Mosfet switch design help

[madstudio]

Hey guys. I need to design simple mosfet (or other) switch able to discharge capacitor and desired threshold point (check the pic). Unfortunately I'm not experienced with mosfet devices design. Can u help about it?
thanks

 

[Arouse1973]

Mr Tesla.

I think designing your own mosfet might be behond this forum But you could look at logic level MOSFETS if you are planning on using a micro to switch it on. You dont show the value of capacitor or mention how quick you want it to be discharched in. This will make a difference.
Adam

 

[madstudio]

Thanks Adam. Yes, I'm sure that this can be searchable by google but honestly I don't know where to start :/ So help is appreciated. Capacitor is 10uf , discharge time is not critical, but have to be reasonable fast.

 

[Arouse1973]

http://www.digikey.com/catalog/en/partgroup/n-channel-logic-level-gate-fets/16119

 

[madstudio]

http://www.digikey.com/catalog/en/partgroup/n-channel-logic-level-gate-fets/16119

Cool. thanks. Now when I buy some of Mosfets from Digikey how to arrange them to make this cap discharge at voltage lower than 10v?

 

[(*steve*)]

madstudio, (may I call you mad?) the first thing we need to know are some details of your circuit.

These things are essential:

1. Is your switching signal 40V? or could it be less? If so, what is the minimum?
   
2. What current can be drawn from the 40V source?
3. How do you define "reasonably fast?" is it 1 second, 1/10 of a second, 1/100th of a second, 1/1000th of a second? or something else?
4. What do you regard as "discharged"? 1V, 0.1V, 0.01V, something else? (don't say 0V)
   
5. Does the current source really have a max voltage of 6V?
6. Can the current source be safely shorted to ground?
7. Are the common grounds shown on your initial diagram correct?
8. Is the polarity of the charge on the capacitor always as shown (i.e. positive with respect to ground)?

 

[madstudio]

Hey Steve, thanks for chiming in. Of course, Mad is my description and nick
1. 40v is the voltage what have to be monitored. Lowering is possible but the switch have to look at the changes and react at Vin lower than 10v.
2. Here is the trick, it has to be low as possible, but few ma's is possible. Lets say 3ma should be limit.
3. 1/10 sec is ok
4. Yes, important. It should be below 1v. Everything below 1v is OK.
5. Yes, 6v 5ua
6. Yes.
7. It has no separate grounds, if u ask is the current source isolated.
8. Yes.

I really appreciate your help guys cheers, Mad

 

[(*steve*)]

I think everything is clear except point 1.

Is the input voltage between 10 and 40 volts, and any voltage in this range should activate the mosfet when it is present?

Or is it a continually varying voltage that you want it to trigger the mosfet at some specific threshold?

 

[madstudio]

I think everything is clear except point 1.

Is the input voltage between 10 and 40 volts, and any voltage in this range should activate the mosfet when it is present?

Or is it a continually varying voltage that you want it to trigger the mosfet at some specific threshold?

Hey Guys. No, Max Voltage is 40V but it's variable and Fet have to ground the capacitor when Vin fall under 10V.
So, yes, Mosfet should be triggered and be in the same state when the Vin is less than 10V on Vin.
What u think about this simple solution?

 

[(*steve*)]

Hey Guys. No, Max Voltage is 40V but it's variable and Fet have to ground the capacitor when Vin fall under 10V.
So, yes, Mosfet should be triggered and be in the same state when the Vin is less than 10V on Vin.

OK, that means you'll need a more complex circuit then. I was also under the impression that the presence of the voltage would turn the mosfet on, shorting the capacitor. If the absence of the voltage turns it on, there will be a problem because the circuit will need some power, so eventually it will turn off again (unless you use a rather more exotic depletion mode mosfet I guess).

What u think about this simple solution?

That circuit won't work well.

The four problems are that:

1. Any adjustment that allows the gate to go higher than 12 to 20 volts (it depends on the mosfet) will damage or destroy the mosfet
2. There is no exact voltage that the mosfet turns on and off at.
3. With the gate near the threshold, the mosfet may be partially on resulting in the capacitor being partially discharged, or very slow to discharge.
4. If it did work, it would short the capacitor as the voltage rises, not as it falls.

You will probably need some external power to operate a comparator circuit.

Do you have some other source of power, or are you limited to around 3mA from the trigger source?

 

[madstudio]

OK, that means you'll need a more complex circuit then. I was also under the impression that the presence of the voltage would turn the mosfet on, shorting the capacitor. If the absence of the voltage turns it on, there will be a problem because the circuit will need some power, so eventually it will turn off again (unless you use a rather more exotic depletion mode mosfet I guess).



That circuit won't work well.

The four problems are that:

1. Any adjustment that allows the gate to go higher than 12 to 20 volts (it depends on the mosfet) will damage or destroy the mosfet
2. There is no exact voltage that the mosfet turns on and off at.
3. With the gate near the threshold, the mosfet may be partially on resulting in the capacitor being partially discharged, or very slow to discharge.
4. If it did work, it would short the capacitor as the voltage rises, not as it falls.

You will probably need some external power to operate a comparator circuit.

Do you have some other source of power, or are you limited to around 3mA from the trigger source?

Thanks Steve. That was just mine desperate attempt to design. As I mention before I have no enough skills but really need this circuit. Well, maybe I can spend few more mA's I was talking about 100% safe consumption with 3ma. So, can some logic in front help? So the fet will see just on/off?

 

[(*steve*)]

Perhaps you need to describe exactly what this is meant to achieve.

Also why the 10V, and is this critical (what should happen at 9.99V or 10.01V -- does it have to be different?)?

 

[madstudio]

Hey, this circuit has to discharge soft start buck converter capacitor

 

[madstudio]

Perhaps you need to describe exactly what this is meant to achieve.

Also why the 10V, and is this critical (what should happen at 9.99V or 10.01V -- does it have to be different?)?

And yes, the 10v is not critical. It can be in the +/- 0.5v tollerance

 

[(*steve*)]

Oh, OK, so you want the soft start to happen again after a brief power outage (which presumably it's not doing).



Try this. I'll have to figure out component values later, but here is what it does. Tell me if you think it's right.

Rx and Cx stand in place of your soft start capacitor and current source.

When the power is on and the input voltage exceeds Vz (D1) by some margin, Q1 is turned on pulling the gate of Q2 low, turning it off (and allowing Cx to charge).

When the input voltage falls below Vz, Q1 turns off, allowing Q2 to turn on, shorting the capacitor Cx.

In more detail...

As the input voltage starts to drop, the voltage across R1 drops. This causes the current through D1 to fall. Eventually there is insufficient current to maintain both 0.7V across R2 and excess for the base current of Q1. At this point Q1 begins to turn off. This allows the gate of Q2 to be pulled high. If the input voltage falls very fast, C1 contains enough charge to turn on Q2 (which need not be a logic level mosfet). In this case, D2 prevents C1 from discharging prematurely.

Note that because the input voltage can exceed Vgs(max), which is typically around 20V, the zener diode D3 between the gate and source of Q2 exists to clamp the gate voltage well under Vgs(max), but well above what is required to turn the mosfet on sufficiently.

As the voltage rises from zero, Q1's base is held to the emitter voltage keeping it turned off. This will allow the voltage at the gate to rise. If C1 was discharged, this will begin to charge C1 and at about 5V the mosfet will be substantially turned on. (if C1 was still at or above 5V, the mosfet would already be on. So the soft start capacitor is still held at 0V. As the input voltage rises above Vz (D1) a current starts to flow through D1. This causes a voltage drop across R1. When the input voltage is sufficient that the voltage drop across R2 reaches about 0.65V (see the equation below) the Base-emitter junction of Q1 becomes forward biased and any additional current flows through the base of the transistor. This allows the transistor to turn on. As the voltage rises further, more and more current flows through the base until at Vcc of 40V, the base current is close to 2mA.

R1 - 10k
R2 - 5k6
R3 - 39k
C1 - 4.7uF 63V
D1 - 9V1 Zener
D2 - 1N4148
D3 - 9V1 zener
Q1 - BC548
Q2 - 2N7002 (note that Vgs of 5V is plenty to turn this on)

With the component values shown, the device will switch in the region of 10V to 11V.

Q2 will be turned off when the voltage drop across R2 (Vbe) reaches about 0.65V to 0.7V This is about Vbe + Vz + (R1/R2)*Vbe. For the component values chosen it will be calculate as close to 11V. However it is important to note that the zener voltage will be a little low at low currents, so the switching point may be lower.

At 40V, the current drawn will be: (40 - 9.1)/15600 + (40 - 0.6)/22000 = 0.002 + 0.001 = 0.003 A = 3 mA

Note that the mosfet may remain turned on for a substantial time after power is removed as it is limited (mostly) by leakage through Q1 and D3 and the self discharge of the capacitor. As soon as the input voltage hits about 10V, the mosfet will be turned off again

If you're lucky the datasheet for whatever zener you choose will give you V vs I curves. As used here, R2 means that you're interested in Vz at about 100uA. Note that in addition to Vz varying with current, zener diodes have a significantly wide tolerance, the actual zener voltage (even at a higher current) can vary +/- 5% (although you rarely see this amount of variation in practice).

 

[madstudio]

Oh, OK, so you want the soft start to happen again after a brief power outage (which presumably it's not doing).

View attachment 14607

Try this. I'll have to figure out component values later, but here is what it does. Tell me if you think it's right.

Rx and Cx stand in place of your soft start capacitor and current source.

When the power is on and the input voltage exceeds Vz (D1) by some margin, Q1 is turned on pulling the gate of Q2 low, turning it off (and allowing Cx to charge).

When the input voltage falls below Vz, Q1 turns off, allowing Q2 to turn on, shorting the capacitor Cx.

In more detail...

As the input voltage starts to drop, the voltage across R1 drops. This causes the current through D1 to fall. Eventually there is insufficient current to maintain both 0.7V across R2 and excess for the base current of Q1. At this point Q1 begins to turn off. This allows the gate of Q2 to be pulled high. If the input voltage falls very fast, C1 contains enough charge to turn on Q2 (which need not be a logic level mosfet). In this case, D2 prevents C1 from discharging prematurely.

Note that because the input voltage can exceed Vgs(max), which is typically around 20V, the zener diode D3 between the gate and source of Q2 exists to clamp the gate voltage well under Vgs(max), but well above what is required to turn the mosfet on sufficiently.

As the voltage rises from zero, Q1's base is held to the emitter voltage keeping it turned off. This will allow the voltage at the gate to rise. If C1 was discharged, this will begin to charge C1 and at about 5V the mosfet will be substantially turned on. (if C1 was still at or above 5V, the mosfet would already be on. So the soft start capacitor is still held at 0V. As the input voltage rises above Vz (D1) a current starts to flow through D1. This causes a voltage drop across R1. When the input voltage is sufficient that the voltage drop across R2 reaches about 0.65V (see the equation below) the Base-emitter junction of Q1 becomes forward biased and any additional current flows through the base of the transistor. This allows the transistor to turn on. As the voltage rises further, more and more current flows through the base until at Vcc of 40V, the base current is close to 2mA.

R1 - 10k
R2 - 5k6
R3 - 39k
C1 - 4.7uF 63V
D1 - 9V1 Zener
D2 - 1N4148
D3 - 9V1 zener
Q1 - BC548
Q2 - 2N7002 (note that Vgs of 5V is plenty to turn this on)

With the component values shown, the device will switch in the region of 10V to 11V.

Q2 will be turned off when the voltage drop across R2 (Vbe) reaches about 0.65V to 0.7V This is about Vbe + Vz + (R1/R2)*Vbe. For the component values chosen it will be calculate as close to 11V. However it is important to note that the zener voltage will be a little low at low currents, so the switching point may be lower.

At 40V, the current drawn will be: (40 - 9.1)/15600 + (40 - 0.6)/22000 = 0.002 + 0.001 = 0.003 A = 3 mA

Note that the mosfet may remain turned on for a substantial time after power is removed as it is limited (mostly) by leakage through Q1 and D3 and the self discharge of the capacitor. As soon as the input voltage hits about 10V, the mosfet will be turned off again

If you're lucky the datasheet for whatever zener you choose will give you V vs I curves. As used here, R2 means that you're interested in Vz at about 100uA. Note that in addition to Vz varying with current, zener diodes have a significantly wide tolerance, the actual zener voltage (even at a higher current) can vary +/- 5% (although you rarely see this amount of variation in practice).

Hey Steve, you are the man Thanks. I will test your design in next few days and come back to report. Cheers

 

[madstudio]

Hey Steve. Thanks again. What you think about changing the Q1 for darlington type (B =2000 or more)? It can decreese base current and in that manner I can increese R1 to approx 22-33k and lower the zener current. It can save me few more ma's

 

[(*steve*)]

yes you could try that but there will be other issues such as the higher BE voltage required and perhaps a softer turn on and off. You may be able to simply increase the values of R1 R2 &R3 by a factor of 10.

I picked values that are reasonably safe. Higher resistances might mean you have some issues with speed and leakage but a factor of 10 might be ok

 

[(*steve*)]

oh, i assume that means the circuit works as it is?

 

[madstudio]

Hey Steve. No I still didn't try it on the bench. Yes, I will first try it with your values and if it works (i cant see any reason not to work) i will try with resistor factor 10.
Thanks again. Cheers