Dear Pengyu,

The solution is simple! You need to insert a diode at the output of the low voltage source to prevent the back flow of the current. This best type of the diode is the low voltage drop Schottky diodes. Please select the appropriate diode rating from the semiconductor data sheets. If you want that the source conducts a current when exceeds certain voltage then you add a zener diode or a series connection of forward biased diodes whose cut in voltage is equal to the on voltage.

If you give a circuit drawing one can give an elaborated solution.

Best wishes

Dear Pengyu,

The solution is simple! You need to insert a diode at the output of the low voltage source to prevent the back flow of the current. This best type of the diode is the low voltage drop Schottky diodes. Please select the appropriate diode rating from the semiconductor data sheets. If you want that the source conducts a current when exceeds certain voltage then you add a zener diode or a series connection of forward biased diodes whose cut in voltage is equal to the on voltage.

If you give a circuit drawing one can give an elaborated solution.

Best wishes

Thanks for your kindly reply.

Here is my dc-dc circuit. C9 is the storage capacitor . The left side are the rectify circuit consists of diode and capacitor. I am not major in EE so that it will be appreciated if you could give a more detailed description.

I can understand the way using zener diode but why do I need to prevent the backflow of the current?

Many thanks

I had found a paper in which they also use tps61200. However, with my limit knowledge, I can't understand the sub-circuit at the input of the converter. Could you please explain it?

"The R1-R2-VD3-R3-VT1-R4-VT2 circuit turns the control circuit

into an enable input of the DA1. Voltage divider R1–R2 sets

the upper threshold of the control circuit; the positive feedback

loop of the control circuit is formed by circuit R3-VD3. Diode VD3

blocks voltage from the output of divider R1–R2 to the input EN

and resistor R3 along with R1 and R2 sets the lower threshold of

control circuit operation. Due to the use of field effect transistor

with induced channel and also large value resistors the current is

about 1.5 A at an upper threshold voltage of 1.2 V."[1]

[1] Baranov A M, Akbari S, Spirjakin D, et al. Feasibility of RF energy harvesting for wireless gas sensor nodes[J]. Sensors and Actuators A: Physical, 2018, 275: 37-43.

After thinking, I doubt if the zener suits my application? For example, using a 3v zener, when the voltage of the capacitor reaches 3V, it will breakdown and allow the current pass through. But as the current flows out, the voltage of the capacitor will soon drop below 3V, what will happen next? Will the zener recovery as soon as the voltage drop below 3V? Or the zener will only recovery after the voltage of the capacitor drop to 0V?

I expect the capacitor can release enough energy to the DC-DC. For instance, the voltage drop from 3V to 1.8V can provide enough energy so that the zener should keep open with 3V-1.8V after the breakdown.

How about using a voltage monitor? For example, when Vin>2V, the monitor enable the EN pin of the dc-dc.

If I got it right (between the lines), your issue is that at very low input voltages, the circuit "pumps":

1. when the voltage is >250 mV, the DC/DC converter starts.

2. resulting from the converter starting, the voltage drops and the converter stops.

3. back to 1.

One solution would be to either use a zener diode between the voltage input and the EN input (could as well be 1-3 normal diodes in series), or a resistive divider (mid-point connected to EN).

Both schemes are expected to work as desired - delaying the converter's start and (to some extend) preventing the pumping.

U. Dreher Abdelhalim Zekry I had tested with the zener, although it can block the voltage but the capcitor can not fully discharged. For example, using 2.8V zener, it will stop discharge as soon as the voltage low down to 2V. Is that because the zener recovery much more faster than the discharge speed?

The zener is quite fast. Be aware the zener voltage somewhat depends on the current passing.

Do I get it right? you want the DC/DC converter deplete input capacitor down to the 0.25 V (or so) - just starting operation at a higher voltage ?

If so, you might be able to short the zener once the DC/DC converter has started. This way, the delay would be single-sided (delayed turn-on, turn-off at min input voltage).

But please don't as me where to derive the necessary control signal from: a monoflop might work - provided you can get some monoflop operating at such low voltages.

Dear Pengyu Li ,

what is the (critical) type[1] of the C9 storage[2] capacitor ?

1. What is the impedance vs frequency characteristics (Z-spectrum[2]) of the C9 storage capacitor ?

2. Who charges faster: Supercapacitor or Capacitor ? https://www.researchgate.net/post/Who_charges_faster_Supercapacitor_or_Capacitor

Dear Pengyu,

The colleague Dreher gave a possible solution. May be a practical solution is to use a DIAC with a turn over voltage equals the voltage at the required start of conduction. Then when it is on it will be be back to low voltage state and thereby you can discharge more charge from the capacitor. If you do not have a proper diac with the required turn over voltage you may use a unijunction transistor. Its turn over voltage can be programmed.

Best wishes

Abdelhalim Zekry So, if using DIAC whos Vth is 2V, once the Vcap reach 2V, it will discharge to certai value(1V fore example) where the DIAC will shutdow? I had find some information which shows that it might not practical to fully discharge.