freewheeling diodes to speed up solenoids?

This is the original drawing micahg tried to post:

micahgORIG.jpg


But I'm having troubles trying to understand how the capacitor voltage and the output voltage are added to get to 48V?
 
Sorry about the delay getting back. The capacator charges to 24 volts using the resistance of the coil and the resistor. The charge rate needs to be fairly slow so it requires an off time long enough for the charging to occur. When the output turns on it puts +24 vdc on the negative end of the capacitor pulling that voltage up to +24 volts. Since the cap is charged to 24 volts it acts as if it were a 24 volt battery in series with another 24 volts giving 48 volts until the cap starts to discharge. So you would get a 48 volt pulse decaying to 24 volts. After the cap is discharged, the diode starts to conduct supplying 24 volts to the coil, (minus about 0.7 volts for the drop across the diode.) Since the current charging the cap in very small, it has no effect on the solenoid. And as mentioned before, the output would have to be able to take -24 vdc when off.
 
If that drawing is correct I do not see it either.

If this is dc and the -24v is there all the time the coil will just see approximately 24vdc across it all the time....I think.

Again, I think, the + and - 24 will negate each other in referece to the common. Please explain if I am thinking wrong.
 
hiii

It will shut off the voltage which was created by magnetic field once u made it off.I think it doesnot increase the response time.
 
When the output is turned off there is basicly no connection there, the cap charges to 24 volts with the negative end facing the output and the positive end facing the coil. Then think of it as a battery in this polarity. Then the output turns on, think of this as another battery referenced to the common of the coil. The output is +24 volts connected to the negative end of the cap (battery) and the +24 volts of the cap is in series with the 24 volts of the supply. At the time the output turns on there is +48 volts on the coil. After the capacitor discharges (battery goes dead) there is +24 volts on the coil, supplied thru the diode,(minus the diode turn on voltage 0.7v). For this to work the recycle-off time of the solenoid would have to be longer than the recharge time of the cap. Also, during the time the output is on there would be 48 volts across the resistor, so the resistance would have to be high enough to prevent excessive current there.
 
micahg said:
...Then think of it as a battery in this polarity. Then the output turns on, think of this as another battery referenced to the common of the coil...
RIGHT! Now I've seen the light (should feel a bit embarrassed having forgotten so much basic electronics).
If anyone is still interested:
CapCoilSmall.JPG



Thanks for your 'enlighting' efforts, micahg.

micahg said:
...For this to work the recycle-off time of the solenoid would have to be longer than the recharge time of the cap. Also, during the time the output is on there would be 48 volts across the resistor, so the resistance would have to be high enough to prevent excessive current there...the output would have to be able to take -24 vdc when off.
Too bad there are (too) many IFs involved before one can implement this setup.
 
wrong thread

Rats! with all the registering and authenticating, I did not get back to the correct thread. This is about suppressing solenoid spikes without slowing the action so much.
 

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