0
\$\begingroup\$

I am looking to power a circuit using a 5.5V, 5F supercapacitor. I want to charge it to approx. 5V but I don't want it to discharge into my circuit until it has charged to 5V. I can do this by simply using a switch or disconnecting/connecting the appropriate wires of the switch, but I want to control this automatically. Is there any way to do this? Thank you for your help!

Edit 12/8/18: I built the following circuit, kindly suggested by user Transistor. For my SCR, I am using a NTE5402 and for the Zener diode I am using a 4.7V 1N4732A. The resister is 1k ohm and the circuit is a simple buck booster (Adafruit Powerboost 1000) connected to an Arduino.

enter image description here

\$\endgroup\$
1
\$\begingroup\$

This is an incomplete answer and may not be suitable for one or many reasons but it is simple and give you some ideas.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. SCR (thyristor) latching switch.

How it works:

  • SCR1 is normally off.
  • As the voltage on C1 rises above about 4.7 V Zener diode, D1, begins to conduct in reverse-breakdown.
  • When the voltage across R1 gets high enough (about 0.7 V) the SCR will trigger and turn on. "Circuit" will now be powered. SCRs, once triggered, remain on until the current through them falls below the holding current (mA).
  • When C1 is discharged sufficiently the SCR will turn off and the process can begin again.
  • The OFF switch allows you to interrupt the circuit to reset it early, if required.

Someone else may be able to correct the values for D1 and R1. It's bedtime for me.

\$\endgroup\$
  • \$\begingroup\$ Can you please give me some pointers on how to select an appropriate SCR and Zener diode? A lot of the SCRs I've seen were rated at 40V - is that the voltage at which the SCR would be "triggered" to switch the circuit on? \$\endgroup\$ – alpha26 Dec 4 '18 at 22:48
  • \$\begingroup\$ No, that would be the maximum reverse voltage the SCR could take. I suggest you find a good SCR tutorial - it doesn't have to be complicated, study it, select an SCR, read the datasheet - there will be a load of new terminology and then ask another question adding in a reference to this one. (If you paste a question URL into your question it will display the question title. You can also include the URL of my schematic into your new question for reference. \$\endgroup\$ – Transistor Dec 4 '18 at 22:52
  • \$\begingroup\$ I want to double-check my understanding please: So the capacitor would charge until it gets to the reverse-breakdown voltage of D1, in this case 4.7V. Up until this point, it can't connect to the circuit because the switch is open. When it does this, the current passing through D1 triggers the SCR, which then allows the circuit to get powered. Is the reason for R1 to make a current that is "acceptable" for the SCR to trigger without overloading it? And when the circuit is turned on, would any current flow through the Zener diode pathway, or is this pathway "turned off" by the SCR? \$\endgroup\$ – alpha26 Dec 4 '18 at 23:39
  • \$\begingroup\$ Your understanding is pretty good. R1 pulls down the gate to prevent false triggering. When it turns on the voltage from the capacitor will fall a little due to its (small) internal resistance so the Zener current will reduce to zero very quickly. \$\endgroup\$ – Transistor Dec 5 '18 at 7:11
  • \$\begingroup\$ I built this circuit and did some testing. The capacitor doesn't begin to discharge until it gets above approx. 4.7V or so, but then once it begins discharging and the voltage decreases again, it seems to stop discharging when it gets to the same ~4.7V. Could this be because of the Zener diode? Is there a way to make it "bypass" the diode when it's in the discharge phase? \$\endgroup\$ – alpha26 Dec 8 '18 at 3:16
0
\$\begingroup\$

The logic design in simple specs might be ;

  • If Vc > 5V turn on switch
  • If Vc < ?V turn off switch or if load current drops below ? mA , same turn off switch
  • Switch resistance < __ milliohms
    • or define load = __ ohms + __ uF and max voltage drop = __ Volts on switch at __ Amps max

One simple solution is a comparator with hysteresis that has an RdsOn and hysteresis that satisfies all the above specs.

If that is acceptable, then fill in the blank specs, then search ... how to design two thresholds in a comparator with hysteresis. a.k.a. Schmitt trigger and how to choose an NFET or PFET with RdsOn with "logic level drive"

Another might be an SCR with Vc scaled down to the trigger voltage if the holding current threshold voltage is acceptable but these have a bigger voltage drop than a HEXFET

\$\endgroup\$
  • \$\begingroup\$ I've got a lot of googling to do on that one! My knowledge of electronics can best be described as slightly better than n00b but not by much. \$\endgroup\$ – alpha26 Dec 3 '18 at 23:17
  • \$\begingroup\$ Hi Tony, so I learned more about MOSFETs and it seems that this may work. I tried the SCR circuit above but the voltage drop was more than I was comfortable with. If I use a NFET, would I have to keep the gate voltage higher than the threshold? Or could I make it so that the MOSFET stays on once the gate voltage goes past the threshold? \$\endgroup\$ – alpha26 Dec 20 '18 at 1:43
  • \$\begingroup\$ No, You need a boost voltage for high side NFET like those used with Half Bridges with low side PWM to drive a diode to the boost cap. So just use a PFET. Voltage references are low current. ...The MAX6035 typically draws only 73µA of supply current and can source 10mA or sink 2mA of load current \$\endgroup\$ – Sunnyskyguy EE75 Dec 20 '18 at 1:51
  • \$\begingroup\$ Fill in the blanks above and add how your charger connects and disconnects, CC or CV etc until then good luck \$\endgroup\$ – Sunnyskyguy EE75 Dec 20 '18 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.