Help needed with TPS61040 boost circuit

[daisizhou]

I designed a boost circuit for the Fluke 187, effectively abandoning the use of the alkaline battery compartment.

I chose a 4.2V lithium battery and used the TPS61040 to build a boost circuit, stepping up the voltage from 4.2V to 5.4V to power a Model 187 digital multimeter.

I wanted the boost circuit to stop operating when the 187 digital multimeter is switched to the "OFF" position, so I designed a sleep circuit—specifically, the section composed of the SS8050 transistor and the 100kΩ and 200kΩ resistors; the inductor and capacitor following this stage serve a filtering function.

Actual testing revealed that the circuit consistently fails to enter sleep mode—specifically, the EN pin of the TPS61040 never activates—and I do not know why.
I would be very grateful if someone could help me.


Circuit simulation:
https://www.falstad.com/circuit/circuitjs.html?ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEECsqYIiSRuALAOwBMtAbLQJy1JIAcTS2rqEACNERbKgAOIhGNQA3CKNQBbTKICmAWk4IAfACgoUYNAAeo6tSj0iTKEUtQAzEy6p4CcVEXJCBBAD0BkYmUOYyji5c9o7UXF4eXj70fuJBhsZoYYi02Fa22PbM9kwJOJIAdojaqELqFP5QygD2iAAm6jAAhgCuADZo8nXVeERIrBNERFxO2JxOrIxD5Aiao9T0TvRcSNTYXJbjgcHGAO7ZEVZxhQ75lu7l6SF9F7n5pVDM0QUPnqh9OHwfhQUDI5VBgL8UyQ9HocSItFoO2w9HYxwyoFerhKhS+1j2vy8KzwqRBYDkkLmqAwPi8cjaohoCJY3CIrEovAO6JC53Ct3xhWoTCscLKfyeZwu-JsdmlBNgjxOwAA5hc9oVRc5saLCdzJXzHPyjcxdRLgLyLPlisyikxTUqLTJrYbitRWHaFeKHVLItihVYovaMY7-QKoOqBUGQhI1cKtdFQzrPU4VPU-lAMMTUBA03xUmklQAlKWu91QFj0cPu35ID2nRIqLqmCkIO1mx14gpQA7fUpR4xmRA9sPDn6epINPx680XYeaxbRJMN9uzuI4qALnH94Ax8KbrubpeIFNNNNeTPgnOT1LTx1y3Ha+XL73hPGaudPxX6YABcAQAxAA

[tszaboo]

It doesn't work, because your ground is on top.
And therefore you incurred the wrath of the god of electronics, who cursed your circuit.

[MariuszD]

It is possible that it is drawing some current in the OFF state. Connect a resistor in parallel to B-E.

[daisizhou]

It is possible that it is drawing some current in the OFF state. Connect a resistor in parallel to B-E.

Yes, the circuit cannot be completely turned off.

For the Fluke 187, I estimate the maximum operating current to be 200 mA.

I would like the output to shut off when there is no current flowing.

How should I adjust this circuit?

Thanks

[MariuszD]

Through the resistors, 20uA flows, so to turn it on, a base current of about 0.1uA can be sufficient.

I don't know how much this meter consumes at a minimum, but if you bypass the B-E junction with a 470Ω resistor, it won't turn on below 1mA.

[daisizhou]

Through the resistors, 20uA flows, so to turn it on, a base current of about 0.1uA can be sufficient.

I don't know how much this meter consumes at a minimum, but if you bypass the B-E junction with a 470Ω resistor, it won't turn on below 1mA.

The operating current of the Fluke 187 digital multimeter is 20–80 mA(At a voltage of 5.4 V).

Do you mean adding a 470-ohm resistor to the circuit, as shown in the figure below?

Instead of adding a 470-ohm resistor, would changing the 100k resistor in the circuit to 200k or 300k result in an improvement?

[MariuszD]

Like this:

[daisizhou]

Hi MariuszD

As shown in the image below, this is a step-up circuit board built using the MC34063; it takes a 4.2V lithium battery input and provides a 9.0V output.
In the output section, an SS8550 transistor is also used for shutdown control; why, then, does this circuit not require a 470-ohm resistor? The difference lies in the output voltage—9.0V versus 5.4V—even though the SS8550 transistor is used for the same shutdown control function.

To be honest, the output shutdown circuit for my TPS61040 design—using the SS8550—was based precisely on this circuit.
You can see that in the MC34063 circuit, the emitter, base, and collector of the SS8550 are not connected to any resistors; however, in my TPS61040 circuit, I specifically added 100kΩ and 200kΩ resistors.


Thanks

[langwadt]

It doesn't work, because your ground is on top.
And therefore you incurred the wrath of the god of electronics, who cursed your circuit.

maybe it is an Australian design

[MariuszD]

...even though the SS8550 transistor is used for the same shutdown control function.

That's not the same, it's a completely different circuit, there through the transistor, a different current can flow. I don't have the schematic and I don't know how it's made. I don't want to waste time on reverse engineering the board with MC34063. If you do it and show me the schematic, I'll explain how it works and calculate the parameters for you.

[daisizhou]

...even though the SS8550 transistor is used for the same shutdown control function.

That's not the same, it's a completely different circuit, there through the transistor, a different current can flow. I don't have the schematic and I don't know how it's made. I don't want to waste time on reverse engineering the board with MC34063. If you do it and show me the schematic, I'll explain how it works and calculate the parameters for you.

I have attempted to draw a circuit schematic based on the PCB; please see the attachment.

I have personally tested this board; it is highly energy-efficient and does not require a dedicated power switch. Typically, simply turning the digital multimeter's rotary dial to the "OFF" position is enough to stop the MC34063 from operating.

The only drawback is that the MC34063 is a rather old chip, and its power conversion efficiency is not high.

[MariuszD]

The schematic makes sense, it just lacks a capacitor connected to the cathode of D1. From the schematic, it follows that Q1 powers the MC34063, meaning the collector current will be around 4mA, and for the circuit to turn off, the base current must drop below 20uA.

[ftg]

Interesting circuit and idea.
If I can get my Agilent U1253A fixed, I'll build something like this for it as well.
Even the 34063A might be acceptable, because it has one or two of those already on board. SO one more in the case would not be the end of the world.
Following this thread with interest.

[daisizhou]

Interesting circuit and idea.
If I can get my Agilent U1253A fixed, I'll build something like this for it as well.
Even the 34063A might be acceptable, because it has one or two of those already on board. SO one more in the case would not be the end of the world.
Following this thread with interest.

Yes, what sets this circuit apart is that it omits the power switch; while previous designs required a switch connected in series with the battery, this circuit eliminates the need for one—and that is precisely what makes it unique.
The current consumption is merely in the microampere range, which is virtually negligible for a lithium battery.

[daisizhou]

Like this:

Hi MariuszD

I connected a 510 ohm resistor, but even after disconnecting the load, the output still has a voltage of 3.74V, which means the boost circuit is still working

Is this normal?

[MariuszD]

I connected a 510 ohm resistor, but even after disconnecting the load, the output still has a voltage of 3.74V, which means the boost circuit is still working
Is this normal?

Through the disabled boost converter, the current flows from the input to the output through the diode.
The converter is not working, it is wasting only a few dozen microamperes through the resistors.

If there were no voltage on the output, re-enabling through load on the output would be impossible.