I am in the process of creating a High power Mosfet based SSR. The idea is using a photovoltaic MOSFET driver such as VO1263 (SIlabs have similar IC but it uses CMOS coupling to create the isolated power). However these type of ICs are pretty slow since they provide about 10uA of drive current at approximately 10v.

During my research on the subject I have found only one circuit without any explanation that should serve to increase the switching speed (without using external DC/DC converter). Shown in the figure below:

enter image description here

I am not quite sure that I understand the circuit completely and I would like your help in sizing some of the components. off-course this circuit is supposed to work in high side configuration that means Vcc (1-60v DC) is supplied at pin4 and load at pin3.

My logic:

  1. D4,R6 (around 1k ?) work as charging path for the bootstrap diode C1. T1,T2 (normal BJT such as BC537- BC527)are normal push-pull driver they work at turn on Only to provide high current to charge the mosfet gate (Sourced by C1).
  2. Z1 and R7(around 100 ohm ?) are used to protect the gate ( however I think they must be reversed order or there should be zener diode across C1 to keep voltage < Vgs max.
  3. R4 and R5 are pull downs (they should in 1MegaOhm range)?? however they cause power loss as R4 is voltage divider with the VO1263 output impedance. (in simulation VO1263 is modeled as voltage source with Series Impedance of 500k) .
  4. D3 and D5 work to keep the mosfet ON after initially being turned on by bootstrap C1 (which will lose its charge if the relay is kept on for sometime).
  5. The PV mosfet driver is not providing enough current for BJTs , So I tried using Darlington instead . The circuit worked after using 2 diodes instead of D5 and 2 diodes instead of D3 (4 diodes in total) donot fully understand why ?

Is my analysis correct ? any suggestion on the resistor value range ? any comments on the circuit is welcomed since I have not found any white paper , appnote or a development board that does something similar to get me started.

best regards

  • 1
    \$\begingroup\$ You're asking as if there are "oh just use this and that value and it will work" kind of answers. There are not, there are bound to be issues with this circuit, issues that you could find if you put the circuit in a simulator (I suggest LTSpice). Also, MOSFET based SSRs already exist. Explain why you need to make your own (what are you trying to achieve?). \$\endgroup\$ Feb 20, 2021 at 21:43
  • \$\begingroup\$ @Bimpelrekkie , Yes offcourse there is no majic value , and it will need tuning to get precise properties. I am just asking about ranges ( ohms , few k's , mega etc..) . 2nd thing is that i am trying to find out if the concept is right ? I am worried I did not find an appnote or evaluation board that use does something similar. 3rd thing , yes you are correct that such products (crydom for example) exist but they are very expensive (>100$) and i want it to be integrated in on the same PCB (not panel mount). \$\endgroup\$
    – ElectronS
    Feb 20, 2021 at 23:17
  • \$\begingroup\$ Check VOM1271, it has integrated fast turn-off circuit. \$\endgroup\$
    – user263983
    Feb 21, 2021 at 1:07
  • \$\begingroup\$ intersting question , since a similar one was asked here : electronics.stackexchange.com/questions/346198/… . the only solution presented was to use a dc dc converter . its nice to see how this circuit will perform . \$\endgroup\$
    – Sarah
    Feb 21, 2021 at 12:51
  • \$\begingroup\$ I could maybe answer one thing regarding the darlingtons , you need to diodes to bais the base with correct value since it contains 2 transistors , Vbe is higher than a single transistor. \$\endgroup\$
    – Sarah
    Feb 21, 2021 at 12:54

1 Answer 1

I could only imagine this circuit working if a resistor is added between T1 collector and base in order to bias the base of T1 (let's call it Ry), and PC being a phototransistor.

Given this assumption, then R6, D4, and C1 act as a "voltage" buffer.

When no signal is given to A1, PC is open, T1 conducts biased by the added resistor Ry. Z1 protects the gate against over-voltage and the current is limited by R6. D3 and D5 avoid current flow back to PC. T2 base voltage is high so it does not conduct. D5 flows current to T2 base and the R5 so it does not conduct. So the MOS gate would be high.

When a signal is given to A1, T1 base is low as shorted to the minus, current is limited by R6 and Ry so T1 does not conduct. T2 base is low as current can flow through R5, with D5 avoiding flowback to PC, so it brings the MOS gate to low. Of course, a simulation would help.


As ElectronS pointed out, when active, the MOSFET voltage drops to 0 thus will remain in that state.

The whole point is PC, which seems to be some sort of Photovoltaic device comprising many diodes in series, generating a voltage / current high enough to trigger T1.

This device could be a candidate with a voltage output up to 8V and a short circuit current of a few uA, however you need to consider the actual current at the required voltage. T1 would need to have a fairly high gain.

  • \$\begingroup\$ thank you for the effort , your analysis has 1 flaw only . that is it will work in low side configuration . However in high side configuration as soon as the mosfet becomes High . the battery voltage will pass from the drain to the source of the mosfet. So the voltage across the push-pull circuit & capacitor C1 becomes approximately Zero. For this reason in Half-bridge circuits using bootstrap technique cannot be on 100% of the duty cycle ON, unless they use a charge pump. \$\endgroup\$
    – ElectronS
    Feb 24, 2021 at 16:59
  • \$\begingroup\$ However the proposed circuit overcome this by using the photovoltaic driver to generate a "floating" voltage that keeps the mosfet ON indefinitely \$\endgroup\$
    – ElectronS
    Feb 24, 2021 at 17:02
  • \$\begingroup\$ @ElectronS you are right. PC must be some special photodiodes in series with a fairly high voltage / current generation capability. \$\endgroup\$
    – Damien
    Feb 25, 2021 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.