1.2 V Power Supply Voltage drop

[vinit2100]

I designed a power supply (Circuit 1) to provide 1.2V (VOUT) to another circuit . that is VOUT is connected to CIRCUIT2 through a switch S1.. But when I switch on S1 . the power supply drop from 1.2V to 500mV range I measured VOUT with a multimeter.. ? But when the switch is OFF , power is 1.2V What could be the reason?

 

[mursal]

The output from the power supply chip, cannot supply sufficient current to drive the circuit. Consider putting an emitter follow as an output stage on the regulator. On a different power supply, measure the current flow into the circuit from the supply to ensure you haven't a short circuit

 

[Miguel Lopez]

The LM317 will supply 1,25V with (I guess) enough current capability to drive the circuit.

 

[vinit2100]

the circuit 2 requires around 20mA.. the operational amplifier provides 4omA source current !!!

 

[(*steve*)]

the circuit 2 requires around 20mA.. the operational amplifier provides 4omA source current !!!

Not sure if that is a typo (4o = 40 or 4???), but at 5V it can source 7 to 10 mA, over the full range it typically guarantees 4mA

See here.

 

[OLIVE2222]

you can use a TS922 if you want to stay with opamps for the 1V2 power supply
http://www.st.com/web/en/resource/technical/document/datasheet/CD00001188.pdf
Concerning the circuit 2, depending on both step-up operating and start-up voltages it may happen that you only have the V_I/O&V_core or only VCC_OTP, this can be not allowed by the "target" circuit . This guessing that the circuit will be 1V2 battery supplied

Olivier

 

[vinit2100]

Sir, the circuit 2 consumes around 10-11mA maximum.. what if more current passes to circuit 2?

 

[vinit2100]

the circuit 2 requires around 20mA.. the operational amplifier provides 40mA source current !!!

 

[(*steve*)]

the operational amplifier provides 40mA source current !!!

No it doesn't. At the supply rail you have chosen it can supply 4mA.

 

[vinit2100]

Sorry Sir, I changed the op amp its not LM324, its TLV2464. op amp at 3.3V, it can supply 40mA. Sorry for my mistake. When i connect circuit 2 to circuit 1, the voltage (1.2V) drops to 0.5V..? the circuit 2 consumes 10mA maximum..

 

[Harald Kapp]

The TLV2464 is not suitable for driving large capacitive loads (datasheet page 25ff.). Do you have an oscilloscope to watch the output signal of the OpAmp? Chances are you will observe ringing.

Apart from that: While the mean current may well be ~10mA, the switch mode regulators may draw current spikes with alot higher amplitude while switching. use a small series resistor between OpAmp and load circuit and monitor the currrent flow.

 

[BobK]

Just add a pass transistor and get it over with.

Bob

 

[vinit2100]

What to do if i want to drive large capacitive load using that opamp?

 

[vinit2100]

This the waveform i get when i connect circuit 1 to circuit2 and the switch is ON.. 1.2V drops to 500mV. and its not a perfect wave form.. So in which side is the actual problem. in Circuit1 or in Circuit 2?

 

[gorgon]

What types of ICs are the U4 and U1 in circuit 2?

I can't really see why you feed circuit 2 with 1.2V.

Any specification of the requirements of this circuit?

 

[vinit2100]

U4 is TPS61010DGS
U1 is TPS61220
Circuit 2 is actually a part of Hearing Aid that operates in 1.2V supply..

 

[gorgon]

Ok, then you should go through your selection of components around U4 and U1.
If you look into the datasheet for the tps61010, you'll see the R15/16 on your design is way too low in value, the recommended value of the R15 is 500k and then the R16 should be 1M for a 1.5V output.

The same goes for R19/20 where R19 should be 500k and R20 calculated from the threshold low voltage.

If U1 is tps61220 you need a resistor network to set the output voltage. The schematic show the use of tps61221, a fixed 3.3V output device.

If your circuit1 'power supply' intends to simulate a 1 cell battery, you need to make it with a very low output impedance, to simulate the low internal resistance of a battery.

 

[vinit2100]

You are right ,Circuit 1 actually is replaced by Battery.. what additional change should i make in circuit 1. so that it can function same like a battery.