Hi, can anyone describe the operation of this circuit for me, in particular what the 10 ohm resistor is for.
It is meant to achieve a very high level of current through the LED.
Many thanks,
John
0
Add a comment
|
3 Answers
\$\begingroup\$
\$\endgroup\$
2
The time constant of the 10\$\Omega\$ resistor and the 10uF capacitor is about 100usec, so it will recharge the capacitors to about 63% of the lost voltage between pulses.
The intention is to have most of the energy to create the flash coming from the capacitors, without the added complexity of a series switch from the power supply. Another couple A will flow from the power supply.. if we assume the capacitor is discharged from 20V to 10V during the 100ns, the energy is about 1.5mJ from the capacitor and a negligible amount from the power supply. Without the resistor the energy would be hard to predict - probably limited by the inductance of the wires running to the power supply.
If, as stated in your comments, the actual current spike exceeds 50A (any inductance in the source circuit can cause measurement errors, so I have doubts on that number) it's exceeding the maximum current of the MOSFET and probably the LED.
answered Mar 14, 2016 at 15:40
-
1\$\begingroup\$ Actually, it will recharge much more than that. The pulse duty cycle is only 0.1%, so a nominal 20A pulse current will only require 20 mA average, and a 0.2 volt drop across the 10 ohms, which is also supported by i/c = dv/dt, So the caps will remain charged to within better than 0.5 volts of the supply voltage. \$\endgroup\$ Mar 14, 2016 at 17:54
-
\$\begingroup\$ @WhatRoughBeast Yup that's in line with the typical Id with 10V Vgs. \$\endgroup\$ Mar 14, 2016 at 18:04
\$\begingroup\$
\$\endgroup\$
4
The resistor may be intended to damp LC oscillations due to power supply [lead] inductance and the capacitors shown here.
Alternatively, maybe the power supply is not stable into what is essentially a dead short, but a 10 ohm resistive load is acceptable.
There may be other reasons as well, but it is hard to know which is the real justification without knowing more about the physical realization of the circuit and its intended application.
-
\$\begingroup\$ The intended application is to achieve a very high current pulse from the capacitor (Pulse = 100ns). A current of 50A has been measured through the 0.1m ohm resistor. \$\endgroup\$– RossMar 14, 2016 at 15:19
-
\$\begingroup\$ @John - This circuit is a disaster waiting to happen. If you're getting 50 amp peak currents, you're well outside the data sheet maximum currents for both your LED and your MOSFET. Plus, of course, you'd be well advised to get a high-speed photodetector to see how your optical output really behaves. The literature I've seen indicates LED response times of >100 nsec, and phosphor decay times of >1 msec. So I'm not at all sure the 100 nsec pulse width and low duty cycle make any sense. Plus, the LED (blue) will turn off much faster than the phosphor (yellow). \$\endgroup\$ Mar 14, 2016 at 18:33
-
\$\begingroup\$ @WhatRoughBeast The circuit was used as it is the simplest way to generate a high optical output. I am using a high-speed BPX65 photodetector which measured an optical power of around 30W/mm^2 \$\endgroup\$– RossMar 14, 2016 at 19:42
-
\$\begingroup\$ @John - What was the optical pulse width like? \$\endgroup\$ Mar 14, 2016 at 20:27
\$\begingroup\$
\$\endgroup\$
The 10 Ohm resistor is there to prevent the short circuit (Caused by turning the LED on) from tripping the over-current protection circuitry inside the 30V PSU.
The 10m Ohm resistor is a current sense resistor, as i'm sure you know.
The 50A is sourced during the pulse from the two capacitors - ceramic and tantalum can definitely supply this much during a short circuit, they have ultra low ESR and can respond fast enough to the short.
The 10 Ohms should allow enough recharge of the capacitors between pulses to allow for a full voltage (or close) between each pulse.