Need help,how to generate this driver wave?

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hurrison said:


Best Regards!

hi,
You require a switched 50mA constant current source and a second adjustable constant current source that can give a linear charge to a capacitor upto a maximum of 150mA, then switch both off.

A timer with an adjustable period and mark/space ratio can do the CC switching.

Does this help?
 
You need a 10khz oscillator to get the 100us. Use a LM555 in astable mode. I would set the duty cycle pretty short. Less than 0.1uS.

Top trace is pin 3, bottom trace is pin 6.

Create a linear ramp from 0 to 2/3 supply. The 0.2 – 1uS is set by C, Re and the voltage at R1+R2. I would set R1,2 so there is about 1 volt across Re. To make the thing more temperature stable add a silicon diode in series with R1. This way the voltage across R1 is nearly the same as the voltage across Re and the 0.6 volts across the diode compensates the B-E voltage of the transistor.

Top trace is the output from the 10khz oscillator driving the trigger input. The middle trace is pin 3. Bottom trace is pin 6. Adjust Re to get a 0.2 to 1uS ramp. In your case the flat part will be around 100uS.

The ramp voltage is from 0 to 2/3 supply. If supply is 10 volts then 0 to 6.66 volts. The resistor values are approximants! The 100k and 20k should be adjusted to divide the 6.66 volt ramp down to 1 volt p-p. The value of 400k and 20k are set to start out the ramp at 0.5 volts and end the ramp at 1.5 volts.

Here is a missing piece! You need to design part of this. The ramp needs to be inverted and referenced to the 10 volt supply.

The last part is a current source. If the input is a ramp that starts out at supply-0.5 volts and ends at supply-1.5 volts then this voltage is also across R4 is a ramp from 0.5 to 1.5 volts. This will cause a current ramp to flow from Q1.

I know there are missing values everywhere. I hope this gets you started.
 

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Last edited:
I forgot one thing. I planned to use pin 3 of the second LM555 to turn off the current source. Low=off High=on.
OR
Now that I think about it again: Sum the voltage a pin6 with the voltage at pin 3. The voltage at pin 3 will add the 0.5 volt offset while the voltage on pin 6 will add the ramp. To make that work we need to divide the voltage on pin 3 down to 0.5 volts and divide the ram voltage to 1 volt and sum them, three resistors.
 
This is all very well, and all very clever, but the question is really very lacking in details - he might not need anything like that?.
 
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