R6 and R7 could not both be 7.4k and the output be 100-110v; as the junction between R6/R7 is where the regulator gets its' feedback. Once that junction reaches ~1.24v, the IC knows that the output voltage has reached the target voltage, and to decrease the ON-time to the MOSFET.

 

Sorry, here is the real circuit: (worked on this alot)

Can you simulate it for me please (guess the unknown values).
R7 can't be measured. (I dont know why, but it showed me R7=7.4 like R6 because of the multymetter that measured R7 parallel with R6).
How can I measure The capacitors???

Thank you.

 

There's something wrong with the way that R4,R5 and C3 are connected. Placing a cap where you have it will cause strange things to happen.

There needs to be a bypass capacitor between Vin and GND - that's where C3 was in the previous simulation.

I'm not going to try to simulate that.

 

you right!

Can you simulate this one?

 

There must be another capacitor attached to Vin and GND somewhere close to the LTC1871.

 

Anyway, here's the simulation without a bypass cap on Vin. It's not accurate, because the power supply in this case is "ideal". There must be a capacitor attached to Vin that you are not showing.

In the prior simulations, C3 was a 100uF cap from Vin to ground. Now it's a 0.1uF cap from the junction of R4/R5 to ground, and causes the regulator to delay a bit before starting up. As you don't have a method to test the caps, I'm leaving most of them at the values they were in the example in the datasheet, as that's the only sound reference I have at the moment.

 

@JK-flipflop - What kind of solenoid are you driving? What does it do on your robot?
Wouldn't it be easier to get a solenoid that is rated at your robot battery voltage?

 

This driver stores energy in the caps to get greatly increased amperage through the solenoid on the initial firing, thus a much stronger stroke.

If he changed to one that was rated for the battery voltage, it would be anemic by comparison, or very large, heavy and expensive.

With solenoids, it's all ampere-turns. You need x number of amperes through y number of turns to generate the necessary EMF to do the work. Going from a 100v solenoid to a 12v solenoid would mean far larger wire diameter, etc etc.

 

There is another capasitor in the lower board C16 between IC ground to vcc.

I finally found R7 ressistance: 600K.
I found Q1 too: STP40NF10L http://www.datasheetcatalog.org/datasheet/SGSThomsonMicroelectronics/mXyuxss.pdf
Can you simulate it now?

Thank you very much for your effort!

@JK-flipflop - What kind of solenoid are you driving? What does it do on your robot?
Wouldn't it be easier to get a solenoid that is rated at your robot battery voltage?

I am using pushh solenoid of 0.64Ω, it's for kicking IR ball, this solenoid is for 24V and my battery is 12V anyway it's need 15A-20A current to kick the ball strong enughe, and my battery can't suplly such current.

 

Changing R7 to 600k results in about 105v output, which is within the range you say you were observing previously.

As before, CC is far smaller than what's on your board; that's just to make the simulation finish quickly.

As before, the capacitors which you don't know the value of, I simply left as-is from the example in the datasheet, or made a SWAG (Scientific Wild-Arsed Guess) for their values.

I don't think your MOSFET is the STP40NF10L, as that MOSFET is only available in a TO-220 package. Your MOSFET appears to be a SMD (surface mount device), and the package dimensions are quite different.

 

Q1 is TO-220 package they just cut off the heat-sink and the name is written on it (p40nf , 10L ,ST) so it must be STP40NF10L.

R4 is not connected to Vin.
SENSE connected to ground.
The value of the ressistors is right.

Why Have you changed it??

We are so close!!

 

Oops. You show R4 connected to Vin.

I changed Sense to ground instead of the source terminal/L1 junction.
I don't have a model for the STP40NF10L, so I'm leaving it as-is for now.

I simulated it, and the output voltage is nearly 300v, current through L1 is around 270A. Double check that sense line connection.

 

I think it may be like that, because of the low value resistor that considerd as wire.

I tripel checked it... I don't know why it's connected to the ground.
thank you.

 

R1 is new.

Are you certain that R1 is 2 Ohms?
Short your meter leads on the lowest Ohm setting, and wait several seconds for it to settle out. Remember that reading.
Then measure the resistor again, waiting the same amount of time.
Subtract the 1st reading from the 2nd reading.

If it actually is 2 Ohms, then it will dissipate 4 Watts of power, and would be burned right off the board.

 

When I short the probes it shows me between 2.8Ω-3.1Ω, R1 = 3.1Ω , with what you have told me its need to be as wire value..

what is this R1?

*I did it again and when I shorted the probes it showed me 1.5 and not 3...
** his name is "R10"

look what I found!!:

It's connected as I said!! (look SENSE connected to Q9 source and than 0.005 resistor)

Please simulate this and analyze the schematics I gave.
Thank you.

 

I sure am taking a lot of time on this thread.

I'm going to give it a rest for awhile.

 

Please can you just simulate it for the last time with this tiny resistor???

Thank you (you really helped me a lot)

 

Feel free to download LTSpice from Linear Technology's website, and simulate it yourself.

It's not fair for me to be spending so much time on your project, when other people need help, and I have so little time to help people.

 

OK you right Thank alot for your help!! you are PRO.

* just one question how do I decide what voltage to simulate on the graph?
and can you send your file??

 

The ASC file is attached.

LTSpice download page: http://www.linear.com/designtools/software/

I changed the MOSFET to one that's in the default LTSpice library, so that won't cause you a problem when you try to run it.