drop out of 0.15 volts down to a Vin of around 6 volts. This is powering a motor, so the current probably won't change as fast as I've sim ulated.
st is minimal even if the parts count is higher than with a regulator.
Tspice XVII. I wonder why it went from IV to XVII...?
7 volts max with minimum drop out voltage as the input drops to 10 volts? I t needs to have a shutdown input as well and draw very low idle current, > >ng
g about an output dropping once the input drops below the output plus some drop out. That's my point, regulators are spec'd for a minimum drop out wh ere they meet full specs. I'm interested in the drop out once they are out of regulation and the pass transistor is turned on fully.
ning the input DC @ 15V with the battery voltage through diodes to feed the main power rail. The motor is the only heavy draw. I'm wondering if it m akes sense to use two FETs back to back to prevent reverse current flow bet ween the two circuits. One pair for the main power source and a second pai r for the battery resulting in a lower voltage drop so the battery can be u sed more effectively.
ow to the main power input. So two diodes and a regulator or four FETs?
much lower power levels. Not sure which way I want to go.
he raspberry pi to enable using a psu or USB power and not backfeed the USB
In a simulation I see one potential problem. I have a battery where the US B jack is and the +5 is our +15 supply after passing through a real diode. When the voltage from supply is coming up as it crosses the 14 volts (or whatever) from the battery there is a range of about 0 to 100 mV across the FET with the gate still driven on. I'm seeing a peak of about 2.5 amps in to the battery lasting for about a quarter ms. I suppose this is not going to hurt anything unless the PSU has something that is tripped by this curr ent added to the normal use current.
My concern is that with the battery potentially being charged up to near th e supply voltage minus the diode, we may see this state persist for some lo ng time. 15V - 5% tolerance = 14.25. Battery at 14.15 and we will see 2 amps flowing into the battery. Or is this current surge just a transient effect from the large resistors and the gate capacitance? Yep, slowing the transition makes it go away. Since the brief transient will have no effec t on anything this should work fine.
I've balanced up the differential amp and I think the circuit is ready to g o.
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tput plus some drop out. That's my point, regulators are spec'd for a mini mum drop out where they meet full specs. I'm interested in the drop out on ce they are out of regulation and the pass transistor is turned on fully.
If one has a curve tracer, it's easy to resistor-load a three terminal regu lator and test it just as any two-terminal device. Rising ramp and flat (fixed curr ent after regulation is achieved) out to the full V rating are what you'd be looking for.
Lots of old 7805s work just like that (I don't recall any oscillation probl ems, but that's something to consider).
output plus some drop out. That's my point, regulators are spec'd for a mi nimum drop out where they meet full specs. I'm interested in the drop out once they are out of regulation and the pass transistor is turned on fully.
gulator and
rrent after
blems, but
Yeah, I understand the concept. I was thinking someone might know of some parts that do this rather than my having to test them myself. Doesn't matt er at this point. The circuit is designed, simulated and seems to work ver y well. The P-channel FET, two 2N7002's and a third to provide shutdown. Drop out is no more than 150 mV at 4 amps. Works for me. The same pFET is being used in the pair of low voltage drop diode circuits that will be use d. One on the battery voltage and another on a 5V power source used for de bugging/programming without the main power supply. Seems the MCU can be po wered and programmed over a USB port with no other power on the board.
I'm really getting tired of the project though. I can see so many issues t hat are being half thought out with virtually no planning. This made me th ink of another one. The power will be the 12V rail which feeds a 5V linear and a pair of 3.3V linears. The 5V and one of the 3.3V linears can be dis abled by the CPU. There are some heavier loads on the 5V rail, 100+ mA. T he MCU will be powered on the always on 3.3V rail, but I'll bet no one has looked at the details to make sure the USB port will be functional with the other supplies turned off. If they need to be on for any reason, the supp ly in the USB programmer probably won't be sufficient. So the always on su pply has to power enough of the circuitry so the MCU can be programmed.
Oh well...
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