Than

It is indeed frustrating when you claim a 50% understanding, but you don't take advice. Yes, a BJT can switch the gate, but so can a SPST switch. Try the switch with a resistor from gate to source as I suggested and see if you can get that to work. Then you are free to enhance the switch on the gate anyway you want. BTW, what prompted you to change the gate-source resistor to 200 Ω? Do you understand that connecting the gate to ground will result in a 50-mA current or 0.5W dissipated in that resistor? Be sure to use a resistor that will handle that power.

Going step-by-step is not "reading articles and application notes;" it is a common method to troubleshoot a problem. In the past, you have complained about people at AAC not trying to help you. Please try to learn from that.

John

Thank you so much .

I always take advice of AAC people and their suggestions. I have learnt a lot from this forum . I also read article posted here. I can see the difference .

We can not connect SPST switch so let me drive it by NPN BJT or NMOS with controller signal.

One more thing I need to know that how do calculate resistor value and it's power.

Rb= 4.7K ...that I can calculate by looking the datasheet of BJT by considering min ib .

Rb= (Vs-Vbe)/ibmin

200E-How to calculate that 200E and it's power.

Please guide me.

 

E(voltage) = I*R
W(power) = I^2 *R

Oops, I used the old value of 10V in my earlier calculation. I will edit that post.

5V/200Ω = 25 mA
(25 mA)^2 * 200 Ω = 0.125W

 

E(voltage) = I*R
W(power) = I^2 *R

Oops, I used the old value of 10V in my earlier calculation. I will edit that post.

5V/200Ω = 25 mA
(25 mA)^2 * 200 Ω = 0.125W

I think in BJT,

Ic is dependent up on Beta times ib..

Ic= beta×Ib

So how do you calculate the above one..!!!

Let me understand biasing of BJT.

Is that the way to lower the Ic more . Means can it be 10mA to avoid extra battery power consumption.

Thanks...!!!

 

My calculation was based on grounding the resistor directly. If you use the BJT, the calculation is obviously different.

At this point, make either circuit (that is, the one with or the one without the BJT) and see of that switches the output as you it. If it doesn't work with the BJT, try it with a mechanical switch.

Replacing that mechanical switch with the microcontroller can be done after you get the power supply switched. Since the microcontroller can probably sink about 25 mA, you can probably use it instead of the BJT. I would use a larger resistor in that event as there is no need for such a high current in your circuit.

John

 

If a 3 V drive signal works with a NPN drive transistor, but does not work with an N-channel drive transistor, the problem is the drive transistor. Since it works with the NPN, isn't that the solution? Go back to a 10K resistor between the power MOSFET gate and source, and increase the NPN base resistor to 10K. Ib plus Ic will be less than 1 mA. You can increase the power MOSFET gate-source resistor to 100K, but this might slow the turn-off transition a bit,

ak

 

Thank

My calculation was based on grounding the resistor directly. If you use the BJT, the calculation is obviously different.

At this point, make either circuit (that is, the one with or the one without the BJT) and see of that switches the output as you it. If it doesn't work with the BJT, try it with a mechanical switch.

Replacing that mechanical switch with the microcontroller can be done after you get the power supply switched. Since the microcontroller can probably sink about 25 mA, you can probably use it instead of the BJT. I would use a larger resistor in that event as there is no need for such a high current in your circuit.

John

Thank you so much..!!!

I will check the both way and get back to you.

The circuit attached in post #19 works fine. So I do not think I have to go for any alternate solutions because I need to stick with cost as well.

Now the only concern is power consumed by this circuit . It should be very minimum during ON mode.

As sugested by AnalogKid -

Base resistor -Rb= 10K
And Rsg = 10K - 100 K

Please let me how to calculate the total current consumption.

If I can calculate ib=(3V-Vbe)/10K
How to calculate ic theoretically.

 

Please let me how to calculate the total current consumption.
If I can calculate ib=(3V-Vbe)/10K How to calculate ic theoretically.

i = e / r
i = (5.0 V - Vcesat) / 100K
i = 4.95 / 100K

ak

 

i = e / r
i = (5.0 V - Vcesat) / 100K
i = 4.95 / 100K

ak

Sorry to interrupt you...!!

If we are talking about the Source to Gate resistance of PMOS i. e. Rsg=100K.

According to you , i=49.5uA.

Is that the current flow to Collector to emitter of NPN BJT while it turns ON.
I do not understand this calculation.

Please underestimate if somewhere I am missing..!!!

You may think that it's better to teach a infant child rather than me...!!

 

DC through the gate-source resistor is defined by a permutation of Ohm's Law as stated in post #27. The voltage across the resistor is the voltage at one end (5.0 V) minus the voltage at the other end. When the drive transistor in on (saturated), the voltage from its collector to emitter (connected to ground) is called Vcesat (or Vce(sat)), the saturation voltage. For a small signal transistor, this usually is 0.1 V or less. My estimate for such a low collector current is 0.05 V. So the total voltage across the resistor is 4.95 V, and the current through it is 49.5 uA. That is the current through the series string of the gate-source resistor and the saturated transistor, and hence the current through each element in the string.

ak

 

DC through the gate-source resistor is defined by a permutation of Ohm's Law as stated in post #27. The voltage across the resistor is the voltage at one end (5.0 V) minus the voltage at the other end. When the drive transistor in on (saturated), the voltage from its collector to emitter (connected to ground) is called Vcesat (or Vce(sat)), the saturation voltage. For a small signal transistor, this usually is 0.1 V or less. My estimate for such a low collector current is 0.05 V. So the total voltage across the resistor is 4.95 V, and the current through it is 49.5 uA. That is the current through the series string of the gate-source resistor and the saturated transistor, and hence the current through each element in the string.

ak

Moving further to this thread...
I have got some efuses from texsas which can do the same job.

Let's have a look of TPS2592Zx series by enabling and dissabling EN we get the same results.

Or else we can control external NFET by BFET pin of chip ...