The only difference between those two 555's is the type of package. Both have the 200 mA totem pole output. Note that the circuit as drawn draws spikes exceeding this from the 555 during startup, although off-datasheet the short duration of these means it's almost certainly ok but to guarantee reliability it might be worth adding a 10Ω resistor between the junction of the three 10μ pumping capacitors and the 10k to the driver.

 

assuming those are ceramic caps 10u and not the polarized ones that look like nano pop cans lol sorry for the terminology.

So a trace between c2, c4, c6 to between r10 and q4 with a 10ohm?

 

That's it. Whilst it would work with electrolytics I was imagining surface-mount ceramics. The 100u power supply decoupling is probably excessive and that could easily be a 10u ceramic of the same type.

 

Surface mount is fine once I nail it all down I can make a board and this whole think will be compact.

Did I miss what r5 50m was doing? is that a simulated load taking the place of the coils ?

Please forgive my noobness I actually understand whats going on for the most part but as I said b4 I know enough to build it and troubleshoot it just not enough to concept a working circuit on my own.

Building this to test with surface mounts is going to be a pain to put on a project board lol

also I am amusing the switch to turn this all on will go on the 555 Vcc leg?

 

The 50mΩ is a simulated load. Since your loads are coiled and even a low inductance may store significant energy* at these currents it's probably worth putting a diode in reverse (so in the direction where it doesn't conduct) across the load to catch any voltage transients.

There's nothing wrong with prototyping with leaded components, I would be inclined to throw something together on stripboard to get it all working but with a singe lower current MOSFET as there's no real need to test switching high currents. You could add capacitance to the MOSFET gate to bring it up to whatever the datasheet says the gate capacitance for the ones you'll eventually use is and check that the drive waveforms still look ok.

You can switch the supply to the whole 555 circuit or just switch the reset pin - pin 4. What you shouldn't do is disconnect just the VCC pin and leave the other pins connected to the power supply. It is perfectly fine to leave the positive connection to your loads though.

*Analogous to the capacitor, the stored energy is 0.5*L*(I^2)

 

C8 = 1n so on mouser a .001uF

is that a typo
Volt rating on those matter @ all?
Schematic does not state cap voltage.

 

C8 is 1n but fairly non-critical. It merely decouples an internal divider inside the 555.

C1, C8 C9 see only the 4.2V battery voltage. 6.3V parts would be ok but 10V is a bit more generous. The multiplier capacitors see higher voltages, 25V rated parts would be good here (actually the earlier stages have fewer volts but there's no need to make things complicated).

 

C

TY for all the help Richard This should work great.
Now 1 last question if we install a 2s battery that puts out 7.4 volts
Everything here will work the same as long as all components are rated high enough?

I just am unsure about the Schottky Diodes boost area.
If I understand it is boosting up and the 12volt z diode is busting it back down to provide the current to drive the Qs and Ms.
It should if I am not totally noob work with 7.2 volts with no change. to anything.

Replacing R7-8 with a 100k trimmer would allow my to trim the cycle to put out a solid 4volts not loose the top end of the 4.2s to the 90% duty cycle loss th idea is to fire @ 3.7V as long as possible.

If this site had a donate button I would hit it yall are great

 

I just am unsure about the Schottky Diodes boost area.
If I understand it is boosting up and the 12volt z diode is busting it back down to provide the current to drive the Qs and Ms.
It should if I am not totally noob work with 7.2 volts with no change. to anything.

The booster multiplies the input voltage by 4, so with 4V in the output is 16V - 6*V_Schottky ~= 14V, with 7 volts in the output would be around 26V. That's probably ok for many power MOSFETs (check Vgs_max) but it's higher than needed, you could safely remove a multiplier stage. Taking off the last two diodes and capacitors would give you around 20V for 7.2V in and that's plenty.

You should ensure that with maximum battery voltage you never exceed Vgs_max, but also that with minimum battery voltage (under load!) you still get enough gate drive to turn the MOSFETs on hard. If you can't do that by tweaking the number of multiplier stages you could add a zener shunt regulator.

The zener in the circuit now doesn't regulate the gate voltage, it provides a threshold to enable gate drive. This ensures that the output MOSFETs are fully off until there's enough drive to switch them hard into conduction. Without it they would dissipate lots of power in the 10ms or so it takes for the gate supply to pump up. It might not actually be necessary depending on the MOSFETs chosen. If you need it (i.e. if your MOSFETs can't cope with dissipating a couple of hundred Watts for 10-20 ms) then you should chose the zener voltage apropriately, if you don't need it you can remove R12, R13, Q4 and D9 and connect R10 directly to the collector of Q2.

 

The booster multiplies the input voltage by 4, so with 4V in the output is 16V - 6*V_Schottky ~= 14V, with 7 volts in the output would be around 26V. That's probably ok for many power MOSFETs (check Vgs_max) but it's higher than needed, you could safely remove a multiplier stage. Taking off the last two diodes and capacitors would give you around 20V for 7.2V in and that's plenty.

You should ensure that with maximum battery voltage you never exceed Vgs_max, but also that with minimum battery voltage (under load!) you still get enough gate drive to turn the MOSFETs on hard. If you can't do that by tweaking the number of multiplier stages you could add a zener shunt regulator.

The zener in the circuit now doesn't regulate the gate voltage, it provides a threshold to enable gate drive. This ensures that the output MOSFETs are fully off until there's enough drive to switch them hard into conduction. Without it they would dissipate lots of power in the 10ms or so it takes for the gate supply to pump up. It might not actually be necessary depending on the MOSFETs chosen. If you need it (i.e. if your MOSFETs can't cope with dissipating a couple of hundred Watts for 10-20 ms) then you should chose the zener voltage apropriately, if you don't need it you can remove R12, R13, Q4 and D9 and connect R10 directly to the collector of Q2.

Picking the Mosfet you listed in your schematic

http://www.mouser.com/ds/2/200/irfh6200pbf-24138.pdf

Absolute Maximum Ratings section
VGS Gate-to-Source Voltage ±12

would that not require just 2 boost sections 4.2 * 3 12.6v

 

Picking the Mosfet you listed in your schematic

http://www.mouser.com/ds/2/200/irfh6200pbf-24138.pdf

Absolute Maximum Ratings section
VGS Gate-to-Source Voltage ±12

would that not require just 2 boost sections 4.2 * 3 12.6v

You are correct

I kinda just picked a MOSFET at random but it actually looks very suitable for your task. The two things to look at are Vgs_max=12V and also at Figure 12. Designing for low battery voltage you can see that 3V wouldn't be enough gate drive, one boost section might just about be enough 3V*2 - 0.3*2 - 0.6V = 4.8V (the final 0.6V being from Q2) which would get the on-resistance about as low as it goes, but only for the "typical" mosfet.

Two boost stages would give a lowest gate voltage of 3V*3 - 0.3*4 - 0.6V = 7.2V which is good and a highest of 4.2V*3 - 0.3*4 - 0.6V = 10.8V which is below the max rating, though a 12V zener between Vhigh and GND would do no harm. I would change D9 to be a 6V2 giving a threshold of around 6.8 Volts.

Something important I've just spotted - the threshold voltage for that MOSFET is specified as 0.5-1.1V and the gate driver as drawn only pulls down to 0.6V because of Q3, therefore it's possible that the MOSFETs may not be fully off. To fix this add another schottky diode between the Gate_Drive node (diode anode) and the collector of Q1 (diode cathode). That lets Q1 pull the last of the gate charge out through the diode. Not necessary if you end up using a MOSFET with a higher threshold.