On Sun, Dec 20, 2015, at 03:45 PM, Gene Heskett wrote: > On Sunday 20 December 2015 04:27:39 Gene Heskett wrote: > > [... back to readable size, on my charge-pump detection to run the SSR's] > > > Has anyone else concocted such a circuit, and were you able to get > > enough pump power out of it running the BoB on 5 volts, with the pump > > at 500 hz, to run an SSR without resorting to an external current > > booster? > > Now I am both bumfuzzled and confused. There really ought to be a formula > for calculating the size of the capacitors needed, but there is > absolutely zip in ITT #5, nor in the hambook of about the same vintage. > > So, I guess its build it, and test it by the old fashioned cuss & cry > method. Seems to me there really ought to be, in the year 2015, a plug > in the blanks formula for that. Sigh... >
My gut feeling is that you won't be able to do it. First, lets make sure we are talking about the same circuit. A google image search found this schematic of a charge pump, which is the circuit I normally use. http://www.zen22142.zen.co.uk/Theory/diode_pump/Image152.jpg (refer to the circuit on the left) Assume that the input is swinging from 0.0 to 5.0 volts. Also assume that the loading is negligible. When the input is low, the voltage on the left side of Cp is 0.0, and the right side is -0.7V due to the drop in D1. When the input goes high, the left side of Cp is 5.0V, and the right side is 4.3V, assuming zero change in the voltage across Cp. Because of the drop across D2, the output is 4.3-0.7 = 3.6V. Pretty marginal. And that is with zero load, and an output that swings a full 5V. No real output will swing all the way down to 0.0 or up to 5.0V, so you lose another couple tenths of a volt. The above assumes that there is NO CHANGE in the voltage across Cp. That means no current flows into or out of Cp, and the circuit delivers no output current (I said "assume loading is negligable"). To deliver any load current, the voltage across Cp has to change, and that will further drop the output voltage. Say you need a load current of 10mA. That is 10 milli-coulombs per second. If the input frequency is 500Hz, then each cycle needs to deliver 0.010/500 = 20 micro-coulombs. If you need an output voltage of 3V, and the ideal no-load output is 3.6V, then the voltage across Cp can only be allowed to drop 0.6V when 20 micro-coulombs of charge is extracted from it. C = Q/V = 0.000020 / 0.6 = 33uF. That is a HUGE pump cap, and will badly load down whatever is driving it. When I make a charge pump, I usually think of something like Cr = 0.1uF, and Cp = 0.01uF. Charge per cycle at 0.6V droop is 0.006 micro-coulombs, and I run it at 10kHz instead of 500Hz, for a total load current of 60 micro- amps. Put a 100K load resistor as a load and drive the gate of a logic level MOSFET and it works fine. But getting milli-amp currents out of a low-powered charge pump is much much harder. John Kasunich ------------------------------------------------------------------------------ _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
