Pull-up resistor is running slowly (inverting MOSFET circuit)

Question

There's a problem with my fan control circuit.

It works well, but when I first power up the circuit, the fan runs for about 0.2 seconds. The reason seems to be the time until the pull-up resistor is activated. How can I fix it?

[EDIT]

I'm sorry it took so long.

The purpose of the circuit I'm going to make is this.

A/C & H/T controller

1. Fan speed controlling
2. Relay on/off
3. Temp ADC

I was trying to using a microcontroller but I changed.

I opened some of the products that control the speed of the fan and I found that most products use 555 timer IC.

Of course, my circuit also works normally but I thought using 555 timer IC was simpler and cheaper.

This is my circuit:

This is cope:

With this 555 timer circuit, the FET does not heat at all.

Answer 1

I agree with user @sai, who suggests the problem is caused by the +5V rail rising way too slowly, leaving Q2 switched off, and allowing the MOSFET Q4 to be switched on by R10 to +BATT.

To fix that, we need to make Q2's default state, in the absence of +5V or a PWM signal, correspond to the motor being off.

Also, your current scheme inverts the "meaning" of the PWM signal, so that low duty cycle corresponds to the motor being powered most of the time, and vice versa.

Both of these problems can be fixed by adding another inverter stage between Q2 and Q4:

^{simulate this circuit – Schematic created using CircuitLab}

At this high PWM frequency of 20kHz, the delay between the input's falling edge and Q4's gate following suit, will be a significant chunk of the entire cycle. This is caused by the BJTs taking forever to recover from deep saturation:

This can be mitigated by preventing the transistors from saturating so deeply, which we can achieve with a little emitter degeneration:

^{simulate this circuit}

You might notice base and gate resistances R9 and 22Ω are gone, we don't need them any more. I made some changes to R2 and R3, and added R4 and R5, which have the effect of reducing gain and saturation, allowing the transistors to recover much more quickly:

If I got those values right (and I believe I did), the circuit will work well from 5V or 3.3V PWM input, and any +BATT potential from +20V to +30V. I do not think this design is suitable for battery voltages outside that range.

Also, since R6 and R7 divide the battery voltage by two, you can get rid of D1. The IRLR7843's maximum \$V_{GS}\$ (±20V) will not be exceeded.

Answer 2

R6 has to charge up the base capacitance before Q2 turns on. In the mean time, R10 is charging up the gate capacitance 10x faster and turning on Q4. Then after Q2 turns on it has to discharge Q4 gate capacitance.

Make R6 smaller and R10 bigger to get the timing closer and prevent turn on.

Answer 3

I misunderstood your problem initially. After reading your question again, it seems you have a power sequencing problem. I am guessing that your BATT comes up first and the +5V comes up only 0.2s later. Thus, base of Q2 is low upon powerup and this makes gate of Q4 high and hence motor is ON for a duration of 0.2s.

Edit: Solution 1 The supply of the pull-up resistor can be made BATT instead of +5V and the resistor value can be made 100k.

Edit: Solution 2 This is a more complex solution and not so elegant. If you want to retain the +5V as pull-up for some reason, you can add a circuit that detects that +5V is available and keep gate of Q4 pulled down until +5V is ready. Here is one such idea.

I added pull-down M3 to define the gate low. The pulldown will be active until +5V is ready. Thereby, the motor will be OFF even though the +5V supply comes up late.