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I was trying to implement a H-Bridge circuit but had a lot of failures. I'm not really good with electronics, just some basic knowledge.

So I've started by creating an buffer to switch the full voltage of a 12V power supply by using NPN and PNP transistors. This is the switch for the power supply.

enter image description here

It works like a charm. Just as I expected.

Then I moved this switch into the H-Bridge circuit to switch my 12V supply On/Off on each side of the bridge. So I added 2 transistors for the bridge to ground the current. But I'm having a massive voltage drop and I don't know why. I hope you could help me with this. Here is my bridge schema.

enter image description here

The bridge logic works fine but I expect to get 12V instead of 3V on the motor it self.

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    \$\begingroup\$ First use your meter to figure out where the voltage is being dropped. Measure the supply rails under load. Measure the nominally high output to the upper supply rail. Measure the nominally low one to the lower rail. Think about what you find and what could cause it. \$\endgroup\$
    – Chris Stratton
    Commented Jan 7, 2019 at 17:13
  • \$\begingroup\$ R7 will pretty much determine your maximum switch speed. Are you switching or just full on and reversing the direction? If no, then it’s something else. \$\endgroup\$
    – winny
    Commented Jan 7, 2019 at 17:28
  • \$\begingroup\$ @winny I can’t find R7 in my schema. But I’m trying to connect this to Raspberry Pi GPIO to be able to rotate a motor in both directions \$\endgroup\$
    – hmak.me
    Commented Jan 7, 2019 at 17:36
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    \$\begingroup\$ You took some measurements, now think about which nodes show an unexpected difference of voltage. \$\endgroup\$
    – Chris Stratton
    Commented Jan 7, 2019 at 17:54
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    \$\begingroup\$ You have nothing but hFE of the NPN transistors limiting the base current of the PNPs. That is an undesirable situation. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 7, 2019 at 20:08

2 Answers 2

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If R5 & R6 on 2nd example are reduced from 1k to 0, it works.

  • But if both switches are turned on,it fails from push-pull short circuit.
  • These are limiting the Voltage. 10V/1k * hFE/10=mA* DCR of motor.
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  • \$\begingroup\$ This was exactly what I was looking for. Thank you. It would be great if you could give me a hint about this so I can learn further about this kind of problem \$\endgroup\$
    – hmak.me
    Commented Jan 7, 2019 at 19:48
  • \$\begingroup\$ search these keywords... design bridge deadtime cross-conduction shoot-thru FET .... pay less attention to the ads \$\endgroup\$
    – D.A.S.
    Commented Jan 7, 2019 at 19:56
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It looks like Q2 is not turning on fully. I say this because only 3.4 volts of the available 12V is dropped across the motor and Q6. The cause for this is most likely because Q1 cannot turn on fully preventing the necessary base current from Q2 to flow.

The Emitter on Q1 will have a voltage of about 0.7 plus the voltage across R6. This voltage is probably too high relative to the voltage at Q1's base. Tony Skyguy's answer seems to reinforce that assumption. R5 and R6 are there to limit the base current of Q3 and Q6, they are not necessary because R2, R3 and the low bias currents of Q2 and Q4 limit the current already.

Hope this helps

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