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What is the purpose of this resistor network? Why not just use one resistor? or connect the cathode of RA93 diode directly to the mosfet drain (this mosfet is high side of the h-bridge).

It seems to me that pin 30 of the relay is connected to pin 87 through diode and the resistor network. I just am not able to figure out why? The diode is probably to preventing the circuit in case someone plugs in the battery terminals backwards, is this right?

Thanks.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

schematic

simulate this circuit

Alternate schematic (by @Transistor) emphasising bulk capacitor and relay bypass of soft-start. Purpose of D3 is unclear. Operator to review.

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    \$\begingroup\$ You forgot to explain the function of the circuit board. What do you think that part of the circuit is for? There's a schematic button on the editor toolbar. Hit the edit link below the question ... \$\endgroup\$ – Transistor Mar 23 at 18:27
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    \$\begingroup\$ @Transistor, thanks for the tip. I have added the diagram. BTW it's a really nice feature. \$\endgroup\$ – Grepsoft Mar 23 at 18:46
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    \$\begingroup\$ Is it perhaps a soft start that is bypassed by the relay later on? \$\endgroup\$ – Wesley Lee Mar 23 at 19:06
  • \$\begingroup\$ @WesleyLee, Your answer makes more sense actually. I just read an article that talks about how soft start is used to stabilize the circuit and then is bypassed using a switch like relay or transistor. Here is the link: sunpower-uk.com/glossary/what-does-soft-start-mean \$\endgroup\$ – Grepsoft Mar 24 at 14:42
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Your network seems to consist of a series / parallel combination of eight 27 Ω resistors that is equivalent to a single 54 Ω resistor. The significant thing is that these are surface mount and will be placed by machine.

Presumably eight are required for power handling. The circuit could have used one 54 Ω through-hole 1 W resistor but this would require an extra component and, probably, hand installation and soldering. With pick and place equipment this could be avoided for this component (although not for the caps, relay and MOSFETs) and possibly there were some savings by using resistors which had been bought in bulk for elsewhere in the circuit.

Another reason one might see series combinations is because of high voltage, such as dimmer-circuits, etc. The components may have a maximum voltage rating considerably less than the circuit voltage and this is solved by series connection of several resistors but that is not the problem in your application as it is low-voltage battery powered.

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  • \$\begingroup\$ There is a small risk of dumping too much power thru one of the resistors in any parallel pair if the true value of the resistors are not close to equal. Some of this risk is ameliorated because the parallel sections have two in series in each leg. \$\endgroup\$ – Carl Witthoft Mar 24 at 14:20
  • \$\begingroup\$ sorry but why would the power dissipation be 1W? using P = V^2/R gives 10W. \$\endgroup\$ – Grepsoft Mar 24 at 15:02
  • \$\begingroup\$ The 1 W specification was added by @FrancoVS in an edit to my answer. I don't know how it was calculated. Your calculation assumes 24 V across the resistor string but this is unlikely. I don't think the OP has correctly traced out the circuit as the H-bridge would be driven directly from the supply positive, not through a relay coil or a 27 Ω resistor. \$\endgroup\$ – Transistor Mar 24 at 15:34
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    \$\begingroup\$ I just assumed 1/8W for each SMD resistor, times 8 resistors for the same power dissipation (source: they look 0805 to me). About OP's circuit, I am also quite confused, but I can see the relay working as a (over?)current detector (it triggers at 12V and has some hysteresis). \$\endgroup\$ – FrancoVS Mar 24 at 16:14
  • \$\begingroup\$ @FrancoVS: I think Chromatix may be on the right track - a soft start (but the capacitors are missing in the schematic). I removed your MathJAX, by the way, and replaced with HMTL entity Ω which avoids the discontinuity that the MathJAX introduces. You can use μ, ± etc., too. \$\endgroup\$ – Transistor Mar 24 at 16:22
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Since this is a low-voltage circuit, the reason is almost certainly to reduce the power dissipation in each resistor. The maximum power dissipation in the network is probably higher than the rated maximum dissipation for a single resistor, so they spread the power among 8 resistors...each resistor is only required to dissipate 1/8 of the total power.

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Using resistor network instead of a single resistor is to increase power dissipation or maximal rated voltage of the single resistor.

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  • \$\begingroup\$ but is the network really necessary? what purpose does it serve? \$\endgroup\$ – Grepsoft Mar 23 at 18:47
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This looks like a soft-start circuit.

Incidentally, you've drawn the relay symbol the wrong way around. The markings on top clearly denote terminals 87 and 30 as being connected to the switched contacts, but you've drawn the diagram with them connected to the coil. I'll assume they really are the switched contacts.

When the relay is open, power flows to the H-bridge (and the inductor driven through it) only through the resistor network. This is commonly done to reduce the inrush current when starting a motor; once the motor is running, the relay closes and power flows directly.

Startup is the highest current draw condition, so the resistor bank has to be able to handle it. The relay is rated for as much as 70 amps! So a resistor network is used to distribute the power dissipation across more than one individual resistor. These surface-mount devices are very small and can't handle much power by themselves, but as a team that capability is multiplied.

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D3 is your standard back emf absorbing diode that stops collapsing magnetic field in coil from generating a transient and blowing fets.

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They use several resistors for power dissipation reasons but the resistors are effectively shunting the relay forcing a certain level of current from the "H" bridge in order to trigger the relay. If the current that's required to operate the inductor L1 is too high, then the relay will trip. There is no frequency indicated so the actual load from inductor L1 is not known. However, if the load is too high or the frequency is too low the current will then be forced across the resistor network and the relay. This looks like a garage door opener overload circuit.

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Ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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The resistores form a Current Dump! And as they are only Surface mount types can't individually shunt the Power so hence the Parallel in Series configuration. This is a very common way of Dumping Current and or Power in the Professional World.

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