eev blog 262

[bruce273]

suggestion - Is a d flip flop not simpler. perhaps a schmitt trigger for debounce. all easy to get 74 series logic. or will it use more power?

[Psi]

An IC would likely be more expensive. It also wouldn't be very elegant.

[EEVblog]

The whole idea is using passive parts you will likely have elsewhere in your design.
And yes, a chip is likely more expensive, even if it is jellybean.

There is in fact an even simpler refinement to my circuit, and it's blindingly obvious with a bit more thought about how my circuit works, I don't know why it didn't occur to me yesterday while filming.
Let's see if someone can come up with it...

Dave.

[johnnyfp]

A Thyristor in a Forced Commutation circuit?

[electronic_eel]

Good idea to save money on the switch.

I think some users will have a problem with the oscillating when they keep pressing the button. Think elderly people or a device which does need some time to boot up. Most users will keep pressing until they see some reaction.

Do you have some links to circuits which don't do this, but otherwise are simple too?

[amspire]

Dave, you set a challenge for us to find something simpler.    OK, here is my attempt. One transistor, one FET,  one capacitor and four resistors.



I tested it is LTSpice, so that's the reason for the weird switch circuit.

I also added a load as there is one problem with this kind of switch. If the load has filter caps across the supply usually. Those capacitors take a time to discharge. If I increase the load capacitor from 10uF to 100uF, the circuit will not switch off, as at there is enough voltage from the load to turn the circuit back on. I think your circuit has the same problem.

So to make this work, you have to find the worse case decay time of the load, and adjust C1 to suit.

This circuit does not oscillate. If does not matter how long you press the button for.


Richard.

[EEVblog]

Dave, you set a challenge for us to find something simpler.    OK, here is my attempt. One transistor, one FET,  one capacitor and four resistors.

Nice try, but one component too many 

Dave.

[amspire]

Dave, you set a challenge for us to find something simpler.    OK, here is my attempt. One transistor, one FET,  one capacitor and four resistors.

Nice try, but one component too many 

Dave.

You didn't coun't the load by any chance? That is not part of the switch - that is just there to test with capacitance on the output.  My circuit is less then yours by one component (a transistor). 

Edit: If you need to allow for a load with a DC resistance of over a few hundred K, then I do need one extra resistor to make sure it stays off if you hold your finger on the button too long while turning off.

[EEVblog]

You didn't coun't the load by any chance?

Nope.

That is not part of the switch - that is just there to test with capacitance on the output.  My circuit is less then yours by one component (a transistor). 

I have a variation that uses one less component again :-P
I'm sure someone will get it, it was staring me in the face the whole time...

Dave.

[amspire]

You got me Dave. Every way I could think of with one less part has a flaw. I usually needed one extra resistor to slow the capacitor discharge to allow for the voltage decay time of the capacitance in the load.

See if anyone else can come up with the solution.

Richard.

[toggi]

It may not be smaller, but at least it does not oscillate, and it's cheaper 

[dics]

toggi, i really like your version of the circuit ! Thumbs up
Here is a modeled online version of it. I changed the capacitor to a 47nF in this simulation, it works a little better.
Also it needs a load

http://goo.gl/Ua1Yi

There is a small problem with it though : in "on" state, if you keep the push button pressed long enough it turns off at first, stays off as long as you have juice in the capacitor and then it turns on again. If you keep the button pressed just a little bit then it turns off and stays off

[BravoV]

toggi, your version for turning on only, how bout turning off ?

[dics]

It works, capacitor acts like a tank and supplies positive voltage to mosfet's gate.

[amspire]

Toggi, I did look at your solution, but I couldn't get it to work fully when you start adding capacitance to the load.

First you want to make R8 5 or more times the value of R4. Otherwise if you press to turn off but hold the finger on the button too long, it turns on again. I just had trouble find a value of C2 big enough to allow for the load to fully discharge without ending up with such a long recharge time for the R8/C2 time constant that you had to wait for 10 seconds or more before you can switch the  supply again.

The version I posted has the same number of components, except it is easier to get time constants to work well with a load with a capacitive decay.

Dave reckons his answer of how to use seven parts including the switch was in front of his eyes, but I still cannot see it.

Dics, the capacitor has to be greatly increased, not reduced. Your capacitor only works with a resistive load.  A real load has caps across the power rails that take time to decay to zero.

Richard.

[dics]

I lowered the capacitor value for this particular type of simulator (which is like arcade versus real simulation), it is quite slow and because of this the 100nF capacitor was charging really slow

[EEVblog]

Toggi, I did look at your solution, but I couldn't get it to work fully when you start adding capacitance to the load.

Dave reckons his answer of how to use seven parts including the switch was in front of his eyes, but I still cannot see it.

Toggi had it essentially right, except it uses a MOSFET instead of the BJT, thus needing one less resistor.
But as it turns out, it seems it's not new, somebody tweeted the same thing on a Russian site:
http://www.uschema.com/safe_power_switch/

It became really obvious on second thought, because I was obsessed with doing the turn on/off using the gate connected BJT. But when you think about the using the gate of the P MOSTFET itself, it's obvious the existing RC can be used and there is no need for the 2nd latch BJT

Dave.

[dics]

amspire : now that i think about it and after i tested a little, you are right. If the load is capacitive and has a large capacitance then the functionality is quite weird.
If you match the capacitance of the load in the timing capacitor then you can kinda make it work but the timings are quite large.
I really have to start thinking laterally when trying to understand circuits

Later edit: This is what i mean http://goo.gl/RCZUI

[amspire]

The mosfet instead of the transistor does work better then a NPN transistor, but it still does depend on the load dropping to below the MOSFET turn-on voltage before the switching cap discharges. With many loads that can happen. With loads like a microcontroller, they usually go into a very low current mode as the voltage drops, and so it is easy to get the load voltage decaying pretty slowly with modern microcontroller and micro perihperal IC circuits.

It is one of those switches that only works with a cooperative load.

Richards.

[fmaimon]

How about this? It would not depend on the load, as the transistor discharges the capacitor instantly.

[toggi]

I really liked my switch, but as amspire pointed out, it will not work on a capacitive load because the output has to fall to zero before the tank capacitor C2 empties 

[dics]

fmaimon : it works for turning on but if you try to turn in off, C2 in instantly discharged through T3 and is not modifying the voltage on T4's gate

[electroguy]

Dave, as you mentioned towards the end of your video about using a micro, how would you do this for a circuit that has a regulator (4.5-15v in 3.3v out) with EN pin, and a micro in it (has lots of free IO and A-D pins to be used if needed).
I have a couple of designs in my head, but none of them are "elegant"
so far i have 3 resistors, 2 MCU pins (1 IO and 1 A-D) I'll draw it up later and post it if i get a chance.

[george graves]

I've got you all beat.  $0.33 each.



 

[electroguy]

I've got you all beat.  $0.33 each.

yuk

lol