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Soft Start Using a PFET

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wuchy143

Member
Hello,

I'd like to design a soft start circuit for a keyboard which runs by a +5V source. The keyboard will consume between 100mA to 300mA depending on if back lighting is enabled from the +5V source.

When plugging in our keyboard to power I would like to limit the inrush current so the circuit(keyboard) gets powered gradually. I have made a quick PFET version of what I think may work. At least I think it might. Attached is my schematic.

How would I test this circuit to see if it works? Are my values in the ballpark? Or is this a bad design? I can simulate with multisim but am unsure how practical it is.

Thanks a lot!

-mike
 

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Roff

Well-Known Member
That won't work. The FET is configured as a source follower, so you will only get 3 or 4 volts on the output. If you swap source and drain, you will get some turn-on delay, but due to the gain of the MOSFET, you will still get pretty fast risetimes unless you increase the time constant dramatically. What risetime are you looking for?
I designed a circuit that will give you a linear ramp for the output voltage, but it requires an additional NPN and PNP.
 

wuchy143

Member
I will have to check with with sales or the spec and see what type of risetime is desirable. I"m assuming it's as few milliseconds for a ballpark.

Exactly how would you set up the bjt's? And do you keep the source and drain opposite to what I have shown? A linear ramp in voltage would be ideal! Exactly what I'm trying to do.

Thanks

-mike
 

Roff

Well-Known Member
I will have to check with with sales or the spec and see what type of risetime is desirable. I"m assuming it's as few milliseconds for a ballpark.

Exactly how would you set up the bjt's? And do you keep the source and drain opposite to what I have shown? A linear ramp in voltage would be ideal! Exactly what I'm trying to do.

Thanks

-mike
Well, I had a circuit that worked well in simulation until I added a large-value capacitor (10-100uF) across the load. This is typical of an actual keyboard load, and unfortunately, the circuit oscillated.
If you can live with a few milliseconds risetime, and a few tens of milliseconds of delay, then the attached circuit should work. As you probably know, you can tweak the times by changing the cap and resistor values.
 

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wuchy143

Member
Hi Ron,

Well your circuit does indeed work! I made a quick breadboard of it and see that with the values you have shown it takes about 30ms for a full +5V to appear on the drain. Attached is a sketch out the output.(drain output)

So first I will admit that I"m an EE and forget exactly how to figure out time constants and such. For my own knowledge how would I adjust R2 and C2 to get a longer time delay.(the RC network) Would R1 play a role at all either? It looks like I would want to increase the value of C2 to get a longer time delay but am not sure. t=RC. So would the current time delay be t = rc = 10000 * .1uF = .001?

Also what role does D1 play? Is it blocking current so none goes down D1 and R1?

Regards,

-mike
 

Roff

Well-Known Member
Hi Ron,

Well your circuit does indeed work! I made a quick breadboard of it and see that with the values you have shown it takes about 30ms for a full +5V to appear on the drain. Attached is a sketch out the output.(drain output)

So first I will admit that I"m an EE and forget exactly how to figure out time constants and such. For my own knowledge how would I adjust R2 and C2 to get a longer time delay.(the RC network) Would R1 play a role at all either? It looks like I would want to increase the value of C2 to get a longer time delay but am not sure. t=RC. So would the current time delay be t = rc = 10000 * .1uF = .001?

Also what role does D1 play? Is it blocking current so none goes down D1 and R1?

Regards,

-mike
The time constant I posted is 0.1uF*1MegΩ, which is a 100millisecond time constant. The delay and risetimes are not easily calculated, because the gate voltage has to reach its threshold before the FET begins to turn on, but then the voltage gain causes the FET to turn completely on during a small percentage of the remaining gate exponential waveform.
Attached are the gate and output voltage waveforms from a simulation of the circuit I posted. Note the ≈55mS delay before turn-on begins, and the ≈17mS risetime. I don't know why you got the nonmonotonic rise in the output voltage. It looks like stray inductance, or internal inductance in the keyboard.
If you increase the capacitance, the delay and the risetime should both increase proportionally.
The diode and the 10k resistor combine to discharge the cap rapidly when the supply is unplugged, so that it will be ready for being plugged back in immediately.
 

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wuchy143

Member
Wow. Very helpfull. Thanks. I like the addition of the diode and resistor you added. I will be adding that to my design. I didn't think about being able to plug back in immediately. I do that with any electronics I own so I guess it makes sense to design products which are capable of doing it :)

I figured out why my output wasn't monotonic. The keyboard wasn't actually powered up because the +5V wire connecting up to my keyboard was open.(bad crimp!) So, the drain was left open. Which I'm not 100% sure why but any open pins can do weird things. I made a new cable and now I see exactly what your graphs says on my scope. I then tweaked it to a 1uF cap which gave me a 100ms rise time so I now know things are making sense.

For the 17ms the gate voltage levels out then once the FET is completely on(+5V at drain) it then continues on being more of a decaying exponential. Is there a reason for this?

Again thanks. You have been very helpful.

Regards,

-mike
 

crutschow

Well-Known Member
Most Helpful Member
For the 17ms the gate voltage levels out then once the FET is completely on(+5V at drain) it then continues on being more of a decaying exponential. Is there a reason for this?
When the FET starts to turn on you get the Miller effect (Miller effect - Wikipedia, the free encyclopedia) which multiplies the gate-drain capacitance (which is large for a power MOSFET) by the transistor gain. Once the transistor is turned completely on there is no more Miller effect.
 
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wuchy143

Member
Thanks crutschow.

In the beginning of this post I had my FET configured as a source follower.(my schematic is in the beginning post) Roff said that if I hooked it up that way that I wouldn't see +5V at the source and that I would actually see 3 or 4 volts instead. So, I quickly breadboarded it out of curiosity(swapped source and drain on to-220 package) and don't see 3 or 4 but I see 5V. Though, it does not ramp really at all. It might take a few milliseconds for it to get up to +5V. I'm a little confused and was wondering if anyone had some insight into this?

-mike
 

Roff

Well-Known Member
Thanks crutschow.

In the beginning of this post I had my FET configured as a source follower.(my schematic is in the beginning post) Roff said that if I hooked it up that way that I wouldn't see +5V at the source and that I would actually see 3 or 4 volts instead. So, I quickly breadboarded it out of curiosity(swapped source and drain on to-220 package) and don't see 3 or 4 but I see 5V. Though, it does not ramp really at all. It might take a few milliseconds for it to get up to +5V. I'm a little confused and was wondering if anyone had some insight into this?

-mike
My bad. A source follower would have the resistor from gate to +5V, and the cap from gate to ground. Your only problem is that you have source and drain swapped, so the intrinsic body diode is passing current to the output as soon as you plug it in. Since MOSFETs are otherwise symmetrical (except for the body diode), the output immediately goes to ≈4.3V, then goes on up to 5V as soon as the capacitor charges. This all assumes you have a load, either real (the keyboard) or dummy (160Ω paralleled by a large cap, like 100µF). If you don't have the load, the results will be only slightly different.
 

dedjazzgadgetz

New Member
Soft-start with +/- 7 volts supply

How would the same principle be applied for a +7 volts/-7 volts soft-start (in my case: to avoid audio click/pop) ? Please excuse my ignorance: I've only just learned of PFETs reading this !!!

Thanks,
Joël

Edit : my guess would be using a NFET on the -7 volts side, with all other polarised components reversed (mirror-image circuit) ? How's that sound ? Getting warm ?!
 
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Roff

Well-Known Member
How would the same principle be applied for a +7 volts/-7 volts soft-start (in my case: to avoid audio click/pop) ? Please excuse my ignorance: I've only just learned of PFETs reading this !!!

Thanks,
Joël

Edit : my guess would be using a NFET on the -7 volts side, with all other polarised components reversed (mirror-image circuit) ? How's that sound ? Getting warm ?!
What is a a +7 volts/-7 volts soft-start?
 

ChrisP58

Well-Known Member
Most Helpful Member
I think he is looking for two soft-start circuits. One on the +7V rail, and one for -7V rail.

dedjazzgadgetz, Yes, the negative side components will need to be the opposite polarity of the + parts. You should try to match the turn-on characteristics of the two mosfets as close as you can.
 

Roff

Well-Known Member
I think he is looking for two soft-start circuits. One on the +7V rail, and one for -7V rail.

dedjazzgadgetz, Yes, the negative side components will need to be the opposite polarity of the + parts. You should try to match the turn-on characteristics of the two mosfets as close as you can.
Makes sense.
If I were to start from scratch, I would make a soft-start +7V regulator, then operationally invert the +7V output.
Joel, do you have that flexibility, or do ±7V already exist?
 
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dedjazzgadgetz

New Member
Oops... I was being way too vague; sorry about that, guys!
(...first post, you can imagine)

I need to power-up a circuit (Roland GK-3 guitar-synth hex preamps, nestled inside a guitar) fed by the bipolar power-supply (thus the +/- 7 volts) of the "synth" device itself (a Roland VG-99)...

Only, the plugging (and unplugging for say, swapping guitars) of the multi-conductor cable connecting the two causes a KraSh/Pop/bAng! in the audio. Thats why this little circuit "seemed" adequate for ramping-up & down the (+7 V) positive voltage...

If so, how would I go about "soft-starting" the negative (-7 V) side, using the same principle ?

Thanks sooo much for reading on,
(mucho appreciated !)
Joël

P.S.: I intend to build a switch-box selecting 1-of-3 guitar inputs, so: a selected GK preamp circuit would be "soft-started" before the next one is "soft-powered-down"... hopefully: noiselessly ! Am I making any sense, here ? ;¬ ) -"Ah, musicians !"...

EDIT: hey, thanks there, Chris (yes, you got it right)! and thanks to you too, Ron (love the avatar! Is that you ?) for the original neat little circuit; I'll give it a go, first thing tomorrow with a (matched turn-on characteristics) P & NFET pair... oh: and let's not forget wuchy, I mean mike! for the original post... thanks guys! G'night...
 
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Roff

Well-Known Member
Oops... I was being way too vague; sorry about that, guys!
(...first post, you can imagine)

I need to power-up a circuit (Roland GK-3 guitar-synth hex preamps, nestled inside a guitar) fed by the bipolar power-supply (thus the +/- 7 volts) of the "synth" device itself (a Roland VG-99)...

Only, the plugging (and unplugging for say, swapping guitars) of the multi-conductor cable connecting the two causes a KraSh/Pop/bAng! in the audio. Thats why this little circuit "seemed" adequate for ramping-up & down the (+7 V) positive voltage...

If so, how would I go about "soft-starting" the negative (-7 V) side, using the same principle ?

Thanks sooo much for reading on,
(mucho appreciated !)
Joël

P.S.: I intend to build a switch-box selecting 1-of-3 guitar inputs, so: a selected GK preamp circuit would be "soft-started" before the next one is "soft-powered-down"... hopefully: noiselessly ! Am I making any sense, here ? ;¬ ) -"Ah, musicians !"...

EDIT: hey, thanks there, Chris (yes, you got it right)! and thanks to you too, Ron (love the avatar! Is that you ?) for the original neat little circuit; I'll give it a go, first thing tomorrow with a (matched turn-on characteristics) P & NFET pair... oh: and let's not forget wuchy, I mean mike! for the original post... thanks guys! G'night...
Yeah, that's me in my avatar.
We need to know how much current the GK-3 draws from the ±7V power supplies. I looked at the datasheet, but it just assumes you are going to be powering it from a compatible unit. I looked at the VG-99 datasheet, but it doesn't even specify the supply voltages. Where did you get the ±7 volt info?
 

dedjazzgadgetz

New Member
Well, I have an old GK-2 service manual (but the same specs have been used throughout the whole GK/GR/VG series) that shows the +/- 7 Volts supply rating... and I read somewhere the current draw of a typical Roland GK hex preamp box is about maybe 45-60mA, 100mA -tops.

Now, regarding that neat soft-start circuit: how difficult/complicated would it be to modify so it's controlled by a logic level say, a CD4017 1-of-10 output ? (I haven't gotten to buy the N & PFETs to fool around with yet, but I should be able to, later on this evening...)

I'll try to figure it all out while dabbling (but FETs are fidgety, methinks...). Still cruising the net for ideas...

Anyhoo, thanks for the interest, Ron !
Later,
Joël

EDIT: Oops ! Roland DID use +/- 15 volts for their earlier 70's/80's GR-100, GR-300, GR-500, GR-700...
-Got the MOSFETs but haven't breadboarded anything yet, darn...
 
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