I want to make a circuit that outputs three states, please help

kaka009

I want to make a circuit that outputs three states, please help [Copy link]

Oh my god, please help me, thank you

It is difficult to meet both high impedance and low level 2V output at the same time

maychang

Q1 is turned on and Q3 is turned off, and the output terminal (OUT) is at a high level (but cannot reach 28V, only 25V~26V).

Q1 is turned off and Q3 is turned on, and the output terminal is at a low level (can be as low as below 2V).

Q1Q3 are both turned off, and the output is in high impedance state.

Note: If both Q1 and Q3 are turned on, both tubes may burn out. This state is prohibited.

maychang

"It is difficult to meet both high impedance and low level 2V output at the same time"

At any one time, the circuit can only be in one state.

High impedance and low level cannot be "simultaneous" .

kaka009

maychang published on 2022-3-9 11:30 Q1 is turned on and Q3 is turned off, and the output end (OUT) is high level (but cannot reach 28V, only 25V~26V). Q1 is turned off and Q3 is turned on, and the output end is low level...

I registered this forum because I saw your reply on a thread. Oh my god.

kaka009

maychang posted on 2022-3-9 11:32 "It is difficult to simultaneously meet the high impedance state and low level 2V output" At any time, the circuit can only have one state. High impedance state and low level, it is impossible to &ld ...

Now I see that what you said in your reply is basically realized, that is, the output has three states. At first, I thought that when Q3 is turned on, OUT will not be pulled to 2V, but will only be pulled down to 2V when OUT is pulled up. Now it seems that the required function can be basically realized.

The 28V conduction voltage drop of PMOS is due to the diode, right? Also, won't the output OUT of Q3 also be 2V when Q1 is cut off due to the existence of 2V? I want to make an adjustable power supply connected to a low level, 0-2v adjustable, and it is estimated that 0V cannot be reached.

maychang

kaka009 posted on 2022-3-9 13:03 Now I see that you have basically realized what I said in your reply, that is, the output has three states. At first, I thought that when Q3 is turned on, OUT will not be pulled to 2V, but only at o ...

Q1 should not use a P-channel MOS tube, but an N-channel MOS tube, which is more convenient.

If Q1 uses a P-channel MOS tube, the source of the tube should be connected to the positive terminal of the power supply and the drain to the OUT terminal. As shown in the figure, the diode in Q1 is always turned on, and the OUT terminal cannot be pulled down by Q3.

chunyang

There are problems with the original poster's circuit, and maychang has already explained the reasons. As for the best solution, the original poster should first talk about the requirements for the output current, as well as the source and number of the control terminals.

kaka009

maychang posted on 2022-3-9 15:04 Q1 should not use a P-channel MOS tube, but an N-channel MOS tube, which is more convenient. If Q1 uses a P-channel MOS tube, then the source of the tube should be connected to the...

I put the Pmos in the wrong place and I found it later

kaka009

chunyang posted on 2022-3-9 15:38 There is a problem with the original poster's circuit. Maychang has already explained the reason. As for the best solution, the original poster should first talk about the requirements for the output current and the control end...

MCU IO control, no requirement for output current, 1ma is enough

chunyang

kaka009 posted on 2022-3-10 13:35 Single chip microcomputer IO control, no output current requirement, 1ma is enough

What is the input of the next stage? If the current is not large, the resistor voltage divider method can be considered. If the input impedance of the next stage is higher, it is even simpler, and the calculation is greatly simplified. Even if you must directly use the 28V and 2V power supplies (for example, as a monitor for the two power supplies), you can also connect resistors in series for protection, so that the control timing and transient problems can be easily solved.