# Tag Info

28

In a word: Efficiency. You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high. When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not ...

13

Here's an example of a "depletion-mode" NMOS NAND gate circuit I found on (German) Wikipedia: The upper transistor is used in depletion mode to provide a load approximating a current source and balancing the rise and fall times. Due to the higher threshold voltages of early MOS technology, a 12 V supply may have been needed to provide a proper bias for the ...

13

The 8080 used nMOS-only technology (no CMOS = pMOS and nNMOS). When you use nMOS (or pMOS) devices only, you have a couple of choices to build a logic inverter cell (see chapter 6.6 in this document, my answer borrows heavily on this source): nMOS transistor and pull-up resistor. Simple, but not good on an IC because the resistor would take up a lot of ...

12

a better idea for this, if you want to use a PMOS high side switch, would be to have the ATSAMD21 switch a low-side NPN switch, which then switches the PMOS for the SIM530. Example: simulate this circuit – Schematic created using CircuitLab R2 (arbitrarily chosen to be 50x R1) pulls Q1 to ground (off) if D1 is in high impedance state, which it may be ...

10

I'm assuming the type of circuit you are thinking of is this: - For use on your battery powered circuit I see little to say against it. A couple of things though; you need to pick a FET with low $V_{GS(threshold)}$ so that the device is still offering a tiny volt drop at low battery voltages AND you'll need a FET with low $R_{DS(on)}$ so that at 30mA (...

10

Knowing the voltage being switched and max current would greatly improve available answer quality. The MOSFETS below give examples of devices which would meet your need at low voltage (say 10-20V) at currents higher than you'd be switching in most cases. The basic circuit does not need to be modified - use it as is with a suitable FET - as below. In the ...

9

If all the components are identical, or 'symmetrical', this is because for the PMOS the load is connected on the source. The drain current of a MOS transistor is a function of its gate to source voltage, while drain voltage has little effect to it. In the case of the NMOS you are driving the transistor with 3.3V to turn it on, since the source is always ...

8

The load is not the main issue for keeping the Rds as low as possible, it's the Vgs you need to concentrate on. For a PMOS the lower the gate voltage, the lower Rds (as Russell points out, higher absolute Vgs). This means in this case the input signals lowest point will cause the highest Rds (if it's an AC signal) So there are 4 options that come to mind: ...

8

That is called a complementary transistor. BS250 is complementary to 2N7000. Check this list for PFETs with the characteristics you want. However I don't know what popular transistors you'll find there.

8

CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching. From here, be the circuit below for a simple inverter: simulate this circuit – Schematic created using CircuitLab When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a ...

7

Nothing will happen. Neither transistor will be able to turn on. The N-channel enhancement-mode transistor requires its gate to be at a higher voltage than the source (or drain), which can't happen if it's connected to Vcc. Similarly, the P-channel transistor requires a negative voltage on its gate, which can't happen if it's connected to ground.

6

If cost is really such a high priority as you say, then why is reverse polarity protection needed at all? When it's more important that the circuit be cheap than robust, you make it cheap at the cost of robustness. If someone installs the batteries backwards, oh well, that's their problem. Of course the battery polarity needs to be clearly labled, but ...

6

ZVP3306A is not too bad a choice as a complimentary part to the 2N7000, rather cheaper (~30%) than the BS250, but it is still more than six times more expensive than the surface mount SOT-23 BSS84. The 2N7000 is not good for 500mA, so you're not really looking for the equivalent (and you'll probably be pushed out of a TO-92 package into a TO-220 such as ...

6

There are IC's which can provide this functionality. One option would be to use a "Power Multiplexor". An example is the TPS2115. It takes the place of the diodes and switches, and is controlled by a 2V-compatible logic signal. Its internal resistance is 110 milli-Ohm (or less), which give a voltage drop of 0.6 mV (!) at 5mA of current. You can find ...

6

The problem The /OE (output enable) pin of the 74CH595 toggles the output drivers between two states: driving the outputs (high or low) and high impedance (letting the output lines float). It is active low, so driving /OE low causes the 74CH595 to drive the outputs high or low depending on the data you have shifted in, while driving /OE high causes the ...

6

Did you see (and understand) Spehro Pefhany's comment? I take the liberty of turning it into an answer because that's also what I'd suggest. Add a diode in front of your circuit that shorts the voltage if the battery is reversed (=V2). In that case the (resettable) fuse will break the current flow. If the battery is not reversed (=V1) there is almost no ...

6

Yes, it's possible. But just like with a high-side NMOS switch, you'd need to drive the gate beyond the supply rail (below Gnd, in this case) in order to switch it fully on. While there's some motivation to use high-side NMOS switches — they perform better than the equivalent PMOS devices — there's no equivalent motivation to use a PMOS on the ...

5

The residual +1V that you are seeing is being caused by bias on the other signal pins to the uSD card. Current passes from either high levels on the microcontroller I/O pins connected at the SDIO interface or via the 47K resistors that you have on these lines into the controller chip in the uSD card. From there it passes through the input protection network ...

5

Wiki says... In a depletion-mode MOSFET, the device is normally ON at zero gate–source voltage. Such devices are used as load "resistors" in logic circuits (in depletion-load NMOS logic, for example). For N-type depletion-load devices, the threshold voltage might be about –3 V, so it could be turned off by pulling the gate 3 V negative (the drain, by ...

5

The reason the FET turns off so slowly is because you only have 2 kΩ pulling it high. Take a look at the datasheet for that FET. It should show you the actual and effective gate capacitance when switching. The relatively weak 2 kΩ pullup is working against that capacitance. Here is a trick I sometimes use in this situation: The double ...

5

I think the top graph does indicate exactly where instability occurs: - The phase angle is zero degrees and the gain is still +45 dB. In other words it's "gain margin" that is causing oscillation here. The transient graph seems to indicate it oscillating at 250 kHz whereas the AC graph tells me it could oscillate at 100 kHz. This doesn't surprise me given ...

5

One problem with your design is if the overvoltage exceeds M2's maximum drain-source voltage, M2 won't actually disconnect the circuit. Another is that when it's tripped, you'll have significant current and significant voltage in D2 and R1 and they'll get hot, and maybe fail. The canonical solution for non-transient overvoltage protection is a crowbar ...

5

You have sortof the right idea: But the capacitor is in the wrong place. For slew rate control, it should be between the drain and the gate, not the source and the gate as you show it. Putting it between drain and gate causes feedback so that when the drain rises quickly, it turns the FET off more. Just a cap between drain and source can be good enough. ...

4

Unloaded outputs:0.18u process using Thick Oxide (0.35u transistors) Input is a 1us slow ramp and decay to show transitions, other line is the output of the inverter and green line is the strange case with PMOS and NMOS flipped with bulk connection as per normal.

4

I designed for 12 volt NMOS technology some years ago. It uses saturated n-channel transistors for the pull-ups. As described by a previous contributor (List item #2 in this answer), this limits the output voltage to one Vt lower than VDD. The 5 volt supply is used for interfacing with TTL. The -5V supply is used to bias the substrate and bring the Vt to a ...

4

T1 is a simple common emitter inverter. CNTRL_L0 is the input, and the node between T1 and Q2 is the output. If the input is high, the output is low. If the input is low, the output is high. Q2 is just another inverter, but it uses a MOSFET instead of a BJT, and because this is a P-channel device, everything is upside-down. In this circuit, as is ...

4

In The LTSpice help file you can find this table, which I'm too lazy to figure out how to reproduce completely: The table is under LTspice IV -> LTspice -> Circuit Elements -> M. MOSFET in the help file contents tab.

4

Microchip designer here... I've never heard of a "fat" MOS transistor, but there is such a thing as a "thick oxide" MOSFET transistor, sometimes referred to as just "thick" transistors. Typically, a MOSFET transistor is designed such that the electrical performance is good for a particular voltage range. For example, 0.7 to 1 volts. That means that it ...

4

As Ignacio is trying to say, you need to pull the gate of the FET up to the 350V supply rail to turn it off. This makes the VGS ~= 0V. Which means the FET is off. To turn it on, you need to reduce the voltage on this gate, perhaps the full 350V is not healthy for it. Perhaps drop the voltage at the gate to +330V to turn it on. This will make the VGS = -20V ...

4

If you are happy with the dual diode arrangement and the only issue is voltage drop, I think you can get by with one PMOS on each battery. You need to have the body diode forward biased during discharge. Essentially, you are putting the FET in "backwards." When the FET is on, the drop will be very low. I drew up the idea below. This is a quirky circuit in ...

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