How this circuit works?

The function of this circuit is to generate pulses for driving Solenoid.

What the function of those two diodes?

Thanks.

D2 is to stop the solenoid supply going back into the supply line.

D3 is to clamp the reverse voltage spike that will be produced when the solenoid is turned off. The magnetic energy stored in the solenoid coil will be released when it turns off and will produce a voltage across the coil terminals in the reversed polarity to when a voltage was supplied to it. The diode D3 normally doesn't conduct but when the reverse voltage spike meets it, it will conduct and protect the FET Q1 from damage.

Brian.

Just to add a little extra information, Diode D3 is usually called a flyback diode or freewheeling diode. The back emf from the solenoid can fry the FET Q1 if the diode is not present.

D2 is to stop the solenoid supply going back into the supply line.

Click to expand...

I would rather call it a start-up circuit. To engage the solenoid, a higher current respectively voltage is needed than to hold it's state.
The raised voltage is supplied from the capacitor, that's charged by the DC/DC converter. In operation, the capacitor voltage drops
until the coil current is supplied by D2. It's basically a method to save energy.

It's basically a method to save energy.

Click to expand...

It can be also means to make solenoid engage faster. If DC/DC converter supplies higher voltage than nominal coil voltage, current needed to operate coil will be reached in shorter time. After tank capacitor is discharged, D2 supplies current to solenoid to maintain solenoid engaged.

Hi,
MOSFET Q1 will not turn ON any Time, Opto coupler collecter is grounded. even opto is ON it will not be able to give Gate Charge

MOSFET Q1 will not turn ON any Time

Click to expand...

Right. The question was focussing on the right circuit part, I think. Actually, there isn't any purpose of an optocoupler in the circuit,
because left and right side have a common ground. So I simply ignored it, supposing a drawing error.

It can be also means to make solenoid engage faster.

Click to expand...

Yes, another important application of the start-up circuit.

Thank you all for suggestions.

U1 circuit is a boost DC/DC ,that is output larger than Vin.

i have data for step down /up converter .

Huzuhong525, I am interested in the data on the step down/up converter. Please go on

send a email to huzuhong525@yahoo.com.cn,and i will relpy with data .

Hi,
Why is it pls inpossible to see for us all!?
The question & your 1. answer are full public, but not the data? :-(
K.

TPS61041 is a switch mode LED driver/controller. The datasheet is available from ti.com: https://www.ti.com/lit/gpn/tps61041

It is a pretty standard design for a SMPS, looks like boost to me; matches with the controller which is a boost controller.

The output voltage of the supply is 1.233V * (1 + (R1 / R2)), which is 1.233×(1+(618÷160)), or 5.995V, close enough to call it 6 volts especially since the TPS61041 probably only has a limited accuracy, a figure which is oddly missing from the datasheet. Though that's all we can really see the rest is cut off.

Not sure what a Smith trigger is... perhaps the designer means 'Schmitt trigger'? The IC, U3, appears to be some kind of hex Schmitt inverter, though there's no part code so this is only an educated guess based on the pin out and the label. Possible candidates include the 7414 or the 40106 IC.

I'm not sure what capacitor C8 is doing. It appears to have a value of 0.1nF, or 100pF, placed close to the unknown IC U3. If it is for smoothing, it should be a 10nF to 100nF capacitor, preferably a low-ESR, non-polar (e.g. ceramic.)

Resistor R3, value 30 ohms, power rating 1 watt, is probably a current sense or overload protection device (to prevent a shorted coil from frying the FET, perhaps?), though I'm not sure, because IC U2 is also unknown.

Hi Tom,
R3 is not 30, bur430 Ohm & is in my opinion with U2`s function bad to understand...
Seems to be similar as the not needed optocoppler too...
I think, these is a montage of some circuits & as less logically designed some thing :-(.
K.

Resistor R3, value 30 ohms, power rating 1 watt, is probably a current sense or overload protection device (to prevent a shorted coil from frying the FET, perhaps?), though I'm not sure, because IC U2 is also unknown.

Click to expand...

You should review the above posts (from sinisa and me), that explained the operation of the solenoid start-up circuit...

The initial voltage can't be read from the circuit, however. U2 can be imagined as just another boost converter, supplying - may be - 12 V or a bit more. But I don't think, it's too important to know it exactly, neither some other arbitrary circuit details and artefacts. The question was about the circuit's principle operation, no one wants to reproduce it as is, I think.

how calculate sr of op amp

Hi,
SR is on the datasheet...
Others you have to take your oscilloscope pls, if you have a pulse on that to see (an edge) you will check that it take an amplitude i.e. 0.5V in 2 microsec(full, form bottom to top)_your circuit has an SR=0.25V/us.
Its all.
K.

Here is what I saw before I read the thread:
A time-varying signal is presented to the input of a Schmidt Trigger, which "cleans' the waveform by removing many potential glitch hazards.

In the "0" state of 1Y a current is drawn through the actuating l.e.d. of an isolated, opto-coupled transistor, ultimately switching the power transistor Q1. (An error in the drawing places a ground where +V should be, behind the 10K resistor.)

When 1Y is at "low", Q1 is "off". When 1Y is "high" Q1 conducts, and a current is drawn through Header 3, determined by a load not shown on the diagram. The presence of the diodes D2 and D3 lead me to suspect an inductive load, but D2 is a puzzle since it really has nowhere to go.
Power for the unknown load is supplied by a converter through a solid current limiting resistor.

I don't see anything I feel I understand in the top-left corner of the drawing.

the Rated Solenoid Voltage (Vdc) is 6V.

If this mean the input voltage to Solenoid can't be greater than 6V?

when Solenoid is in hold state, what is the current consumption?

Thank you all again.

Listen there's still a bit more to know at our end please! We're not magicians!
I hope you'll let me, a junior member, have another go now I've read the thread.
The FET's job is indeed to initiate a magnetic "grab".
The hold current of a real-world industrial solenoid is much, much less than the operate current. This is because the air-gap is (presumably) closed by the magnet by the time the initiating pulse has been pulled through by the FET, and the magnetic resistance has dropped by several orders of magnitude.
What more can be said while you keep mum about the resistances and voltages involved in the circuit?
At least tell us what the circuit is part of.