# Help me to understand this circuit

Discussion in 'General Electronics Discussion' started by alexrp, Oct 17, 2016.

1. ### alexrp

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Oct 17, 2016
Hi! I was wondering if anyone knows how this part of the circuit works... I was looking on it but I don't understand the zener and mosfet part...
Thank you!!

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2. ### duke37

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Jan 9, 2011
Do not see a zener diode, surely D1 is a Schotky diode.
Is this a buck convertor?

3. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Everything to the right of the inductor, the MOSFETs, the Schottky diode, etc. are part of a monitoring circuit used to identify the presence of the input voltage. Since the output of this buck converter circuit is regulated by feedback voltage-divider resistors R3 and R4, introducing a diode to isolate Q7, Q8, D2 and D3 will result in a forward-biased voltage drop across the diode and have an effect on the 5V voltage-regulated output. By using a Schottky SS34 diode with its much lower forward-biased voltage drop for D1, the diode forward voltage drop is minimized while still allowing the diode to perform its isolation function.

4. ### dorke

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Jun 20, 2015
The MP2307 is a Buck step down converter.
Like @hevans1944 (Hop) said above:
The DC voltage regulated output is set by R3 and R4 to be about 5.22V.
The diode D1 is a Schottky diode(although the symbol in the schematic is that of a Zener) .
Depending on current and temperature, the max voltage drop on it will be 0.5V.
So although the "output" is marked " +5V " it will probably be around 4.8V .

Although there are lines marked as " IN_DET ",
they actually detect the presence of the output voltage of the MP2307 converter.
(obviously there is an input voltage as well in that condition).

Do you have the part numbers of Q7 and Q8?

Last edited: Oct 18, 2016
5. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Thank you for the extended analysis, @dorke. I don't know what V_808 is supposed to do. And I failed to notice that Q8 is N-channel while Q7 is P-channel MOSFET. So, if power is present, Q8 turns on and grounds the gate of Q7, presumably turning Q7 on and allowing voltage to appear at the anodes of D2 and D3 with significant current capability. Don't ya just love schematics with "mystery" components?

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6. ### alexrp

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Oct 17, 2016
yes, thank you! but i still don't know the function of the 2 mosfet (Q8 & Q7), when there is an input, the voltage in the Q8 gate is ~ 5.3V in the case of Q7 is ~ -5.3. If this is correct, Q8 is saturated and its D is the same point of G(Q7); Therefore, at that moment the gate of Q7 is in GND or -5.3 (V-INT)??
The Q7 as you said is a P-channel so it is activated when the voltage in the gate is <0 isn't it?
I don't know the Q values.. sorry
Thank you!!

7. ### dorke

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Jun 20, 2015
I have removed part of my original post ,here it is again:

Here is my view:
The 2 Mosfets are shown to be enhancement mode :

Q8- -N channel ,Q7- P channel.
Q7 is a "high side switch"

The voltage marked as " V_808 " will be present only after the voltage of the MP2307 output reached above Vth(thershold) of Q8.
In that case Q8 will be on.
In that time the Vgs voltage of Q7 will go lower than it's Vth and Q7 will be on as well.

The " V_808 " output will be a "step like" one, after power up/enable of the converter MP2307 .

That is different from the gradually rising(ramping) output of the converter.

The 2 diodes (D2,D3) in parallel suggest the output " V_808 " is relatively "high current",
so Q7 is probably a power Mosfet.

Last edited: Oct 20, 2016
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8. ### alexrp

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Oct 17, 2016
So as I can guess, in the next image which is the continuation of the circuit, the first one shows the continuation of the power supply circuit, so u can select the way you introduce power to the circuit, by V_808 or Vbat.
As dorke says above, the 2 diodes D2 &D3 it could be possible that they are there to reduce the voltage to 3.6 v (0.7 + 0.7 in direct mode)

As the Q4 is a p-ch which is activated when there is no MP2307 output. the first question is, why there is a jumper which connects v808 and vbat? in the case im right, the v808 is 0 when Q4 is active.. so there is no sense to put there a connection (there are two different ways to obtain the 5v. The first one introducing Vin (first image) and the second one by a battery (second image) so the second question is why the v808 needs to go through vbat & vccl to power the pt1301 to get 5v if there is already 5V in the circuit (got it in the first part)

thanks

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9. ### dorke

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Jun 20, 2015
Well,
The diodes D2 and D3 are in parallel,so they would drop the voltage only by a single Vf .
That Vf would depend on the specific diode and the current through it.

You raise some good questions.
Sometimes it is hard to get into the mind of a "designer" without knowing is intentions/worries/ limitations etc.
In order to get the "full picture" ,can you please let us now more about the "product".