High frequency signal conditioning

Do anyone have any idea about this signal conditioning circuit?



Signal source: KA3525 PWM controller IC.
Frequency: 39KHz

Signal is driving a N-Channel MOSFET directly.

I've no idea how this circuit is working or conditioning the signal but without this P_N_P transistor MOSFET draws more current and get heated. With transistor Current is low as well as MOSFETs are cool.

What is the explanation of this signal conditioning circuit?

The circuit is sketched incorrectly. Flip BC557 emitter and collector.

10 k seems too high to drive MOSFETs, even with switch-off speeding up PNP transistor. What's the connected MOSFET type?

No. Circuit is drawn correctly as per the running system. This is the main issue of this post. I know its flipped but its working as it is. And I explained it before.

And sorry, for 10K, its 10Ohms.

Show me Vmin and Vmax for pulse input to R1.
Is Vmin negative ?

I know its flipped but its working as it is. And I explained it before.

Click to expand...

Where did you explain the inverted PNP?

10 ohm makes more sense. A (correctly connected) pnp transistor will increase the gate current in switch-off, a solution that can be found quite often in MOSFET gate drive.

An inverted transistor hasn't much higher current gain than unity, the base-emitter will also show breakdown above 7 - 9V. If it actually shows depends on 3525 VC level.

Didn't hear yet about the connected MOSFET.

Mithun_K_Das said:

I've no idea how this circuit is working or conditioning the signal

Click to expand...

Neither do we.
The circuit is an incorrectly designed Mosfet turnoff assist.

I mean incorrectly designed because:
-You don't show a complete circuit.
-The gate resistor is wrong by 3 orders of magnitude, although you have explained that already.
-The transistor's C-E connections are flipped. They might appear to work, but all you are doing is adding a poor anti parallel diode with D1. Because the C-B junction is only working as that, a diode.

This is the complete circuit. This configuration is used in many small inverter circuits. Here a PNP transistor is connected inverted but it has significant effect in the circuit.

Test results: With same load it takes 3.15A at 11.8V supply to run a 24Watt lamp without that PNP transistor. MOSFET get heated quickly and current increases and increases. Finally MOSFET are burnt.


But with the same load connected, if the transistor is used it takes only 1.58A at 11.8V supply. Also MOSFET is totally cold, no increment of load current. Although the lamp is not fully glowed. But there is no change in lamp voltage and intensity in these two different circuits.


It seems like a wrong circuit diagram as I thought before but it has a significant effect in circuit. You can also check and test what happens.

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Lamp used: CFL 24W / 220V / 50Hz

This inverter is used mostly in solar inverter, portable inverters for mobile chargers, portable inverter for CLF lamp. Output is around 170V (after making it DC)

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Lamp is also connected after the bridge rectifier and capacitor filter. I forgot to add that in the circuit diagram.

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schmitt trigger said:

Neither do we.
The circuit is an incorrectly designed Mosfet turnoff assist.

I mean incorrectly designed because:
-You don't show a complete circuit.
-The gate resistor is wrong by 3 orders of magnitude, although you have explained that already.
-The transistor's C-E connections are flipped. They might appear to work, but all you are doing is adding a poor anti parallel diode with D1. Because the C-B junction is only working as that, a diode.

Click to expand...

Yes, I thought like you before. But when I tested the circuit myself, I'm really really confused that how it is working?

I posted the test result. If you want to see, I can post the test circuit and result.

This configuration is used in many small inverter circuits.

Click to expand...

Can you name a specific product? Or show a photo that validates the inverted transistor connection? Does it actually use BC557?

After the 10k issue has been clarified, there are other open points. You are showing 1N4739, a 9.1V Z-diode. Seriously? Or did you mean a schottky rectifier? Whats's 3525 collector supply (VC) voltage, is it 12V?

Did you check if the circuit also works with non-inversed PNP?

A SPICE simulation claims inverse current gain of 2 to 3 in the relevant current range. In so far, we can expect a certain gate current enhancement. But not obvious at first sight why the circuit doesn't use non-inverted configuration. Problem of possible base-emitter junction breakdown is still pending.

It works as follows:
With a high output drive, the mosfet's gate (which behaves as a capacitor) is charged thru R1,D2.
With a low output drive, D2 will be reversed biased and only Q1's base is driven. By transistor gain action, the collector current will now discharge the Mosfet's gate.

The discharge current will be multiplied by the transistor gain, as compared to the charging current....in other words it allows a much faster turnoff, which in some instances may be required.

What I think is happening is that such a fast turnoff is creating a very large voltage spike due to the leakage inductance. Your circuit does not show any type of snubber, You must include one.

Lastly....you are driving the transistor with no means of core reset. Unless it is a properly designed transformer with an air gap, the core will eventually saturate.

Here is some images and I tested on my working table. Please take a look.

























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FvM said:

Can you name a specific product? Or show a photo that validates the inverted transistor connection? Does it actually use BC557?

After the 10k issue has been clarified, there are other open points. You are showing 1N4739, a 9.1V Z-diode. Seriously? Or did you mean a schottky rectifier? Whats's 3525 collector supply (VC) voltage, is it 12V?

Did you check if the circuit also works with non-inversed PNP?

A SPICE simulation claims inverse current gain of 2 to 3 in the relevant current range. In so far, we can expect a certain gate current enhancement. But not obvious at first sight why the circuit doesn't use non-inverted configuration. Problem of possible base-emitter junction breakdown is still pending.

Click to expand...

If you check recent China portable inverters, mobile chargers or Indian small inverter circuits you can find it there.

Yes, 9.1V zener is used, I tested with 6.8V too. And it works too. Nop, No schottky rectifier just a zener diode. Yes, its 12V input to VC of KA3525 IC.

Yes circuit works without Non-inversed PNP but takes extra current and MOSFETs get hot and burnt.

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Here is a video presentation of that circuit.

Yes circuit works without Non-inversed PNP

Click to expand...

Question was, does it work with non-inversed PNP?

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I see that the transistor is BC557 and polarity as stated, thanks.

"Yes, 9.1V zener is used, I tested with 6.8V too. And it works too. Nop, No schottky rectifier just a zener diode."

There is a satirical fable by renowned 18th-century Spanish poet Tomas de Iriarte, called "The donkey and the flute".
Without going into much detail about the fable, a donkey one morning was eating from a bale of hay, where a flute had been accidentally dropped.
The donkey snorted and the flute emitted a musical note.
The donkey went on telling everyone that he was already a musician!

The point I'm trying to drive across is that some designs may appear to work, and "engineers" copy it indiscriminately without understanding exactly its operation.

In this case the zener is operating like a simple switching diode, and a very poor one. Zeners have very wide junctions and therefore are slow switchers. You would be far better with a real switching diode like the 1N4148, or even better a real schotky.
I believe the original "designer" someday saw a zener diode across a B-E junction, and decided to copy it without understanding its purpose. B-E junctions break down at low voltages, usually at about 6 to 7 volts. But in this case it is across a C-E junction, which is extremely robust and withstands the transistor's full rated voltage. A zener is not needed here. You require a real switching diode.

Now that I've educated you, please give me a thumbs up, OK?

It's true that the zener breakdown voltage will be never reached, because a transistor junction is connected anti-parallel.

But as you already mentioned, the switching behavior is different. The diode capacitance is probably higher by a factor of 50. Thus I don't want to guess about the exact effect. Similar problem is in estimating the effect of BC557 reversal which has been verified at least for the circuit shown in post #10. The transistor still increases the available gate current, but current gain is low, as explained. I'm not presently sure about the dynamic effect. For a substantiated answer, I would need to make real measurements.

Please check the video and if you have any explanation please post.

The explanation has been already given. The transistor, though having low current gain in inversed configuration, speeds up MOSFET switch-off.

It would be more informative to see the bare oscilloscope screenshots where one could derive quantitative information like switching time. I guess the oscilloscope can export it. Instead of almost identical gate and base waveforms, I would prefer gate and drain voltage.

Besides the clear information that the circuit is effective in speeding-up MOSFET switching, the experiment can't clarify why the circuit is specifically designed so.

hmm, the circuit have no clear explanation still now. I want to test some more. Today I'm making another oscillation circuit like such configuration and will test the result.

Using transistor(s) in common collector circuit as gate driver is just standard. In this case you have asymmetric diode-transistor circuit (only switch-off current amplified) and yet unclear effect of inversed transistor configuration.

It's important to look at the whole switching circuit. The expectable large leakage inductance caused voltage peak leakage during switch-off multiplies losses.